Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2

In this article, we will see some problems of derivatives a fundamental concept in calculus, and mathematical analysis that measures how a function changes as input changes.

Exercise 6.3 focuses on the application of derivatives to find the rate of change of quantities. This exercise builds upon the concepts of differentiation and explores how derivatives can be used to solve real-world problems involving rates of change in various contexts, such as physics, geometry, and economics.

What is Derivative

A derivative is a fundamental concept in calculus and mathematical analysis that measures how a function changes as input changes. In more practical terms, the derivative of a function at a given point describes the rate at which the function's value is changing at that point.

The content of this article has been merged with Chapter 6 Application of Derivatives -Exercise 6.3 as per the revised syllabus of NCERT.

Question 14. Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x⁴- 6x³ + 13x² - 10x+ 5 at (0, 5)

(ii) y = x⁴ - 6x³ + 13x² - 10x + 5 at (1, 3)

(iii) y = x³ at(1, 1)

(iv) y = x² at(0, 0)

(v) x = cos t, y = sin t at t = π/4

Solution:

(i) Given curve

y = x⁴ - 6x³ + 13x² - 10x + 5

Given point, (0, 5)

dy/dx = 4x³ - 18x² + 26x - 10

dy/dx = 4(0)³ - 18(0)² + 26(0) - 10

dy/dx = -10,

-dx/dy = 1/10 

Now, with the help of points slope form 

y - y₁ = m(x - x₁)

y - 5 = -10(x - 0)

y + 10x = 5 is the required equation of the tangent 

For equation of normal,

y - y1 = \frac{-dx}{dy}(x-x_1)

y - 5 = \frac{1}{10}(x-0)

10y - x - 50 is the equation of normal.

(ii) Given curve: y = x⁴ - 6x³ + 13x² - 10x + 5

Given point is (1, 3)

dy/dx = 4x³ - 18x² + 26x - 10 

dy/dx = 4(1)³ - 18(1)² + 26(1) - 10

dy/dx = 4 - 18 + 26 - 10 = 2

dy/dx = 2       

-dx/dy = -1/2  

Using point slope form, equation of tangent is 

y - y₁ = dy/dx(x - x1)

y - 3 = 2(x - 1)

y - 2x = 1 is the equation of tangent.

Using point slope form, equation of normal is

y -  y₁ = -dx/dy(x - 1)

y - 3 = -1/2(x - 1)

2y - 6 = -x + 1

2y + x = 7 is the equation of normal.

(iii) Given curve : y = x³

Given point is (1, 1)

dy/dx = 3x²

dy/dx = 3(1)² = 3

dy/dx = 3 & -dx/dy = -1/3

Using point slope form, equation of tangent is y - y₁ = dy/dx(x - x₁)

y - y₁ = dy/dx(x - x₁)

y - 1 = 3(x - 1)

y - 3x + 2 = 0 is the equation of tangent

Using point slope form, equation of normal is 

y - y1 = \frac{-dx}{dy}(x-x_1)

y - 1 = \frac{-1}{3}(x-1)

3y - 3 = -x + 1

3y + x = 4 is the equation of normal.

(iv) Given curve: y = x²

Given point (0, 0)

dy/dx = 2x

dy/dx = 0

dy/dx = 0 & -dx/dy = not defined is 

y - y₁ = dy/dx(x - x₁)

y - 0 = 0(x - 0)

y = 0 is the equation of tangent.

Using point slope form, equation of normal is 

y - y₁ = \frac{-dx}{dy}(x-x_1)              -(1)

-dx\dy is undefined, so we can write eq(1) as 

\frac{-dy}{dx}(y-y_1)=x-x_1

Now putting dy/dx = 0 we get

0(y - 0) = x-0

x = 0 is the equation of normal.

