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Class 12 Applied Maths Chapter 6 Solutions
Application of Derivatives
EXERCISE- 6.2
Q.1 What is the slope of the tangent to the following curves:
(i) y = 3x^4 – 4x at x = 4?
(ii) x^2 + 3y + y^2 = 5 at the point (1, 1)?
(iii) y = x^3 – 3x + 2 at the point whose x-coordinate is 3?
Ans.(i) y = 3x^4 – 4 at x= 4
dy/dx = 12x^3 – 4
(dy/dx) [at x = 4] = 12 x (4)^3 – 4
= 12 x 64 – 4
= 764
(ii) x^2 + 3y + y^2 = 5
2x + 3 dy/dx + 2y (dy/dx) = 0
dy/dx (3 + 2y) = -2x
(dy/dx) (1,1) = -2x/(3 + 2y)
dy/dx = -2/(3 + 2)
dy/dx = -2/5
(iii) y = x^3 – 3x + 2
dy/dx = 3x^2 – 3
(dy/dx) (x = 3) = 3 x (3)^2 – 3
dy/dx = 27 – 3
dy/dx = 24
Q.2 What is the slope of the normal to the following curves:
(i) y = x^3 – 5x^2 – x + 1 at the point (1,-4)?
(ii) y = 2x^2 + 3e^x at x = 0?
Ans.(i) y = x^3 – 5x^2 – x + 1
dy/dx = 3x^2 – 10x – 1
(dy/dx) (1, -4) = 3(1)^2 – 10(1) – 1
dy/dx = 3 – 10 – 1
dy/dx = -7 – 1
Slope of tangent = -8
Slope of Normal = -1/slope of tangent
Slope of Normal = -1/-8
Slope of Normal = 1/8
(ii) y = 2x^2 + 3e^x
dy/dx = 4x + 3e^x
(dy/dx) (x = 0) = 4 x 0 + 3 x e^0
dy/dx = 0 + 3 x 1
Slope of Tangent = 3
Slope of Normal = -1/3
Q.3 Find the point on the curve y = x^2 – 2x + 3 at which the tangent is parallel to x-axis.
Ans. y = x^2 – 2x + 3, slope of tangent = 0
dy/dx = 2x – 2
dy/dx = 0
2x – 2 = 0
2x = 2
x = 1
y = x^2 – 2x + 3
y = 1 – 2 + 3
y = -1 + 3
y = 2
Therefore, Point (1, 2)
Q.4 Find the slope of the tangent to the (x – 1)/(x – 2) at x = 10.
Ans. y = (x – 1)/(x – 2)
dy/dx = [(x – 2) – (x – 1)]/(x – 2)^2 [Using Quotient Rule]
dy/dx = [x – 2 – x + 1]/(x – 2)^2
(dy/dx) (at x = 10) = -1/(x – 2)^2
dy/dx = -1/(10 – 2)^2
dy/dx = -1/64
Q.5 Show that the tangents to the curve y = 7x^3 + 11 at the points where x = 2 and x = -2 are parallel.
Ans. y = 7x^3 + 11
dy/dx = 21x^2
(dy/dx) (at x =2) = 21 x (2)^2
dy/dx = 21 x 4 = 84
(dy/dx) (at x = -2) = 21x^2
dy/dx = 21 x (-2)^2
dy/dx = 21 x 4
dy/dx = 84
Q.6 Find the points on the curve x^2/4 + y^2/25 = 1 at which the tangents are
(a) parallel to x-axis (b) parallel to y-axis.
Ans.(a) x^2/4 + y^2/25 = 1
1/4 (2x) + 1/25 (2y) (dy/dx) = 0
2y/25 (dy/dx) = -x/2
dy/dx = -25x/4y
dy/dx = 0
-25x/4y = 0
-25x = 0
x = 0
x^2/4 + y^2/25 = 1
0/4 + y^2/25 = 1
y^2/25 = 1
y^2 = 25
y = ±5
x = 0, y = 5 [point 0, 5)
x = 0 y = -5 [point 0, -5]
(b) x^2/4 + y^2/25 = 1
1/4 (2x) + 1/25 (2y) (dy/dx) = 0
2y/25 (dy/dx) = -x/2
dy/dx = -25x/4y
When parallel to y-axis dy/dx = Not defined
1/0 = -25x/4y
4y = 0
y = 0
x^2/4 + y^2/25 = 1
x^2/4 = 1
x^2 = 4
x = ±2
x = 2, y = 0 [Point 2, 0]
x = -2, y = 0 [Point -2, 0]
Q.7 Find the points on the curve y = x^3 – 3x^2 + 2x at which the tangent lines are parallel to the line y – 2x + 3 = 0.
Ans. y = x^3 – 3x^2 + 2x
dy/dx = 3x^2 – 6x
Equation of line = -2x + y + 3 = 0
Slope = -A/B = -(-2)/1 = 2
Therefore, tangent is parallel to line.
