CHAPTER – 7
INTEGRALS
FORE-VIEW:
Integral Calculus is motivated by the problem of defining and calculating the area of the region bounded by the graph of the functions. Also a natural question arises that given derivative at each point of an interval, Can you determine the function? The answer is very much positive and such functions which has given functions, as derivative are called anti-derivatives or primitive of a function. These anti-derivatives are called Indefinite Integral and process of finding it is called integration.
Some of the situations which warrant integral calculus are:
Position of the object at any instant
Problems under normal probability curve.
Applications in economics
Area bounded by the graph of function under certain conditions.
Conclusion:- Integration is inverse process of differentiation.
Derivative of a function is unique, but a function can have indefinite Integrals.
For example
as = 0
In general
Or +2
Or - 5
Or + C where C is an arbitrary constant.
is known as indefinite integral.
is called integrand and x is variable of integration.
COMPARISON BETWEEN DIFFERENTIATION AND INTEGRATION
Both have wider applications in various fields.
EXPOSITION OF THE SUBJECT MATTER
INDEFINITE INTEGRALS
1.
2. Cos x Sin x + c.
3. Sin x - Cos x + c
4. Sec2 x Tan x + c
5. Cosec2 x - Cot x + c
6. Sec x Tan x Sec x + c
7. Cosec x Cot x - Cosec x + c
8. Tan x
9. Cot x
10. Sec x
11. Cosec x
12. Sin–1 x + c
13. Tan-1 x + c
14. Sec-1 x + c
15. e x e x + c
16. a x
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
Tips
1. Use
Put
2. (Use I LATE)
3. Split using
and then put
PARTIAL FRACTIONS
Form of rational expression Form of the partial fraction
1.
2.
3.
4.
5.
6. where
cannot be factored in to linear factors.
7. If degree of Nr is greater than the Dr divide and get deg Nr < deg of Dr and proceed further for partial fractions.
8. Whenever the partial fractions of the type. , to find A, B, C, D compare the co-efficients of x, otherwise put x = a if (x – a) is a factor and x = b if (x – b) is a factor and so on.
Note : To find an anti-derivative of a given function search intuitively for a function whose derivative is the given function. This method of intuitive search is called method of inspection.
Various methods used for integration
Type – 1 : Integration by substitution
Type – 2 : Integration using trigonometric identities.
Type – 3 : Integration using special integrals.
PROBLEMS:
1. I = = =
=
= =
I1 =
Put Cos x – Sin x =t, (- Sin x- Cos x)dx = dt, - (Sin x + Cos x)dx = dt
I1 = =
I =
2. (a) I =
I = dx
= dx
=
=
2(b) I = put log Sin x =t, , Cot x dx =dt
= = =
3. =
put Sin x =t, Cos dx =dt
= =
= = =
4.
=+=+
= = Sec x –Cosec x +C
5. =
Sin (A + B) – Sin (A- B) = 2 Cos A Sin B
= =
= =
6. =
==
= =
7. I =
x+2 =
x+2 = 4l – 2l x + m
Comparing the co-efficient of x
1 = -2l ⇒ l = - ½
Comparing constants 2 = 4l +m =
m = 2 + 2 = 4
I = = +
= I1 + I2
I =
Put 4x – x2 = t2, (4 - 2x)dx = 2t dt
= = =
I2 = = =
= = =
I =
8. 5x + 3 =
5x + 3 = 2lx + 4l + m
Comparing co-efficient of x
5 = 2l ⇒ l =
Comparing constants 3 = 4l+m
= 4 × +m
m = 3 – 10 = -7.
I ==
= I1 + I2
I1 = Put x2 + 4x +10 = t2
(2x+4)dx =2tdt
= = = 5t =
I2 = =
=
I = +C
9.
=
=
=
=
=
10. =
=
=
11.
=
x2 = A(x+1)(x-1)2 + B(x+1)(x-1) + C(x+1)+D(x-1)3
Put x = 1
1 = 2C ⇒ C =
Put x = -1
1 = - 8D ⇒ D = -
Put x = 0
0 = A – B + C – D = A – B + +
= A – B → (1)
Put x = 2
4 = 3A + 3B + 3C+ D
4 – 3 X = 3A + 3B
= 3A + 3B
= 3A + 3B
= A + B → (2)
= A – B → (1)
(2) + (1) ⇒ = 2A ⇒ A =
(2) - (1) ⇒ = 2B ⇒ B =
I =
=
12.
=
=
=
=
= log
13.
(Multiply by x3 in Nr and Dr)
=
=
=
=
14. Evaluate
=
=
=
I1 = =
= =
I =
15.
=
= - x Cos x +
=
16.
This is of the form
=
= Sin x
(x) = Cos x
=
17.
= , (x) =
I = =
18.
x +3 = A .
= - 4A – 2Ax + B
Comparing Co-efficient of x
- 2A = 1⇒ A =
Comparing Constant
3 = - 4A + B
B = 3+4 (- ½ ) = 1
I=
=
= I1 + I2
I1 =
= =
= =
I2 = =
==
I =
REVIEW
Evaluate reducing to standard form :
Evaluate By Suitable Substitution Method:
x > 0
Evaluate by making a perfect square:
Evaluate by partial fractions(using suitable substitutions if needed):
Evaluate using the formula: =
Evaluate using integration by parts:
Answers:
Hint : Write as ie
Ans:
.
22.
+
Definite Integral:
The definite integral has a unique value. A definite integral is denoted by where a is called the lower limit of the integral, and b is called the upper limit of the integral. The definite integral is introduced either as the limit of the sum or if it has an anit-derivative F in the interval [a, b], then its value is difference between the values of F at the end points, i.e., F(b) – F(a).
Integration as a limit of a sum:
where
Important results:
1. 1 + 2 + 3 + ……..+ (n – 1) =
2. 12 + 22 + 32 + ……..+ (n – 1) 2 =
3. 13 + 23 + 33 + ……..+ (n – 1) 3 =
4. a + a r + ar2 + ……+ a rn-1 =
Integration as a limit of a sum:-
1. Evaluate
Here f(x) = x + 1 Where h =
f(a) = f(0) = 0 + 1 Here h = nh = 5
f(a + h) = f(h) = h + 1 When
=
= = =
=
2. Here h = = = nh = 1
= =
=
= 5 + 1 = 6
3. Here h = = nh = 2
=
=
=
=
=
=
= =
4. Here h = = = 🡺 nh = 2
=
= =
= = =
5. Here h = nh = b – a
∴ =
=
=
=
= =
=
=
Properties of Definite Integral: -
:
i.e. integration is independent of the change of variable.
: . In particular
i.e. If the limits of definite integral are interchanged then its value changes by minus sign only.
:
:
:
: =
:
:
Evaluate The Following By Using Properties Of Definite Integrals:
1.
Let I =
= by
I =
2I = + =
2I = ⇒ I = Ans.
2.
Let I = =
2I = = ⇒ I = Ans.
3.
Let I = =
I =
2I = = ⇒ I = Ans.
4. = = = I (say) ……….[1]
I =
= ………………………………..[2]
[1]+[2] ⇒ 2I = = ⇒ I = Ans.
5. Let I = …………………………………[1]
=
I = ………………………………………………[2]
[1] + [2] → 2I 2I = =
2I = ⇒ I = 0 Ans.
6. Let I = ………………………………………[1]
I =
I = ………………………………………[2]
[1] + [2] → 2I 2I = =
2I = 0 ⇒ I = 0 Ans.
7.
Let I = = =
= = -
2I = = = I = Ans.
8. = =
I = =
2I = = =
= =
= I = Ans.
9.
I = = = 0
10. I =
I = =
2I = = =
I = Ans.
11.
I = ………………………………[1]
=
I = …………………………………………..[2]
2I = = = 0
I = 0
12.
I =
I = 0
13.
I = =
= 2I = 0 ⇒ I = 0 Ans.
14. Let I = =
I = ⇒ 2I =
2I = =
= =
= =
=
= = =
2I = ⇒ I =
Evaluate each of the following integrals
1.
= =
==
=
= =
2. dx = = =
Let t = tan-1x, dt =
Change of limits :
When x = 0; t = tan-1(0) = 0,
x = 1 ; t = tan-1(1) =
3. = =
= =
4.
=
∴
5. =
= = t = 1 – Cos x
= = =
= =
6.
=
=
=
= =
= =
7. I = ;
=
=
=
=
= I =
8.
=
=
==
= = =
9. = =
= =
= = = 2
10.
=
=
= tan-1(1) – tan-1(0)
I =
11. =
=
= = = = 5
12.
=
= =
= = = 12.5
13. = =
=
= = 2 =
14.
= =
= =
= 37
15. =
= =
= (1 – 0) – (0 – 1) = 1 + 1 = 2
REVIEW
Evaluate as limit of sums
1.
2.
3.
4. dx
5.
Evaluate the following:
1.
2.
3.
4. If find the value of k
5.
Evaluate using properties:
1.
2.
3.
4.
5.
Answers: 1. 33/ 2 ; 2. 27/2; 3. ; 4. 26/3; 5. 15;
(1-log2); 2. ; 3. ; 4. k=1/2 5.
1. 17 2. 3. 4.0 5. 9
POINTS TO CONCENTRATE
Indefinite integrals
Identification of type of integrand
Use of correct formulae, proper substitution and trigonometric identities.
Use of correct form to integrate functions of special type.
Definite Integral
Use of correct properties of definite integrals.
Change of limits according to substitution.
Use of correct formula for limit of sums.
formula usage correctly.
Test -I
as limit of sums
as limit of sums.
as limit of sums
Answers:
12
0
+c
23/6
–cot(log x)+c
27/2
******
CHAPTER- 9
DIFFERENTIAL EQUATIONS
FORE-VIEW:
Differential equation: An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation.
Ordinary differential equation : A differential equation involving derivatives of the dependent variable with respect to only one independent variable is called an ordinary differential equation,
Order of a differential equation: Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation.