(v) Equation of curve: x = cos t and y = sin t

Point t = π/4

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}          -(1)

\frac{dx}{dt}=\frac{d(\cos t)}{dt}=-\sin t

\frac{dy}{dt}=\frac{d(\sin t)}{dt}=\cos t

On putting these values in eq(1), we get

\frac{dy}{dx}=\frac{\cos t}{-\sin t}=-\cot t

\frac{dy}{dx}=-\cot \frac{π}{4}=-1

dy/dx = -1 & -dx/dy = 1  

Now for t = π/4,

y₁ = sin t = sin(π/4) = 1/√2

x₁ = cos t = cos(π/4) = 1/√2

The point is (1/√2, 1/√2) 

y - y₁ = \frac{dy}{dx}(x-x_1)

y - (1/√2) = -1(x - 1/√2)

y - 1/√2 = -x + 1/√2

x + y = √2 is the equation of normal is 

y-y_1=\frac{-dx}{dy}(x-x_1)

y - 1/√2 = 1(x - 1/√2)

x = y is the equation of normal.

Question 15. Find the equation of the tangent line to the curve y = x² - 2x + 7 which is 

(i) Parallel to line 2x - y + 9 = 0 

(ii) Perpendicular to the line 5y - 15x = 13

Solution:

Given curve: y = x² - 2x + 7

On differentiating w.r.t. x, we get

dy/dx = 2x - 2         -(1)

(i) Tangent is parallel to 2x - y + 9 = 0 that means,

Slope of tangent = slope of 2x - y + 9 = 0

y = 2x + 9

Slope = 2            -(Comparing with y = mx + e)

dy/dx = slope = 2

2x - 2 = 2

x₁ = 2

Corresponding to x₁ = 2,

y₁ = x₁² - 2x₁ + 7 

y₁ = (2)² - 2(2) + 7

y₁ = 7

The point of contact is (2, 7).

Using point slope form, equation of tangent is 

y - y₁ = \frac{dy}{dx}(x-x_1)

y - 7 = 2(x - 2)

y - 2x = 3 is the equation of tangent. 

(ii) Tangent is perpendicular to the line 5y - 15x = 13

That means (slope of tangent) x (slope of line) = -1

For, slope of line 5y - 15x = 13

5y = 15x + 13

y = 3x + 13/15

Slope = 3

(Slope of tangent) x 3 = -1

Slope of tangent =-1/3 

Now, y = x² - 2x + 7

dy/dx = 2x - 2

On comparing dy/dx with slope, we get

2x - 2 = -1/3

6x - 6 = -1

6x = 5

x₁ = 5/6

For x₁ = 5/6,          

y₁ = x₁² - 2x₁ + 7 

y₁ = (5/6)₁² - 2(5/6)₁ + 7 

y₁ = 217/36 

Now using point slope form, equation of tangent is 

y - y₁ = m(x - x₁)

y-\frac{217}{36}=\frac{-1}{3}(x-\frac{5}{6})

\frac{36y-217}{36}=-1\frac{(6x-5)}{6.3}

36y - 217 = -12x + 10

36y + 12x = 227 is the required equation of tangent.

Question 16. Show that the tangent to the curve y = 7x³ + 11 at the points where x = 2 and x = -2 are parallel.

Solution:

Given curve: y = 7x³ + 11

On differentiating w.r.t. x, we get

dy/dx = 21x²

dy/dx = 21(2)² = 84

The slopes at x - 2 & -2 are the same,

Hence the tangent will be parallel to each other.

Question 17. Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinates of the point.

Solution:

Given curve: y = x³

On differentiating w.r.t. x, we get

dy/dx = 3x²          -(1)

Now let us assume that the point is (x₁, y₁)

dy/dx = 3x₁² 

Also, slope of tangent at (x1, y1) is equal to y₁.

So, 3x₁² = y₁         -(2)

Also, (x1, y1) lies on y = x³ x3,so

y₁ = x₁³           -(3)

From eq(2) & (3)

3x₁² = x₁³  

3x₁² = x₁³ = 0

x₁²(3 - x₁) = 0 

For x₁ = 0, y₁ = x₁³ = (0)3 = 0

One such point is (0, 0)

For x₁ = 3, y_1. = (3)3 = 27

Second point is (3, 27)

Question 18. For the curve y = 4x³ - 2x⁵, find all the points at which the tangent passes through the origin.