3x^2 – 6x + 2 = 2
3x(x – 2) = 0
x = 0, x = 2
If, x = 0, y = 0 [Point 0, 0]
If, x = 2, y = (2)^3 – 3 (2)^2 + 2(2)
y = 8 – 12 + 4 = 0 [Point 2, 0]
Q.8 Find the equations of the tangent and the normal to each of the following curves at the given point:
(i) y = x^3 at (1, 1)
(ii) y = x^4 – 6x^3 + 13x^2 – 10x + 5 at (1, 3)
(iii) y = x^4 – 6x^3 + 13x^2 – 10x + 5 at (0,5)
(iv) y = x^2 at (0,0)
Ans.(i) y = x^3
dy/dx = 3x^2
dy/dx (1, 1) = 3 (1)^2 = 3
Therefore, slope of tangent = 3
Equation of tangent, y – 1 = 3 (x – 1)
y – 1 = 3x – 3
3x – y – 2 = 0
Slope of Normal = -dx/dy = -1/3
Equation of Normal, y – 1 = -1/3 (x – 1)
3y – 3 = -x + 1
x + 3y – 4 = 0
(ii) y = x^4 – 6x^3 + 13x^2 – 10x + 5 at (1, 3)
dy/dx = 4x^3 – 18x^2 + 26x – 10
dy/dx (1, 3) = 4 – 18 + 26 – 10
dy/dx = 30 – 28 = 2
Therefore, slope of tangent = 2
Equation of tangent, y – 3 = 2 (x – 1)
y – 3 = 2x – 2
2x – y + 1 = 0
Slope of Normal = -1/2
Equation of Normal, y – 3 = -1/2 (x – 1)
2y – 6 = -x + 1
x + 2y – 7 = 0
(iii)
(iv)
Q.9 Find the equation of the normal to the curve ay^2 = x^3 at the point (am^2, am^3).
Ans. ay^2 = x^3
2ay (dy/dx) = 3x^2
dy/dx = 3x^2/2ay
dy/dx (am^2, am^3) = (3a^2m^4)/(2a x am^3)
dy/dx = 3m/2
Slope of Normal = -2/3m
Equation of Normal,
y – am^3 = -2/3m (x – am^2)
3my – 3am^4 = -2x + 2am^2
2x + 3my – 3am^4 – 2am^2 = 0
2x + 3my – am^2 (3m^2 + 2) = 0
Q.10 Find the equations of the tangent and the normal to the curve x^(2/3) + y^(2/3) = 2 at (1, 1).
Ans. x^(2/3) + y^(2/3) = 2
2/3x^(-1/3) + 2/3y^(-1/3).(dy/dx) = 0
y^(-1/3) (dy/dx) = -x^(-1/3)
dy/dx = – x^(-1/3)/y^(-1/3)
dy/dx = – (y/x)^(1/3)
dy/dx [at 1, 1] = -1
Slope of tangent = -1
Equation of tangent,
y – 1 = -1 (x – 1)
y – 1 = – x + 1
x + y – 2 = 0
Slope of Normal = 1
Equation of Normal,
y- 1 = 1 (x – 1)
y – 1 = x – 1
x – y = 0
Q.11 Find the equations of all lines having slope 2 and being tangents to the curve y + [2/(x – 3)] = 0.
Ans. y + [2/(x – 3)] = 0
y = – 2/(x – 3)
dy/dx = 2/(x – 3)^2 [Quotient Rule]
ATQ, 2/(x – 3)^2 = 2
(x – 3)^2 = 1
x – 3 = ±1
If, x – 3 = 1, x = 2
If, x + 3 = 1, x = 4
When x = 4, y + 2/(4 – 3) = 0, y = -2 [Point 4, -2]
Equation of tangent, y + 2 = 2 (x – 4)
y + 2 = 2x – 8
2x – y – 10 = 0
When x = 2, y + 2/(2 – 3) = 0, y = 2 [Point 2, 2]
Equation of tangent, y – 2 = 2 (x – 2)
y – 2 = 2x – 4
2x – y – 2 = 0
Q.12 Find the equations of the normals to the curve 3x^2 – y^2 = 8 which are parallel to the line x + 3y = 4.
Ans. x + 3y = 4
3y = -x + 4
y = -1/3x + 4/3
[From y = mx + c]
m = -1/3
Now, 3x^2 – y^2 = 8
6x – 2y (dy/dx) =0
dy/dx = -6x/-2y = 3x/y
Slope of Normal = -1/(dy/dx)
-1/(3x/y) = -4/3x
ATQ, -y/3x = -1/3
x = y
Now, 3x^2 – y^2 = 8
3x^2 – x^2 = 8
2x^2 = 8
x^2 = 4
x = ±2
If, x = 2, y = 2
Equation , y – 2 = -1/3 (x – 2)
3y – 6 = -x + 2
x + 3y – 8 = 0
If, x = -2, y = -2
Equation, y + 2 = -1/3 (x + 2)
x + 3y + 8 = 0
Q.13 Find the equation of the normal to the curve y = x + 1/x, x > 0 perpendicular to the line 3x – 4y = 7.