Degree of a differential equation
By the degree of a differential equation, when it is a polynomial equation in derivatives, we mean the highest power (positive integral index) of the highest order derivative involved in the given differential equation. sometimes if the differential equation is not a polynomial equation in derivative (y’ )the degree of such a differential equation may not be defined.
Order and degree (if defined) of a differential equation are always positive integers.
General Solutions of a Differential Equation
The function y=f(x) which satisfies the given differential equation is called the solution of the differential equation.
The solution which contains arbitrary constants is called the general solution (primitive) of the differential equation.
Particular Solutions of a Differential Equation
The solution free from arbitrary constants i.e., the solution obtained from the general solution by giving particular values to the arbitrary constants is called a particular solution of the differential equation.
Formation of a Differential Equation
To form the Differential equation of the given equation
(i)Differentiate the given equation as many times as there are arbitrary constants.
(ii)Eliminate the arbitrary constants from the given equation by substituting the values of the arbitrary constants in terms of derivatives obtained by differentiation.
The order of a differential equation representing a family of curves is same as the number of arbitrary constants present in the equation corresponding to the family of curves.
First Order, First Degree Differential Equations
(i) Differential equations with variables separable
A first order-first degree differential equation is of the form = F(x, y)
If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (A) is said to be of variable
separable type.
Method of Solving Variable separable type
The differential equation has the form = F(x, y) ... (A)
= h (y) . g(x) ....................... (B)
If h (y) 0, separating the variables, (B) can be rewritten as
dy = g(x) dx ...................... (C)
Integrating both sides of (C), we get
dy = g(x) dx …………..(D)
Thus, (D) provides the solutions of given differential equation in the form
H(y) = G(x) + C
Here, H (y) and G (x) are the anti derivatives of and g (x) respectively and C is the arbitrary constant.
Note: Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing x should remain with dx.
Steps:
1. Separate the variables.
2. Integrate and attach arbitrary constants.
(II) Homogeneous Differential equation.
A differential equation which can be expressed in the form =f(x,y) or= g(x, y) where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation.
Method of solving Homogeneous type:
Put the differential equation in the form =f(x,y) or= g(x, y) as the case may be.
(ii) Put y= v x and = v + x or x = v y and =v + y
(iii) Apply variable separable method.
(III) Linear Differential Equation:
A differential equation of the form +Py= Q , where P and Q are constants or functions of x only is called a first order linear differential equation.
[Another form. A linear differential equation can also be put in the form +Px= Q ,
where P and Q are constants or functions of y only .)
Method of solving Linear Differential Equation:
1. Put the equation in the form +Py= Q, where P and Q are constants or functions of x only.
2. Find the integrating factor I.F =e
3. Solution is given by y. (I.F) =
Focus on the concept
The following Examples illustrate the method of solving the Problems on differential equations.
1. Find the order and degree, if defined, of each of the following differential equations:
(i) - sinx = 0 (ii) xy -x()-y= 0 (iii) y”’+2y2 +ey’ = 0
Solution:
(i) The highest order derivative present in the differential equation is , so its order is one. It is a polynomial equation in y’ = and the highest power raised to is one, so its degree is one.
(ii) The highest order derivative present in the given differential equation is so its order is two. It is a polynomial equation in y’ and the highest power raised to is one, so its degree is one.
(iii) The highest order derivative present in the differential equation is y’’’ so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.
2. Determine the order and degree of the equation. ,
Solution: Order =2, Degree = 2
3. Form the differential equation representing the family of concentric circles is x2 + y2 = r2
Solution:
Equation representing the family of concentric circles is x2 + y2 = r2 …(1) whose centre is at origin .
By differentiating, we get the differential equation 2x + 2y = 0
or x + y = 0 ..(2),which represents the family of concentric circles given by equation (1).Here r is the arbitrary constant and the given equation is differentiated once only and equation (2) is free from Arbitrary constants.
4. Form the differential equation representing the family of straight lines y = mx+c
Solution:-
By differentiating ,we get = m .Again differentiating we get =0 which is the differential equation representing the family of straight lines y = mx+c
5. Form the differential equation by eliminating the arbitrary constraints y = aebx , where a and b are arbitrary constants.
Solution:
y = aebx ………. .(1) differentiating with respect to x
⇒ = abebx = by (by using (1))
diff again with respect to x
⇒ y ′′ = b
∴ y ′′ = i.e. y y ′′ = (y ′)2 which is the required differential equation.
6. The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4 Solution: (D) 4
7. The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3 (B) 2 (C) 1 (D) 0 Solution: (D) 0
Solve
Solution; This problem is of Variable separable Type.
, , ,
y = (Integrating by parts)
=
∴ y = is the required solution.
9. Solve
Solution:
⇒ , ∴
- log (1 + Cos y) = log (1+ Sin x) + log c
log (1 + Cos y) (1+Sin x) = - log C
∴ (1 + Sin x) (1 + Cos y) =
10. Solve
Solution:
The Differential equation is
This is a Linear differential equation
∴ I.F =
Hence the solution is
, ⇒
i.e. y = x2 + Cx is the required solution.
11. Solve given that
Solution
Given
⇒
⇒
⇒
⇒
⇒ A = 1.
Hence
y = is the required particular solution.
12. If y = x + tan x. P.T. Cos2 x
Solution: y = x + tan x,
= 1 + Sec2 x,
y ′′ = 2 Sec2 x tan x
cos2x -2(x + tanx)+2x
= 2tanx - 2x – 2tan x + 2x
= 2 tan x – 2 tan x = 0
13. Solve (y + xy)dx + (x – xy2)dy = 0
Solution:
The differential equation can be written as
y (1 + x) dx + x (1 – y2) dy = 0
i.e. dy
⇒ ⇒ log
log written is the required solution.
14. Indicate the order and degree
(y’’)2 + (y’)3 + Siny = 0 ⇒ degree = 2 & order = 2
y1 + ey = 0 ⇒ degree =1 & order =1
y1V + Sin y111 = 0 ⇒ degree - Not defined & order = 4
y111 + y11 + y1 + y Siny = 0 ⇒ degree = 1 & order =3
15. Find the differential equation that will represent the family of curves given by the following equations where a is a parameter.
(a ) y = ax3
Solution
Y=ax3 ---------- 1
Differentiate both sides w.r. to x
⇒ y1 = 3x2.a
a = y’/3x2
∴1 y =
i.e., y = is the required diff. equation of the given family of curves.
(b) x2 + y2 = ax3 _________(1)
Solution
Diff. w.r. to x
⇒ 2x + 2yy’ = a.3x2
i.e., a =
Substitute ‘a’ value in (1)
x2 + y2 = ⇒ 3x2 + 3y2 = 2x2 +
⇒ x2 + 3y2 = 2xyy’ which is the required diff. equation.
(c ) y = eax
Solution
Diff. w.r. to x
⇒ y’ = a eax ⇒ a = i.e., a =
∴y = ⇒ logy = which is the required diff. equation.
16. Find the diff. equation of all circles in first quadrant, which touch the co-ordinate axes.
Solution
Let ‘a’ be the radius of the circle, which touches the co-ordinate axes. Then its center is (a, a).
Hence the equation of the circle is
(x – a)2 + (y – a)2 = a2
⇒ x2 + y2 – 2a(x + y) + a2 = 0 ______(1)
Diff. (1) w.r. to x, we get
⇒ x + y = a
⇒ a = where P =
putting this value of ‘a’ in (1), we get
x2 + y2 –
⇒ (x2 + y2) (1 + P)2 – 2(x + yP) (x + y) (1 + P) + (x + yP)2 = 0
⇒ (x + yP)2 = (1 + P) [2(x + yP) (x + Y) – (x2 + y2) (1 + 0)]
= (1 + P) [(x2 + 2xy – y2) + P(y2 + 2xy – x2)]
where P = is the required diff. eqn.
Solve = x5 Tan-1(x3)
Solution. dy = x5 Tan-1(x3)dx
=
Let x3 = t ⇒ 3x2dx = dt
∴ y =
=
=
=
=
i.e., y = , which is the required solution.
Solve
Solution
The D.E is
Here I.F =
So, the solution is
= [Let t = Tan-1x]
= =
⇒
⇒ y =
⇒ y = which is the required solution.
Solve the following diff. equation
Solution
The given diff. equation can be written as
⇒ ⇒
⇒ logy =
⇒ logy =
⇒ logy + which is the solution.
Solve the following diff. equation
Solution
Let x + y = z ⇒ . So, the diff. equation reduces to
⇒
⇒ ⇒
⇒
[Let t = ]which is the required solution.
21. Verify that each of the following functions as defined below is a solution of the accompanying initial value problems:
(a) φ (x) = Sinx + Cosx : y’’ + y = 0, y(0) = 1, y’(0) = 1
Let φ(x) = y
∴φ’(x) = y’ = Cosx – Sinx
φ’’(x) = y’’ = -Sinx – Cosx = -(Sinx + Cosx) = -y
∴y’’ + y = 0 is the differential equation and hence
φ(x) = Sinx + Cosx is the solution for the diff. equation
y’’ + y = 0.
Further y(0) = 1 i.e., φ(0) = 1 and φ’(x) = 1 hence φ satisfies the initial conditions. So the given function φ satisfies the given initial value problem.
(b) φ(x) = ex + e-x : y’’ – y = 0, y(0) = 2, y’(0) = 0
φ’(x) = ex - e-x
φ’’(x) = ex + e-x
φ’’(x) = φ(x)
⇒ y’’ = y ⇒ y’’ – y = 0
Hence, if we substitute y for φ, the given function φ satisfies the given diff. equation for all x ∈ R.
Further
and
i.e., φ satisfies the initial conditions. So, the given function φ satisfies the given initial value problem.
(c) φ(x) = ex + e2x : y’’ – 3y’ + 2y = 0 y(0) = 1, y’(0) = 3
φ’(x) = ex + 2e2x
φ’’(x) = ex + 4e2x
∴y’’ – 3y’ + 2y = ex + 4e2x –3ex – 6e2x + 2ex + 2e2x
= 3ex + 4e2x – 3ex – 6e2x + 2e2x
= 0
∴ φ(x) is the solution for the given equation y(0) = 1, y’(0) = 3
⇒ φ(0) = e0 + e2 × 0 = 1 + 1 = 2
yٰ(0) = e0 + 2e0 = 1 + 2 = 3
i.e., φ satisfies the initial condition. So the given function φ satisfies the given initial value problem.