Ans. 3x – 4y = 7
-4y = -3x + 7
y = -3/-4x + 7/-4
y = 3/4x – 7/4
m1 x m2 = -1
3/4 x m2 = -1
m2 = -1 x 4/3
m2 = -4/3
Now, y = x + 1/x
dy/dx = 1 – 1/x^2
dy/dx = (x^2 – 1)/x^2
Slope of Normal = -1/[(x^2 – 1)/^2]
x^2/(x^2 – 1) = -4/3
3x^2 = 4x^2 – 4
-x^2 = -4
x = ±2
If, x = -2 [Not possible because according to the question x > 0]
If, x = 2, y = 2 + 1/2, y = 5/2
Equation of Normal, y – 5/2 = -4/3 (x – 2)
(2y – 5)/2 = (-4x + 8 )/3
6y – 1 5= -8x + 16
8x + 6y – 31 = 0
Q.14 Find the equation of the normal to the curve x^2 = 4 which passes through the point (-1,4).
Ans. x^2 = 4
2x = 4 dy/dx
dy/dx = x/2
dy/dx [a, b] = a/2
Slope of Normal = -1/(a/2) = -2/a
Equation of Normal, y – y1 = -1/(dy/dx) (x – x1)
4 – b = -2/a (-1 – a)
4a – ab = 2 + 2a
4a – a x a^2/4 = 2 + 2a
16a – a^3 = 8 + 8a
a^3 – 8a + 8 = 0
Put, a = 2
8 – 8 x 2 + 8 = 0
8 – 16 + 8 = 0
a – 2 is a factor,
So, a = 2
Therefore, 4b = (2)^2
4b = 4
b = 1 [Points 2, 1]
Slope of Normal = -2/2 = -1
Slope of Equation, y – 1 = -1 (x – 2)
y – 1 = -x + 2
x + y – 3 = 0
Q.15 Prove that the curves xy = 4 and x^2 + y^2 = 8 touch each other.
Ans. xy = 4
x = 4/y
Putting value of x in x^2 + y^2 = 8
16/y^2 + + y^2 = 8
(16 + y^2)/y^2 = 8
16 + y^2 = 8y^2
y^4 – 8y^2 + 16 = 0
(y^2 – 4)^2 = 0
y^2 – 4 = 0
(y – 2) (y + 2) = 0
y = 2, y = -2
When, y = 2, x = 4/y = 4/2 = 2 [Point 2, 2]
When, y = -2, x = 4/y = 4/-2 = -2 [Point 2, -2]
Now, xy = 4
1 (y) + x (dy/dx) = 0 [Using Product rule]
dy/dx = -y/x
m1 = -y/x
Now, x^2 + y^2 = 82x + 2y (dy/dx) = 0
y (dy/dx) = -x
dy/dx = -x/y
m2 = -x/y
At (Point 2, 2)
m1 = -1, m2 = -2
At (Point -2, -2)
m1 = -1, m2 = -1
Q.16 Find the condition that the curves 2x = y^2 and 2xy = k intersect orthogonally.
Ans. 2x = y^2
x = y^2/2
Putting value of x in 2xy = k
2 x y^2/2 x y = k
y^3 = k, y = k^(1/3)
If, y = k^(1/3), x = k^(2/3)/2
Therefore, points (k^(1/3), k^(2/3)/2]
Now, 2x = y^2
2 = 2y (dy/dx)
dy/dx = 1/y
m1 = 1/y
Now, 2xy = k
2 [1 (y) + x (dy/dx)] = 0
x (dy/dx) = -y
dy/dx = -y/x
m2 = -y/x
AtQ, m1 x m2 = -1
(1/y) x (-y/x) = -1
x = 1
k^(2/3)/2 = 1
k = 2^(3/2)
k^2 = 2^(2/3 x 2)
k = 8
FAQ’s related to Class 12 Applied Maths Chapter 6 on Application of Derivatives :
Q.1 What are the main applications of derivatives?
Ans. Derivatives have the following applications:
(i) Finding approximations using differentials.
(ii) Finding rate of change of a quantity.
(iii) Determining tangents and normals to a curve.
(iv) Solving optimization problems (maximum and minimum values).
(v) Identifying the increasing or decreasing nature of functions.
Q.2 How do you find the rate of change of a quantity?
Ans. The rate of change of a quantity y with respect to another quantity xxx is given by dy/dx. This represents how y changes as x varies.
Q.3 What is the geometrical significance of a derivative?
Ans. The derivative dy/dx represents the slope of the tangent to the curve at a given point. It indicates how steep the curve is at that point.
Q.4 What are tangents and normals, and how are they calculated?
Ans. A tangent is a line that touches a curve at one point without crossing it.
A normal is a line perpendicular to the tangent at the point of contact.
Slope of normal: m(n) = −1/m(t)
Slope of tangent: m(t) = dy/dx at the given point.
These are a few Frequently Asked Questions relating to Class 12 Applied Maths Chapter 13
In Class 12 Applied Maths chapter 6, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Class 12 Applied Maths Chapter 6 promises an enriching learning experience that will set you on the path to success. Class 12 Applied Maths Chapter 6, we delve deep into advanced mathematical concepts that are crucial for understanding.