(d) φ(x) = xex + ex : y’’ – 2y’ + y = 0 y(0) = 1, y’(0) = 2
φ(x) = y = xex + ex
φ’(x) = y’ = xex + ex + ex = xex + 2ex
y’’ = xex + ex + 2ex = xex + 3ex.
∴ y’’ – 2y’ + y = xex + 3ex – 2xex – 4ex + xex + ex
= 0
∴ φ(x) = y. satisfies the equation y’’ – 2y’ + y = 0
φ(0) = 1, i.e., y(0) = 1 & y’(0) = 2
given φ(x) = xex + ex
φ(0) = 0 + e0 = 1
φ’(0) = 0 + 2e0 = 2
φ satisfies the initial value problem
22. Find a solution of each of the family of diff. quations.
y’ = ex+y + e-x+y
Solution:
Taking integration on both sides after separating the variables
⇒ ⇒
⇒ e-x – e-y = ex + c which is the required solution
y’ = (Cos2x – Sin2x) Cos2y
Taking integration on both sides after separating
We have [Cos2x – Sin2x = Cos2x]
⇒
⇒ Tan y = which is the required solution.
[C] (xy2 + 2x) dx + (x2y + 2y) dy = 0
x(y2 + 2) dx = - y(x2 + 2) dy
i.e.,
⇒
⇒
⇒ (x2 + 2) (y2 + 2) = A
⇒ y2 + 2 = is the required solution.
(c). xy’ = 1 + x + y + xy
xy’ = 1 + x + y (1 + x)
i.e.,
⇒
i.e., which is the required solution.
(d). y2 + x2y1 = xyy1
⇒ ⇒
V – logV = logx + A
Y/x – logy + logx = logx + A
⇒ y = x logy + Ax is the required diff. Equation.
(e) (y2 – 2xy) dx = (x2 – 2xy) dy
Let
∴
=
log x =
x =
i.e., x =
i.e., x3 =
i.e., xy2 – yx2 = A3 = k
. y2 dx + (x2 – xy + y2) dy = 0
i.e.,
i.e.,
i.e.,
⇒
Consider:
1 – V + V2 = A(V2 + 1) + (BV + C) V
V = 0 gives
1 = A + 0 ⇒ A = 1
V = 1 gives
⇒ 1 = 2A + B + C
B + C = -1
V = -1
⇒ 3 = 2A + (-B + C) (-1)
3 = 2A + B – C
B – C = 1
B + C = -1
B – C = 1
2B = 0 ⇒B = 0
C = -1
∴
⇒
Tan-1V – log V = log x + A
Tan-1y/x – log y = A
Logy = Tan-1y/x – A
=
23. Solve
Solution: The given equation is homogeneous differential equation.
On putting V= we have
⇒
⇒
x = =
24. Solve y’ Cos2x = Tan x – y
Solution;
Now, ⇒
dividing by Cos2x, we have
therefore the differential equation is Linear diff. Equation
∴ Solution is dx+ C ,where C is the Constant of integration.
e=
∴ Solution is
Integrating using integration by parts, by taking t = Tan x
dt = Sec2x dx
∴ R.H.S is
=
∴G. Solution is
⇒
Solve the following
(a)
Solution:is a Linear diff. Equation
∴ G. Solution +c
= 1 + Sin x.
∴
=
∴General Solution is
i.e., y =
(b)
Solution:
⇒
i.e.,
[Linear diff. Equation]
∴ Solution is +c
Here I.F =
=
= Tan y + A
∴ Solution is
⇒ x =
x =
(c) (x + 2y3) x’= y
Solution:
i.e.,
dividing by y
⇒
i.e.,
[Linear diff. Equation]
∴ General Solution is +c
i.e.,
i.e.,
=
∴ x = y [y2 + A] is the required solution.
Solve.
(i) y(0) = 0
i.e.,
i.e., Let
y(0) = 0 ⇒ c = 1
∴ The solution is
(ii) Cos (x + y) dy = dx y(0) = 0
Substituting x + y = z
We get
∴
⇒
integrating both sides we get
⇒
x = 0, y = 0
⇒ c = 0
∴ is the solution.
(iii) (x + y + 1)2 dy = dx y (-1) = 0
x + y = z
Diff. w.r. to y ⇒
∴ ⇒
∴ ⇒
Integrating on both sides
⇒ Tan-1 (z + 1) = y + A
⇒ Tan-1 (x + y + 1) = y + A _________(1)
Given y (-1) = 0 ⇒ x = -1 & y = 0
Substitute this value in (1)
⇒ Tan-1 (0) = 0 + A ⇒ A = 0
∴ Solution is Tan-1 (x + y + 1) = y
(or) x + y + 1 = Tan y
26. (x – y) (dx + dy) = dx – dy, y (0) = -1
Solution: Let x – y = z
∴ i.e.,
∴ Given equation is (x – y)
i.e., ⇒
i.e.,
i.e., ⇒ 2z dx = dz (z + 1)
∴
i.e.,
⇒
⇒
⇒
i.e.,
Given y (0) = -1
⇒ 0 – 1 = log 1 + A
⇒ A = -1.
∴ Particular solution is
i.e.,
or
In the above problems we make a substitution to reduce the differential equations to Variable separable Type.
A population grows at the rate of 2% per year .How long does it take to double? Use differential equation for it.
Solution: Let P0 be the initial population and let the population after ‘t’ years be P.
Then
This implies logP = When t=0 , P= P0
log P0= C putting the value of C we have t= 50 log() when P=2 P0,
t= 50 log( ) = 50 loge 2yrs
----
REVIEW
Order of a differential equation is the order of the highest order derivative occurring in the differential equation.
Degree of a differential equation is defined if it is a polynomial equation in its derivatives.
Degree (when defined) of a differential equation is the highest power (positive integer only) of the highest order derivative in it.
A function which satisfies the given differential equation is called its solution.
The solution which contains as many arbitrary constants as the order of the differential equation is called a general solution and the solution free from arbitrary constants is called particular solution.
To form a differential equation from a given function we differentiate the function successively as many times as the number of arbitrary constants in the given function and then eliminate the arbitrary constants.
Variable separable method is used to solve such an equation in which variables can be separated completely i.e. terms containing y should remain with dy and terms containing x should remain with dx.
A differential equation which can be expressed in the form =f(x,y) or= g(x, y) where, f (x, y) and g(x, y) are homogenous functions of degree zero is called a homogeneous differential equation and is solved by putting y = v x
A differential equation of the form +Py= Q , where P and Q are constants or functions of x only is called a first order linear differential equation.
A problem involving differential equations with the given boundary Conditions (initial Value or final value) is called a Boundary value problem. These conditions are given in the problem to get the particular solution of the given differential equation.
Problems for Practice
1. Determine order and degree (if defined) of differential equations given in Exercises.
(i) = 3 sin x (ii) (iii)
(iv) x+y=0.
2. Form the differential equation of the family of curves y2-2my+x2= m2
3. Form the differential equation corresponding to y2= a(b-x)2 by eliminating a and b
4. Show that y = Ax + is a solution of the differential equation
5. show that y = 4 sin 3x is a solution of the differential equation
6. Solve the differential equation: = x5+x2-
7. Solve
8. Solve the differential equation(x2-y2) dx + 2 xy dy=0
.
9. Solve =+tan
11. Solve.=
12. Solve.=
13. Solve. + y = Cos x
14. Suppose the growth of the population is proportional to the number present. If the population of a colony doubles in 25 days in how many days the population becomes Triple.
Ans :- (25() days.
******
CHAPTER - 6
APPLICATION OF DERIVATIVES
FORE-VIEW
Equation of tangent at a point .
Point is known, slope = at the point = m point slope form
Equation of normal at a point:-
Point is known Normal and is perpendicular to tangent
slope
Given tangent to find point
Slope = 0 tangent || x – axis
Slope not defined , Dr = 0 tangent || y-axis
Increasing decreasing functions & derivatives :-
1) A function y = f(X) :A B is increasing in a domain iff for any pair of x1, x2 the domain x1 < x2 f(X1) f(X2)
A function f(X);A B is strictly increasing in a domain iff x1<x2 f(X1) < f(X2) x1, x2 Domain A
2) A Function y= F(X) :A B is strictly decreasing or just decreasing accordingly as :
x1 < x2 F(X1)> F(X2) or F(X1) f(X2) respectively .
Increasing or decreasing functions are called monotonic functions. We say monotonic increasing or monotonic decreasing
Condition In terms of Derivatives :
y = f(x) is increasing in Df
x1, x2 Df f (x1)< f(x2) when x1<x2
Proof :-
Let X be any point. Consider x - or x +
From LHS: x - < x f (x - ) < f ( x )
( x )
From RHS
when a function is increasing at a point
when a function is increasing in a domain the domain
Rolle’s theorem : (3 conditions)
(x) is continuous in [ a,b] ( continuous in closed interval)
(x) exists in (a,b) ( derivable in open interval )
(a) = f(b) (end point values are equal)
If all the 3 above are satisfied ∃ atleast one c ∈ (a,b)
(c) = 0 i.e. tangent at (c, f(c)) is || to x – axis to the curve at c.
Meaning of atleast one c (all 3 conditions are satisfied)
For decreasing functions :
When Rolle’s theorem is not applicable
1st condition not satisfied 2nd condition not satisfied
Derivative doesn’t exist at
interval point
3rd condition not satisfied
1st case
c does not exist
2nd case
3rd condition not satisfied First condition not satisfied Second condition not satisfied
↓ ↓ ↓
at c/ at c/
a c b a c c/ b
f(a) f(b)
In a subinterval all conditions are satisfied. Hence c exists even though in [a,b] the conditions are not satisfied. Hence in problems one cannot conclude immediately C does not exist if any of the three conditions is not satisfied only Rolle’s theorem is not applicable for inference one has to derive f(x) & verify solution is in the interior of interval for eg y = sin x in [0, ] ;sin0 =0, sin =-1. But at x= ; f ' (C)=0.Rolle’s theorem conditions are satisfied in [0,]
Caution in 2nd condition
Saying f ' (x) exists in all interior points does not imply it does not exist at end point. Only Rolle’s theorem is not concerned with these derivatives at end points. They may or may not exist.
Lagranges’s Mean Value Theorem :
Conditions :
f(x) is defined & continuous in [a,b] 2) f’ (x) exists in (a,b)
Then ∃ c ∈ [a,b] where
If f(b) = f(a), f1(c) = 0 reduces to Rolle’s theorem
f(b) – f(a) is slope of the secant joining [a,f(a)] , [b,f(b)]
b-a
f1(c) is the slope of tangent at C[c, f(c)]
⇒ tangent at C is parallel to the secant joining the end points of the curve.
Problem:
Let f defined on [0,1] be a twice differentiable function such that | f ''(x ) 1, for all x [0,1] if f (o) = f(1) ,then show that |f ’( x ) | <1 , for all x [0,1]
Pf : f satisfies conditions of Rolle’s theorem there exists a c (o,1) :
f '(c) = 0
o c A
[Remember differentiability continuity]
let x [0,1].Let x [0,c]
Lagrange’s M.V.T is satisfied on [x,c]
-(c- x)
Maxima Minima Terminology
Maximum – greatest value
Maxima – plural of maximum
Minimum – least value
Minima – plural of minimum
General for both maxima & minima – optimum & optima
In the normal sense of the word maximum & minimum have a connotation of boundedness & also uniqueness since there can be only one highest value or one least value. But then why the plural. Consider the marks obtained by students. Marks vs student is a function. Two or more students may obtain the highest mark. As a value it is unique. But it is attained at more than one point.
Another situation: section wise maximum in a particular class. Each maximum value is different, by right being a maximum in a local neighborhood. But there will be a maximum of all maxima.
We have the following situations :
Domain is bounded and closed
Maximum or minimum Maximum at end point, minimum Maximum & minimum at
At end points at an interior pt. Vice versa interior points
Local maxima, minima Local maxima Local maxima, minima
Nil exist exist exist
Note For closed domain definitely at some point maximum as well as minimum is attained.
In a closed interval maximum & minimum attained at least once
To say it is attained more than 1 time is in the sense of local maxima & minima.
Maximum of all maxima is called absolute maximum
Minimum of all minima is called absolute minimum.
Conclusion : A function defined in a closed interval attains an absolute maximum & absolute minimum once either at end points or at interior points.
Jargon
We say a function attains maximum or minimum at a point of the domain & corresponding functional value is the maximum or minimum value of the function & is in the range.
When the domain is open (open interval (a,b)
Two Possibilities
Neither maximum Local maxima Nor minimum minima
Condition for Maximum or minimum in terms of derivatives
Necessary Conditions : If derivative exists at the maximum or minimum points
Maximum at x0 ⇒ f′ (x) > 0 to the left
In every neighborhood of xo to the left f(x ) f ′ (x) < 0 to the right
& to the right Hence at x0, f ′ (x0) = 0
For minimum :
To the left decreasing => f ′ (x) < 0
To the right increasing => f ′ (x) > 0
At x0, f ′ (x0) =0 i.e tangents at turning point are parallel to x-axis.
f′ (x) = 0 is only a necessary condition but not sufficient
For instance y = x3 f ′ (x) = 3x2 = 0 at x = 0
But x =0 is not maxima or minima
Sufficient Condition :When the 1st and 2nd derivative exist at the turning point
f ′ (x) = 0 and f ′′ (x)< 0 for maxima (f′ (x) decreases from +ve to –ve value)
f ′ (x) = 0 and f ′′ (x)> 0 for minima (f′ (x) decreases from -ve to +ve value)
Sufficient condition independent of derivatives:
for maxima to the left, f(x) increases and to the right decreases.
for minima to the left, f(x) decreases & to the right increases.
Working Rule
f(x) defined in closed interval [a, b] to find absolute maximum & absolute minimum.
Step 1 : Find functional values at end points i.e. f(a) & f(b)
Step 2 : find f′ (x) & solve f′(x) = 0
Let x1, x2 , x3 ,……… be solutions (possible points of local optima)
Step 3 : Find f(x1), f(x2), f(x3) ,…….
Step 4 : Find the highest and lowest of f(a), f(b), f(x1), f(x2), f(x3) etc
Step 5 : Highest is absolute maximum
lowest is absolute minimum.
For finding local maxima & local minima in open interval usually in the set of R
Find points where f′ (x) = 0
Find f ′′(x)(if exists)
f′′ (x) > 0 ⇒ point is local minima
f′′ (x) < 0 ⇒ point is local maxima
If f ′′ (x) does not exist then by first principle determine maxima or minima
There are functions where f 1(x) ≠ 0 at a point but still is a maxima or minima
Eg: | x |
at x =0 no derivative exists , but x = 0 is minimum.
To the left decreasing & to the right increasing.
Important Result
Derivability ⇒ continuity but not conversely
Eg: |x| is continuous at x = 0 but not derivable
Proof: Let f(x) be derivable
⇒ Lt f(a + Δx) – f(a) exists = f/(a) ( say)
Δx🡪 0 Δx
⇒ f(a + Δx) – f(a) = f(a) Δx + ∈
where ∈ 🡪 0 as Δx 🡪 0. f / (a) is finite & exists from both sides
f/ (a) Δx 🡪 0 when Δx🡪 0
⇒ in the neighborhood of a functional values are as close as possible to f(a)
⇒ f(x) is continuous at a
Converse is not true.
| x | is continuous at x = 0
h > 0
Right Limit = =
Left Limit = =
But not derivable :
h > 0
RHS = =
LHS =
Word Problems
Algorithm
1) Identify the function to be optimized say f (x,y,…..)
2 ) Identify the independent variable
Identify the equation connecting the independent variables if they are more than one.
Obtain from this all variables in terms of any one convenient variable (say x) preferably the one whose dimension is focused upon and express the function in terms of this variable.
Derive the function to be optimized i.e. find f1(x) w.r.t the single variable in (4)
Equate f′(x) = 0 & solve . Let x1, x2,…. be solutions.
Find f′′ (x)
f′′(xi) > 0 ⇒ xi minimum
f′′(xi) < 0 ⇒ xi maximum
If it is a closed interval take into consideration values of the function at end points also.
Problem :
1 ) Prove that of all the rectangles of given perimeter square has the maximum area.
(1)Functions to be maximized A(x, y) = xy
(2) & (3) Equations of constraints 2 (x + y) = constant = P
x + y = P / 2
(4) y = P / 2 - x
A(x) = x ( P/2 – x ) = P/2x – x2
(5) A′ (x)
= P/2 – 2x = 0
2x = P/2
x = P / 4
⇒ y = P / 2 – P / 4 = P / 4
x = y = P / 4
A′′+(x) = -2 < 0
⇒ x = P / 4 corresponding to maximum
⇒ It is a square having the maximum area
Maximum area = P2 / 16
2 ) Given the sum of the perimeter of a square and a circle, show that the sum of their areas is least when side of the square = the diameter of the circle
Let x be the side of the square
Let r be the radius of the circles
(1) A(x, r) = x2 + r2
(2) 4x + 2 r = C
r = C - 2x
r = C – 2x
(3) A(x) = x2 + (C – 2x)2
(4) A1(X) = 2x + 2( C – 2x) x – 2
2x - 4C + 8x = 0
(5) x ( + 4 ) - 2C = 0
x = 2C
+4
A11(x) = + 4 > 0
A(x) is a minimum when x =
Now, Radius r = C – 2x
=
Points to remember - REVIEW
Local maxima/minima exist only at turning points i.e where ,’c’ is always on internal point c (a,b)
If , it is not necessary that a local maxima/minima exists if c(a,b)
If , and ( c ) is also zero , c(a,b) is a point of inflexion
Absolute maxima / minima always exist in closed interval
Absolute maxima / minima may be at extremities of the interval or at interior points of the interval
6. Important: Whenever there is a point ‘c’ (ab) when f ' ( C ) = 0 ,the behaviour of #he function (trend) changes .
Increasing to decreasing (local maxima)
Decreasing to increasing (local minima)
Increasing to increasing or decreasing to decreasing (point of inflexion)
7. To be very sure of the trend we get in to second derivative
f′′ (c )
if f′′(c ) is zero no turning point
if f′′ (c )is <0 is local maximum
if f′′ (c ) is >0 is local minimum
APPLICATION OF DERIVATIVES
1) A particle moves along the curve 6y = x3 + 2, Find the points on the curve at which the y- co-ordinate is changing 8 times as fast as x - co-ordinate.
2) The radius of a spherical soap bubble is increasing at the rate of 0.2 cm / sec. Find the rate of increase of its surface area when radius is 4 cm.
3) The side of a square is increasing at the rate of 0.2 cm / sec. Find the rate of increasing of the perimeter of a square.
4) A man 2 metres high walks at a uniform speed of 6 mt / min. away from a lamp post 5 metres high. Find the rate at which length of his shadow increases.
5) An edge of a variable cube is increasing at the rate of 10 cm/sec. How fast is the volume of the cube increasing when the edge is 5 cm long?
6) Find the point on the curve y = 2x2-6x-4 at which the tangent is parallel to x-axis.
7) Find the points on the curve y = x3-2x2-x at which the tangent lines are parallel to the line y = 3x-2.
8) Show that the curves 4x = y2 and 4xy = k cut at right angles, its k2 = 512.
9) Find the equation of the normal line to y = x3 + 2x + 6 which is parallel
to the line 14y + x + 4 = 0.
10) Find the equation of the tangent to the curve x = a Sin3, y = b Cos3 at = /4
11) Verify Rolle’s Theorem for the following :
a) f (x) = x2 - 4x + 3 [1, 3]
b) f (x) = x3 + 3x2 - 24x - 80 [-4, 5]
c) f (x) = x 1/3 [-1, 1]
d) f (x) = Sin 2x [0, 2]
e) f (x) = Cos x
12) Using Rolle’s Theorem, Find the points on the curve y=x2, x [-2, 2] where the tangent is parallel to x-axis.
13) Verify Lagrange’s Mean value theorem for the function
f(x) = (x-1) (x-2) (x-3) in [0,4 ].
14) Verify Lagrange’s Mean value theorem for the function f(x) = x3 - 2x2 - x - 3 in [0, 1].
15) Using Lagrange’s Mean value theorem, find a point on the parabola y=(x-3)2, where tangent is parallel to the chord joining (3,0), (5,4).
16) Verify Lagrange’s Mean value theorem for the function f(x) = in [2, 4 ].
17) Find the intervals in which the function f (x) in increasing or decreasing
f(x) = 2x3 - 12x + 5.
18) Find the intervals in which the function f (x)= x3-6x2+9x+10 is
(a) increasing (b) decreasing.
19) Find the intervals in which the function is increasing or decreasing
F (x) = Sin x – Cos x, O<x <2
20) Find the intervals in which the function
F(x) = x4-4x3+4x2+15 is
(i) Increasing (ii) decreasing.
21) Show that the height of the right circular cylinder of maximum volume that can be inscribed in a given right circular cone of height h is h/3 .
22. Show that a closed right circular cylinder of given surface area and maximum volume is such that its height is equal to the diameter of the base.
23. Show that a right triangle of given hypotenuse has maximum area when it is an isosceles triangle.
24. A wire of length 28 meters is to be cut into two pieces. One of the other into a square. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?
25. Show that the semi-vertical angle of the right circular cone of given slant height and maximum volume is a semi circle.
26. A rectangle is inscribed in a semi circle of radius and with one of its sides on the diameter of the semi circle. Find the dimensions of the rectangle, so that its area is maximum. Also find the maximum area.
27. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be the least when the depth of the tank is half of its width.
28. A window is in the form of a rectangle surmounted by a semicircle. Given the perimeter of window find its dimensions in order that the area may be maximum.
29. Show that the height of a cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/
30. An open box with a square base is to be made out of a given quantity of sheet of area a2. Show that the maximum volume of the box is
a3/ 6
31. Find a point on the parabola y2=2x. which is closer to the point (1,4).
32. Find the point of local maximum and local minimum as well as the corresponding local maximum and local minimum values of the function f(x) = (x -1)3 (x + 2)2.
33. A square tank of capacity of 250 cubic meters has to be dug out. The cost of land is Rs. 50 per sq. mt. The cost of land is Rs. 50 per sq. mt. The cost of digging increases with the depth and cost digging increases with the depth and cost for the whole tank is 400 (depth), rupees, Find the dimensions of the tank for the least cost.
34. Find two positive numbers x and y such that x + y = 60 and x y3 is maximum.
35. Using differentials, find the approximate value of
36. Find the approximate value of
37. If Y=x4+10 and x changes from 2 to 1.99, find the approximate change in y.
38. Find the approximate change in the total surface area of a right circular cone, when the radius r remains constant while the altitude h changes by .
39. It is given that for the function f(x)= x3 + 6x2 + ax + b on (1, 3), Rolle’s theorems holds with C = 2 + (1/ ) Find the values of a and b if f(1) = f(3) = 0
If radius of a circle is increased from 5 cm to 5.1 cm. Find the approximate increase in area.
Question Bank
A circular metal plate expands under heating so that its radius increases by 2%. Find the approximate increase in the area of the plate if the radius of the plate before heating is 10cms.
Show that the triangle of maximum area that can be inscribed in a given circle is a equilateral triangle
An open tank with a square base and vertical sides to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when depth of the tank is half of its width.
The function f(x) = x3 + bx2 + cx – 3 has a maximum at x = 1 and a minimum at x = 5. Then (b, c) is
i) ( 9, 15 ) ii) ( -9, -15 ) iii) ( -9, 15 ) iv) (9,-15) Ans : iii
The points of inflexion of the curve x = (log y)3 are at
(0,1), (8, e2)
(1,0), ( e2 , 8)
(0,-1), (8, -e2)
(-1,0), (e2 , 8) Ans : ii
For what value of ‘a’ for which the roots of the equation ax2 + (a-1)x + 2 = 0 is minimum
ABCD is a rectangle in which AB = 12cm, BC = 5cm. A point P is taken in CD such that PC = x cm. If PA2 + PB2 is minimum. Then x = ?
The equation of the line through the point (3,4) which cuts off from the I quadrant a triangle of minimum area, has x intercept and y intercept _ _ _ _ _ _.
If f(x) = exg(x); g(0) = 2, g1(0) = 1 Then f1(0) is
1 b) 3 c) 2 d) 0
10) The derivative of tan-1 at x = 0 is
a) 1/8 b) ¼ c) ½ d) 1
11) The curve y = x1/5 has at (0,0)
a) Vertical tangent b) horizontal tangent c) no tangent
12) If y = a logx find dy/dx
13) If a < 0, the function f(x) = eax + e –ax is a monotonically decreasing function for values of x given by
a) x > 0 b) x < 0 c) x < 1 d) x > 1 Ans : b
14) The normal to the curve y = f(x) at the point (3, 4)makes an angle 3Π/4 with the positive x-axis, then f1(3) = ?
15) If x = sin-1t, y = log( 1- t2 ) find (d2y/dx2)t = ½
16) Let f(x + y) = f(x) . f(y) for all x and y. If f(5) = 2 and f1(0) = 3 find f1(5)
17) The value of derivative of | x –1 | + | x-3 | at x = 2 is
a)–2 b) 0 c) 2 d) not defined
18) If y = sinx + ex , find d2x/dy2.
FOCUS ON THE CONCEPT
1. An edge of a cube of variable length is increasing at the rate of 3cm per sec. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution:
Let S be the edge of the cube given
Volume of a cube = S3
V = S3
= 3S2 = 3 x 10 x 10 x 3 = 900 cm3/sec.
2. A particle moves along the curve 6y =x3 +2. Find the points on the curve at which the y co-ordinate is changing 8 times as fast as the x co-ordinate.
Solution:
6y = x3 + 2
Given = 8
Diff. 6 = 3x2
6 x 8 = 3. x2
x2 = = 16
x = ± 4
when x= 4 ⇒ 6y = 43 + 2 = 66, y = 11
Point = (4, 11)
when x= - 4 ⇒ 6y = (-4)3 + 2 = - 62, y =
Point =
3. Use differential to approximate : (a) (b)
Solution:
(a)
Choose x = 25; Δ x = 1
So that x + Δx = 26
Function is y = x1/2
∴ = = 5 + = 5 + = 5.1
(b)
Choose x = 125
Δ x = 2 ⇒ x + Δx = 127
Function is y =
∴ = = = 5 +
= 5 + = 5 + 0.027 = 5.027
4. Verify Rolle’s theorem for the function f (x) = x3 – 6x2+11x – 6 in [1, 3].
Solution:
Being a polynomial function.
it is continuous in [1, 3] (ii) it is differential in (1, 3)
(iii) (a) = (1) = 1 – 6 +11 – 6 = 0
(b) = (3) = 33 – 6 . 32 + 11 . 3 – 6 = 27 – 54 + 33 – 6 = 0
∴(1)= (3)
Hence Rolle’s theorem is satisfied
∴ f’(x) = 3x2 - 12x + 11
Since f’(c) = 0 ⇒ 3c2 – 12c +11 =0
C = = =
= ==
= 2.86 or 1.32 ∴ C ∈ (1, 3)
5. Find the intervals in which the function (x) = x3+3x2-105x + 25 is increasing or decreasing.
Solution:
(x) = x3 + 3x2 –105x + 25
(x) = 3x2 + 6x – 105 = 3(x2 + 2x – 35) = 3(x + 7) (x – 5)
Since (x) =0 ⇒ x = -7, 5
The intervals are (-∞, -7), (-7, 5), (5, ∞)
Intervals 3 (x + 7) (x – 5) sign
(-∞, - 7) + – – + ↑
(-7, 5) + + – – ↓
(5, ∞) + + + + ↑
The function is increasing in (-∞ , - 7) and (5, ∞)
and decreasing in (-7, 5).
6. Find a point on the parabola y = (x – 3)2 where the tangent is parallel to the chord joining (3, 0) and (4, 1)
Solution:
Being a polynomial,
(i) it is continuous in [3, 4] (ii) it is differentiable in (3, 4)
Hence LMVT is satisfied.
∴ there exist f’(c) =
(x) = x2+9 – 6x (c) = 2c – 6
(3) = 32 + 9 – 6 x 3 = 0 (4) = 42 + 9 – 24 =1
b – a = 1
∴ 2c – 6 =
2c = 7, c =
i.e. x =
y = =
Point is
7. Find the equations of the tangent and normal to the
curve y = x2 +2x +6 at the point (2, 18).
Solution:
y = x2 + 2x +6, = 2x + 2
Slope at (2, 18) = 2 x2 +2 = 6
Equation of the tangent is
y – 18 = 6 (x – 2), y – 18 = 6x – 12
i.e. 6x – y = - 6
Slope of the normal =
Equation of the normal is
y – 18 = (x – 2)
6y – 108 = -x + 2
i.e. x + 6y = 110
REVIEW
1. Sand is pouring from a pipe at the rate of 12 cm 3 / Sec. The falling sand forms a cone on the ground in such a way that the height of the cone is always one - sixth of the radius of the base. How fast is the height of the sane – cone increasing when the height is 4 cm?
Solution :- Given h = h ⇒ r = 6h & = 12 cm 3 / sec.
Volume = πr2h = π(6h) 2 × h = 12πh3
= 12π.3h2 () 12 = 12 π.3h2 ()
=
∴ | h = 4 = = cm/sec.
2. Find the approximate value of using differentials.
Choose x = 0.04 & ∆x = - 0.003
So that x + ∆x = 0.04 – 0.003 = 0.037
f(x) = = f ′(x) =
∴ f(x + ∆x) = f(x) + f′(x) ∆x
= + × - 0.003
= 0.2 -
= 0.2 – 0.0075 = 0.1925.
3. If y = x2 + 4 and x changes from 2 to 2.1, find the approximate change in y.
Solution :- y = x2 + 4 and x = 2 & x + ∆x = 2.1
⇒ ∆x = 0.1
∴∆y = 2x ∆x = 2 × 2 × 0.1 = 0.4
∴ y| x = 2 = 22 + 4 = 8 y + ∆y = 8 + 0.4 = 8.4
4. Show that the curves 4x = y2 and 4xy = k cut at right angles it k2 = 512.
Solution :- 4x = y2 and 4xy = k
y = 4x =
⇒ k2 =64x3 ⇒ x = =
y = = k.
Point of intersection = (,k)
Diff. 4x = y2 w. r. t x, we get
4 = 2y or =
| y = k =
Diff. 4xy = k w. r. t x, we get
4[x. + y.1] = 0
⇒ =
= =
As the curves cut at right angles, product of slopes = -1
∴ × - = -1 ⇒ 8 = k
Cubing, k2 = 512.
5. Find the points on the curve y = x3 -2x2 -x at which the tangent lines are parallel to the line y = 3x –2.
Solution :- y = x3 -2x2 –x
= 3x2-4x –1
Slope = 3x2-4x –1
y = 3x – 2 Slope = 3
Since they are parallel ⇒ 3x2-4x –1 = 3
⇒ 3x2-4x –1 Or 3x2-4x –4 = 0
(3x+2) (x-2) =0 ⇒ x = 2, -
When x = 2; y =(2)2 – 2(2)2 – 2 -2
Point = (2,-2)
When x = - y =(-)2 – 2(-)2 – = -
Point = (-,-)
6. Verify Rolles’ theorem for the function f(x) = sin 2x in [0,]
Being a trigonometric function
it is continuous in [0, ]
it is differentiable in (0,)
f(o) = sin 0 = 0 f() = sin = 0 ∴ f(o) = f()
Hence Rolles’ Theorem are satisfied hence there exist a constant c(0, ) such that F′(c)=0 2cos 2c = 0 ⇒ 2c = or c = ∴c = ε(0, ).
7. Verify LMVT for the function f(x) = (x – 3)(x – 6)(x – 9) in [3, 5].
Solution: Being a polynomial function
it is continuous in [3, 5]
it is differentiable in (3, 5)
Hence conditions are satisfied, hence there exist cε[3, 5] such that
f′(c) =
f′(x) = (x – 3)(x – 6) + (x – 6)(x – 9) + (x – 3) (x – 9)
f′(c) = c²-9c + 18 + c² - 15c + 54 + c² - 12 + 27 3c² - 36c + 99
Hence 3c² - 36c + 99 = 4 3c² - 36c + 95 = 0
8. Prove that the function f(x) = x² - x + 1 is neither increasing nor decreasing in (-1, 1).
Solution: f(x) = x² - x + 1 and f′(x) = 2x – 1
For ↑ fn., f′(x) > 0 when x > ½ and for ↓ fn., f′(x) < 0 when x < ½
Hence in (-1, 1), the function increases the interval (1/2, 1) and decreases in interval (-1, ½).
9. Find the interval for which the function f(x) = sin3x is increasing or decreasing in [0, ].
Solution: f(x) = sin 3x ⇒ f′(x) = 3cos3x
since f′(x) = 0 ⇒ cos3x = 0 3x = x =
For 0 < x < , f′(x) > 0 ∴ f is increasing in (0, )
For < x < , f′(x) < 0 ∴ f is decreasing in (,)
10. It is given that at x = 1, the function f(x) = x4-62x² + 9x + 9 attains its maximum value in [0, 2]. Find the value of a.
Solution: As the function attains maximum value at x = 1,∴ f′(1) = 0
Now f(x) = x4-62x² + 9x + 9
f′(x) = 4x3-124x + 9 ⇒ f′(1) = 0 ⇒ 4.1 – 124.1 + a
⇒ -120 + a = 0 ⇒ a = 120
11. Show that a closed right circular cylinder of given total surface area and maximum volume is such that is height is equal to the diameter of its base.
Solution: s = 2πrh + 2πr²
∴ h =
Volume = πr²h = πr² =
For maximum,
Since r =
-6π
∴ V is maximum when r = ∴ 6πr² = S.
Hence S = 2πrh + 2πr² ⇒ 6πr² = 2πrh + 2πr²
⇒ 4πr² = 2πrh ⇒ 2r = h or d = h
12. Show that the surface area of a closed cuboid with a surface base and given volume is minimum when it is a cube.
Volume = x.x.y = x²y =v
Surface Area = 2( x.x + x.y + x.y)
= 2x² + 4xy 2x² + 4x.
= 2x² + S = 2x² +
For minimum s,
⇒
For x³ = v, surface area is minimum
Since v = x³. v = x²y ⇒ x³ = x²y x = y
Hence cuboid is cube.
13. A figure consists of a rectangle and a semi-circle on its one side. Given the perimeter of the figure, find its dimensions in order that the area may be maximum.
Solution: In the figure, let OC = x then AB = 2x and BC = AD = y
Let P be the given perimeter p = 2x + y + + y
Y =
Area of the figure
= Area of rect. ABCD + Area of the semicircle CDOE
= (2x)y + =
A =
For maximum s
= p – 4x - 2πx + πx
= p – 4x - πx
0 ⇒ x(4 + π) = p x =
∴ A is maximum when x =
y = =
=
= =
∴ x = = radius
y = = breadth
2x = = length.
*******
MULTIPLE CHOICE QUESTIONS
1. The function f(x) = cos x is
(a) Strictly decreasing on [0, ] (b) Strictly decreasing on [, 2]
(c) Strictly decreasing on [0, 2] (d) None of these.
2. The function f (x) = sin 3x, x [ 0, ] is strictly decreasing on
(a) [ 0, ] (b) [ 0, ] (c) [ , ] (d) [ 0, ]
3. On R, the function f(x) = 7x – 3 is
(a) Strictly decreasing (b) decreasing
(c) increasing (d) Strictly increasing
4. The total revenue Rs.R received from the sale of x units of a product is given by
R(x) = 3x2 + 36x + 5. The marginal revenue when x = 5 is ( marginal revenue is the rate of change of total revenue w.r.to the number of items sold at an instant).
(a) 66 (b) Rs.66 (c) 69 (d) None of these
5. The rate of change of the area of a circle w.r.t. its radius r when r = 5 cm is
(a) 10 (b) 10 cm2/cm (c) (d) None of these
6. Equation of normal to the curve y = sin x at (0, 0) is
(a) x = 0 (b) y = 0 (c) x + y = 0 (d) x – y = 0
7. The angle between the curves y2 = x and x2 = y at (1, 1) is
(a) tan -1 (b) tan-1 (c) 900 (d) 450
8. Then slope of the tangent to the curve y=x3-x at x=2 is
(a) 6 (b) 0 (c) 11 (d) None of these
9. The point on the curve - 1, at which the slope of the tangent is is
(a) (3, 2) (b) (2, 3) (c) (1, 0) (d) None of these
10. The point on the curve y = x2 – 4x + 4 at which the tangent is parallel to x-axis is
(a) (0, 2) (b) (2, 0) (c) (0, 0) (d) None of these
11. An approximate value of f (3.02), where the f (x) = 3x2 + 5x + 3 , is
(a) 45.02 (b) 3.46 (c) 45.46 (d) None of these
12. The function f (x) = x has
(a) only one maximum (b) only one minimum
(c) one maximum and one minimum value (d) no extreme value
13. The function f (x) = x3 – 3x + 3 has
(a) a local maximum at x = 1 (b) a local maximum at x = -1
(c) a local minimum at x = -1 (d) None of these
14. Maximum value of f (x) = sin cos x is
(a) (b) 0 (c) (d) None of these
15. The straight line touches the curve at the point (a, b) for
(a) n = 2 (b) n = 3 (c) any n (d) 0
16. The function f(x) =x3 has
(a) a local maximum at 0 (b) a local minimum at 0
(c) a point of inflexion at 0 (d) None of these
17. The point on the curve x2 =2 y which is nearest to the point (0, 5) is
(a) (2, 4) (b) (2, 0) (c) (0, 0) (d) (2, 2)
18. For all the real values of x, the minimum value of is
(a) 0 (b) 1 (c) 3 (d)
19. The line y = mx + 1 is a tangent to the curve y2 = 4 x if the value of m is
(a) 1 (b) 2 (c) 3 (d)
20. For a strictly increasing function f(x) on D, f ’(x) is
(a) positive (b) negative
(c) neither positive nor zero (d) None of these
ANSWERS
1. (a) 2 (c) 3. (d) 4. (b) 5. (b)
6. (c) 7. (b) 8. (c) 9. (a) 10. (b)
11. (c) 12. (d) 13 (b) 14 (c) 15 (c)
16. (c) 17. (a) 18. (d) 19 (a) 20 (a)
******
CHAPTER - 8
APPLICATIONS OF INTEGRALS
FORE-VIEW: The formulae of elementary geometry help us to calculate areas of many simple figures. But to calculate the area enclosed by curves concepts of integral calculus is essential. We give here a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas and ellipses.
Areas of bounded Regions
2. Area bounded by the curve y = f(x), the y axis and y = c to y = d is given by
Area =
Where y =f (x) ⇒ x = g(y)
Note: 1.If the position of the curve under consideration is below the x-axis then the area comes out to be negative. Therefore, only the numerical value of the area is taken into consideration.
2. If some portion of the curve is above x-axis and some is below x-axis then the area bounded by the curve y=f(x), x-axis ,the ordinates x=a and x=b is given by sum of the two areas.
AREA
1. Using integration, find the area of the region bounded by
(-1, 1), (0, 5) and (3, 2).
Let A (-1, 1), B (0, 5) and C (3, 2) be the vertices of a ΔABC.
[1]Equation of AB is y –1 =
y = …………………[2]
Equation of BC is y – 2 =
y = -x + 3 + 2 = 5 – x ..…………………[3]
Equation of AC is y – 1 =
y = 4x + 4 + 1 = 4x + 5…………………[4]
Required area of ΔABC =
=
=
= = 3 + 10.5 – 6= 7.5 Sq.unit.
2. Find the area of the region enclosed between the two circles
x² + y ² = 1, (x – 1)² + y² = 1
On Solving x² + y ² = 1 and (x – 1) ² + y² = 1,
We get x² + y ² = x² - 2x +1+y²
Required Area = 2
=
=
=
= Sq. unit.
3. Draw a sketch of the region bounded by the curve x² = 4y and x = 4y – 2 and determine its area.
Solve x² = 4y & x = 4y – 2
x² = x + 2
x² -x –2 = 0 ⇒ (x – 2) (x – 1) = 0 ⇒ x = -1, 2
Points of intersection of Parabola x² = 4y the line x = 4y – 2 are
; (2, 1)
Required Area = =
= = = = Sq.unit.
4. Sketch the graph of the following region and find its area.
x + y = 1
y = 1 - x
Required Area =
=
=
= Sq.unit.
Sketch the graph of y = |x – 1|. Evaluate . What does the value of this integral represent on the graph?
Required Area = =
= Sq.unit.
Sketch the region common to the circle
x² + y ² = 16 and
the parabola
y² = 6x. Also
find the area
of the region using integration.
Solve x² + y² = 16 & y² = 6x x² + 6x – 16 = 0
(x + 8) (x – 2) = 0
Required Area =
=
=
= =
= = Sq.units.
Find the area of the
region bounded by the
two parabolas y² = 4x
and x² = 4y by using
the method of integration.
Solution: Points of intersection of y² = 4x and x² = 4y is obtained by solving them.
; y² = 4x becomes ⇒ x4 = 64 x
⇒ x(x³ - 64) = 0 ⇒ x = 0; x = 4
Points of intersection are (0, 0) and (4, 4).
Required Area = =
= Required Area = Sq.units.
Find the area bounded by the curve y = x² - 4 and the lines y = 0 and y = 5
Solution: Required Area =
= = = =
= = Sq.units.
9. Draw the rough sketch of the curve y = sinx and y = cosx as x varies from 0 to and find the area of the region enclosed by them and the x-axis.
Solution: y = sinx;
y = cos x;
Required Area = =
= =
= 2 - = Required Area = Sq.units.
10. Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola y2 = 4x.
Solution:. On Solving y2 = 4x & 4x2 + 4y2 = 9 ⇒ 4x2 + 16x – 9 = 0
We get
Required Area =
11. Find the area of the region bounded by two parabolas 4y = x2 and 4x = y2.
Solution: Solve 4y = x2 & y2 = 4x
x4 = 64x ⇒ x (x3 – 64) = 0 ⇒ x = 0, 4.
Required area =
Required Area =
12. Find the area of the region bounded by two parabolas y2 = 4a2(x – 1) and the line x = 1.
Solution: Points of intersection of y2 = 4a2(x – 1) and y = 4a, is x = 5.
Required Area =
Required Area .
13. Find the area enclosed by x2 = y, the line y = x + 2
and the x axis.
Solution:. On solving x2 = y, y = x + 2
We get
x2 = x + 2 ⇒ x2 – x – 2 = 0
(x – 2)(x + 1) = 0
Required area =
14. Find the area of the region { (x, y) : y2 ≤ 4x, 4x2 +4y2 ≤ 9}.
Solution. Solve y2 = 4x; 4x2 + 4y2 = 9
4x2 + 4(4x) – 9 = 0
4x2 + 16x – 9 = 0
Required Area =
Required Area = Ans.
15. Find the area bounded by the curves y = 6x – x2; y = x2 – 2x.
Solution. Put y = 0 in 6x – x2 = y
x(6 – x) = 0 ⇒ x = 0; x = 6
Put y = 0 in x2 – 2x = y ⇒ x2 – 2x = 0
x(x – 2) = 0 ⇒ x = 0, 2.
6x – x2 = x2 – 2x ⇒ 2x2 – 8x = 0
x = 0,4.
Required Area =
POINTS TO CONCENTRATE
1. Identification of curves (circle, parabola etc from the given function)
2. Identify the enclosed area and limits for x or y as warranted.
3. Idea of symmetry
REVIEW
Find the area of the smaller part of the circle cut off by the line x=.
Find the area of the region bounded by ,x=1,x=4 and the x-axis in the first quadrant.
Find the area cut off from the parabola 4y= by the line 2y=3x+12.
Find the area of the region .
Using integration find the area of the triangle whose sides have the equations y=2x+1, y=3x+1and x=4.
Answers:
(2) 28/3 (3) 27 (4) 1/3 (5) 8
Test yourself :
1. Find the area of the triangle PQR where P(2,1) Q(3,4) and R(5,2).
2. Draw a rough sketch of the region { (x,y):yand }. Also find the area of the region sketched using method of integration.
3. Find the area by integration, bounded between
4. Using integration find the area of the region in the first quadrant enclosed by the x-axis the line y=x and the circle
5. Find the area of the smaller region bounded by the ellipseand the straight line .
Answer :
(1) 4 sq.units (2) sq.units (3) sq.units
(4) sq.units (5) sq.units
******
INVERSE TRIGONOMETRIC FUNCTIONS
(Note:- As per CBSE sample paper 2M (1Q) = 2M. There are chances of giving 1 Mark question also)
1 MARK QUESTIONS
Using principal value, write the value of the
-32
-32
-1
-13
-23
-2
Using principal values, write the value of
12+ 12
12- -12
12+ -12
3- -3
1+ -12
1+ -12
32 +-12
-1 +-2
2--2
1 + -12 + -12
-32+ -3
Using principal values, write the value of
sin 3π5
cos 13π6
tan 7π6
cos 43π5
[tan3π4]
[cos7π6]
[sin(-2)]
cos2π3+sin2π3
cos -680°
2cos 212
sin 3 --12
cos 2-37
Find the value of ‘x’, if
x+ 12 =2
3+ x =2
12+ x =2
2+ x =2
Write the domain and range (principal value branch) of the following inverse trigonometric functions: or write the values of ‘x’ and ‘y’ (principal value)
y=x
y=x
y=x
y=x
y=x
y=x
2 MARK QUESTIONS
Simplify 1-cos x 1+cos x , 0<x<π
Simplify cos x 1- sin x , -3π2<x <2
Simplify cos x – sin x cos x+sin x , 0<x<π
Simplify acos x –bsin x bcos x +asin x ,if abtan x >-1
Simplify 1+ x2-1x , x≠0
Simplify 1x2 -1 , x>1
Simplify xa2-x2 , x<a
THREE DIMENSIONAL GEOMETRY
(NOTE: As per CBSE Sample paper: 1M (2Q) + 2M (1Q) + 5M (1Q) = 9 Marks)
LINE
Write the distance of the point (3, – 5, 12) from x-axis.
Write the direction cosines of x, y, and z axis.
Find the direction cosines of the line parallel to z axis.
What are the direction cosines of a line, which makes equal angles with the co – ordinate axes?
Write the directions cosines of a line equally inclined to the three co – ordinate axes.
If a line has direction ratios 2, – 1, – 2, then what are the direction cosines?
If P(1, 5, 4) and Q(4, 1, – 2), find the direction ratios of PQ.
Find the direction cosines of the line passing through the points (–2, 4, – 5) and (1, 2, 3).
Write the direction cosines of the vector: -2i+j-5k.
Find the direction cosines of the line 4 - x2=y6=1- z3
Write the equation of the straight line through the point (α, β, γ) and parallel to z – axis.
If a line makes angles 90° and 60° respectively with the positive directions of x and y axes, find the angle which it makes with the positive direction of z-axis.
Find the vector equation of a line whose Cartesian equation isx-53=y + 47=6 -z2.
Write the Cartesian equation of the line given in vector form: r = (2i + j – 4k) + λ(i – j – k).
Find the equation of a line parallel to x-axis and passing through the origin.
Find the vector and the Cartesian equations of the line through the point (5, 2, – 4) and which is parallel to the vector 3i+ 2j– 8k.
Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2i –j + 4 k and is in the direction i+ 2 j –k.
Write the Cartesian equation of a line which passes through the points (– 2, 4, – 5) and is parallel to the line x+33=4-y5=z+86 .
Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z
Find the vector and Cartesian equation for the line passing through the points (–1, 0, 2) and (3, 4, 6).
Find the vector and the Cartesian equations of the lines that passes through the origin and (5, – 2, 3).
2 MARK QUESTIONS
The x – coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, –2) is 4. Find its z – coordinate.
If the equation of a line AB is 2x-13=y+22=z-33 , find the direction ratios of the line parallel to AB.
Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear. [Hint: Show that d.r’s are proportional.
Write down the Cartesian and vector equations of a line passing through (2, – 1, –1) which is parallel to the given line.
3 MARK QUESTIONS
The position vectors of points A, B and C are λi + 3j, 12i +µj and 11i – 3j respectively. If C divides the line segment joining A and B in the ration 3:1, find the values of λ and µ.
Find the vector equation and Cartesian equation of the line passing through the points P (0, 1, –2) and Q (–1, –1, –3). Also, prove that it passes through the point R whose position vector is (–3i – 5j – 5k).
5 MARK QUESTIONS
Find the shortest distance between the lines and hence write whether the lines are intersecting or not. x-12=y+13=zand x+15=y-21;z=2.
Show that the lines x-12=y-23=z-34 and x-45=y-12=z intersect. Find their point of intersection.
show that the lines x-13=y-1-1;z+1=0 and x-42=z+13;y=0 intersect each other. Also find their point of intersection.
Find the shortest distance between the two lines whose vector equations are r = (6i + 2j + 2k) + λ(i – 2j + 2k) and r = (– 4i – k) + μ(3i – 2j – 2k).
Find the shortest distance between the lines r = (1– t)i + (t – 2)j + (3 – 2t)k and r = (s + 1)i + (2s – 1)j – (2s + 1)k.
Show that the lines r = 3i + 2j – 4k + λ(i + 2j + 2k) and r = 5i – 2j + μ(3i + 2j + 6k) are intersecting. Hence, find their point of intersection.
Find the vector and Cartesian equations of the line passing through the points (1, 2, – 4) and perpendicular to the lines x-83=y+19-16=z-107 and x+53=y-298=z-5-5
A line passes through (2, – 1, 3) and perpendicular to the lines r = (i + j – k) + λ(2i – 2j + k) and r = (2i – j – 3k) + μ(i + 2j + 2k). Obtain its equation in vector and Cartesian form.
Find the equation of the perpendicular drawn from the point (2, 4, –1) to the line x+51=y+34=z-6-9. Also find the foot of the perpendicular. Find the Image of the point the line.
Find the coordinates of the foot of perpendicular drawn from P(1, 8, 4) on the line joining R(0, –1, 3) and S( 2, –3, –1). Find the image of the point on the line.
Find the image of the point (1, 6, 3) in the line: x1=y-12=z-23.
Find the point on the line x+23=y+12=z-32 at a distance of 5 units from point P (1, 3, 3).
PLANE
1 MARK QUESTIONS
Find the direction cosines of the normal to YZ plane?
Find the coordinates of the point where the line x+33=y -1-1=z-5-5 cuts the XY plane.
Write the vector equation of the plane: 3x – 4y + 2z – 8 = 0.
Write the Cartesian equation of the plane: r. (2i –j + 5k ) = 7
Determine the direction cosines of the normal to the plane 3x + 4y + 12z = 5.
Write the intercept cut off by the line 2x + y – z = 5 on x – axis.
Find the equation of the plane with intercepts 2, 3 and 4 on the x, y and z -axis respectively.
Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.
2 MARK QUESTIONS
Determine the direction cosines of the normal to the plane 2x + 3y – z = 5and the distance from origin.
Find the vector equation of the plane which is at a distance of 629 from the origin and its normal vector from the origin is 2i –3 j + 4k. Also find its Cartesian form.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3i+ 5j – 6k.
Find the direction cosines of the unit vector perpendicular to the plane r.(6i–3j –2 k) + 1 = 0 passing through the origin.
Find the vector equation of the plane that passes through the point (1,0,0) and contains the line 𝑟⃗ = λ 𝑗̂.
Write the length of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 21 = 0.
Find the distance of the plane 3x – 4y + 12z = 3 from the origin.
Find the distance of the point (2, 3, 4) from the line r. (3i – 2j + 2k) = – 11.
Write the distance between the parallel planes 2x – y + 3z = 4, 2x – y + 3z = 18.
Find the distance between the planes r.(2i – 3j +6k) – 4 = 0 and r. (6i – 9j +18k) + 30 = 0
3 MARK QUESTIONS
A variable plane which remains at a constant distance 3p from the origin cuts the co-ordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is 1x2+1y2+1z2=1p2.
Find the equation of the plane determine by the points A(3,–1,2) B (5,2,4) and C(–1,–1,6). Also find the distance of the point P(6, 5, 9) from the plane.
Show that the points (0, –1, –1) (–4, 4, 4) (4, 5, 1) and (3, 9, 4) are coplanar. Also, find the equation of the plane.
Find the equation of the plane passing through the points A(3, 2, 1) , B(4, 2, - 2) and C(6, 5, - 1) and hence find the value of λfor which A(3, 2, 1), B(4, 2, - 2), C(6, 5, - 1) and D(λ, 5, 5) are coplanar.
Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1, 2),B(5, 2, 4) and C(– 1, – 1, 6).
Find the vector and Cartesian equations of the plane which bisects the line joining the points (3, –2, 1) and (1, 4, –3) at right angles.
5 MARK QUESTIONS
Find the equation of a plane through the point (–1, –1, 2) and perpendicular to each of the planes 2x + 3y – 3z = 2 and 5x – 4y + z = 6.
Find the equation of the plane passing through the points (1, –1, 2) and (2, –2, 2) and which is perpendicular to the plane 6x – 2y + 2z = 9.
Find the equation of the plane passing through the points A (1, 2, 3) and B(0, – 1, 0) and parallel to the line x-12=y+23=z-3.
Find the equation of a plane through the point (–1, 3, –2) and parallel to the lines x1=y2=z3 and x+2-3=y-12=z+15.
Find the vector equation and Cartesian equation of the plane passing through the intersection of the planes r. (2i – 7j + 4k) + 3 = 0 and r. (3i – 5j + 4k) + 11 = 0 and passing through the point (–2, 1, 3 ).
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
Find the equation of the plane which contains the line of intersection of the planes r.(i + 2j + 3k) –4 = 0; r. (2i + j – k) + 5 = 0 and which is perpendicular to the plane r. (5i + 3j – 6k) + 8 = 0.
Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 and parallel tox-12=y-34=z-55.
Find the equation of the plane passing through the line of intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios 1, 2, 1. Also, find the perpendicular distance of the point P (3, 1, 2) from this plane.
Find the equation of the plane passing through the line of intersection of the planes r. (i + j + k) = 1 and r. ( 2i + 3j – k ) + 4 = 0 and parallel to x – axis.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1and 2x + 3y + 4z = 5 and twice of its y-intercept is equal to three times its z-intercept.
Find the equation of the plane passing through the line of intersection of the planes r.(i + 3j) – 6 = 0 and r.(3i – j – 4k) = 0, whose perpendicular distance from the origin is unity.
Find the equation of the plane containing the linesx-83=y+19-16=z-107 and x-383=y+298=z-5-5
Find the equation of the plane containing the lines r = (2i + j – 3k) + λ(i + 2j + 5k) and r = (3i + 3j + 2k) + µ(3i – 2j + 5k).
Show that the lines x+3-3=y-11=z-55 and x+1-1=y-22=z-55are coplanar. Also, find the equation of the plane containing the lines.
Show that the lines r = (i + j – k) + λ(3i – j) and r = (4i – k) + μ(2i + 3k) are coplanar. Also, find the equation of the plane containing them.
If the lines x-1-3=y-2-2k=z-32 and x-1k=y-21=z-35 are perpendicular, then find ‘k’, and hence find the equation of the plane containing the lines.
Find the equation of the plane containing the lines r = i + j + λ (i + 2j – k) and r = i + j + μ(– i + j – 2k). Find the distance of this plane from origin and also from the point (1, 1, 1).
Find the equation of the plane passing through the point P (1, 1, 1) and containing the line r = (–3i + j + 5k) + λ(3i – j – 5k). Also, S.T. the plane contains the line r = (– i+ 2j + 5k) + µ(i – 2j – 5k).
Find the equation of the plane containing two parallel lines x-12=y+1-1=z3and x4=y-2-2=z+16. Also, find if the plane thus obtained contains the line x-23=y-11=z-25.
Find the coordinates of the point where the line x+12=y+23=z+34 meets the plane x + y + 4z = 6.
Find the coordinates of the point where the line through (5,1,6) and (3,4,1) crosses the ZX plane.
Find the point where the line joining the points (1, 3, 4) and (–3, 5, 2) intersects the plane r. (2i + j + k ) + 3 = 0.
Find the coordinate of the point where the line through A(3, - 4, - 5) and B(2, - 3, 1) crosses the plane passing through three points L(2, 2, 1), M(3, 0, 1) and N(4, - 1, 0). Also, find the ratio in which P divides the line segment AB.
Find the vector equation of the plane passing through three points with position vectors i + j – 2k, 2i – j + k and i + j + k. Also, find the co – ordinates of the point of intersection of this plane and the line r = (3i – j + k) + λ(2i – 2j + k).
Find the equation of the line passing through the point P(4, 6, 2) and the point of intersection of the linex-13=y2=z+17 and the plane x + y – z = 8.
Find the distance of the point (2, 12, 5) from the point of intersection of the line r = (2i – 4j + 2k) + λ(3i + 4j + 2k) and the plane r.(i – 2j + k) = 0.
Find the co-ordinates of the point where the line r = (– i – 2j – 3k) + λ(3i + 4j + 3k) meets the plane which is perpendicular to the vector n = i + j + 3k and at a distance of 4/√11 from origin.
Find the equation of the perpendicular drawn from point (1,–2,3) to the plane 2x–3y+ 4z+9= 0. Also, find the coordinates of the foot of the perpendicular. Also find image.
Find the length and foot of the perpendicular from the point P (7, 14, 5) to the plane 2x + 4y – z = 2. Also, find the image of the point P in the plane.
Find the image of the point with position vector 3i + j + 2k in the plane r.(2i – j + k) = 4.
Find the image P’ of the point P having position vector i + 3j + 4k in the plane r.(2i – j + k) + 3 = 0. Hence find the length of PP’.
Prove that image of the point (3,–2,1) in the plane 3x – y +4z = 2 lies on the plane x + y + z + 4 = 0.
Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2x = y = z.
Find distance of the point (1,–2,3) from the plane x – y + z = 5 measured parallel to the linex -12=y-33=z+2-6.
Find the distance of the point (–2,3,–4) from the line x+23=2y+34=3z+45 measured parallel to the plane 4x + 12y – 3z + 1 = 0.
Find the vector and Cartesian equations of a line passing through the point P(1, 2, 3) and parallel to the planes r.(i – j + 2k) = 5 and r.(3i + j + k) = 6.
Let P(3, 2, 6) be a point in the space and Q be a point on the line r = (i – j + 2k ) + μ(–3i + j + 5k), then find the value of μ for which the vector PQ is parallel to the plane x – 4y + 3z = 1.
Find the distance between the line x -53=y-43=z-81 and the plane determined by the points A(2, – 2, 1), B(4, 1, 3) and C(– 2, –2, 5).
Find the distance between the line r = 3i + 5j – 2k + λ(3i + j + 3k) and the plane determined by the points A(1, 1, 0), B(1, 2, 1) and C(– 2, 2, –1).
Topic - Probability
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