Solution:

Given curve: y = 4x³ - 3x⁵

Clearly at x = 0, y = 0, i.e the curve passes through origin. 

Now the tangent also passes through origin. 

Equation of a line passing through origin is y = mx.

Now tangent is touching the curve, so 

y = mx will satisfy in curve.

mx = 4x³ - 2x⁵         -(1)

Now dy/dx = 12x² - 10x⁴

Also m is the slope of tangent, so

m = 12x² - 10x⁴          -(2)

From eq(1) & (2),

(12x² - 10x⁴)x = 4x³ - 3x⁵

x³(12 - 10x²) = x³(4 - 2x²)          -(3)

For the first point, x = 0

For x₁ = 0, y₁ = 4x₁³ - 2x₁⁵ = 0 

So, (0, 0) is one such point

Now for other roots of 3

12 - 10x² = 4 - 2x²

8 = 8x²

x² = 1

x = ±1

For x₂ = 1, y₂ = 4x₂³ - 2x₂⁵ = 4(1)³ - 2(1)⁵ = 2 

For x₃ = -1,  y₃ = 4x₃³ - 2x₃⁵ = 4(-1)³ - 2(-1)⁵ = -2   

The other points are(1, 2) & (-1, -2)

Question 19. Find the points on the curve x² + y² - 2x - 3 = 0 at which the tangents are parallel to the x-axis.

Solution:

Given curve: x² + y² - 2x - 3 = 0

On differentiating w.r.t. x, we get

3x + 2y(dy/dx) - 2 - 0 = 0

x+y\frac{dy}{dx}=1 \frac{dy}{dx}=\frac{1-x}{y}

Given that the tangent are parallel to x-axis,

So, dy/dx = slope = 0

1 - x/y = 0

For x₁ = 1,

x₁² + y₁² - 2x₁ - 3 = 0      

(1)² + y₁² - 2(1) - 3 = 0      

 y₁² = 4     

y₁² = ≠2

The points are (1, 2) & (1, -2)

Question 20. Find the equation of the normal at the point (am², am³) for the curve ay² = x³.

Solution:

Given curve: ay² = x³

On differentiating w.r.t. x, we get

2ay.dy/dx = 3x²

\frac{dy}{dx}=\frac{3x^2}{2ay}  & \frac{-dx}{dy}=\frac{-2ay}{3x^2}

\frac{-dx}{dy}=\frac{-2a.am^3}{3.(am^2)^2}=\frac{-2}{3m}

So, by point slope form, equation of normal is,

y-y_1=\frac{-dx}{dy}(x-x_1)    

y-am^3=\frac{-2}{3m}(x-am^2)

3my - 3am⁴ = -2x + 2am²

Hence, 3my + 2x = 2am² + 3am⁴ is the required equation of normal.

Question 21. Find the equation of the normals to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.

Solution:

Given curve: y = x³ + 2x + 6

On differentiating w.r.t. x, we get

\frac{dy}{dx}=3x^2+2

\frac{-dx}{dy}=\frac{-1}{3x^2+2}          -(Slope of normal)

Now, the normal are parallel to x + 14y + 4 = 0

\frac{-1}{3x^2+2}=\frac{-1}{14}

13 = 3x² + 2            

3x² = 12 ⇒  x² = 4

x = ±2

x₁ = 2 & x₂ = -2

For x₁ = 2; y₁ = x_1^3+2x_1+6  = (2)³ + 2(2) + 6 = 18

For x₂ = -2; y₂ = x_2^2+2x_2+6  = (-2)³ + 2(-2) + 6 = -6

Normal through (2,18) is 

y - 18 = \frac{-1}{14}(x-2)

14y - 252 = -x + 2

14y + x = 254 is one such equation.

Normal through (-2, -6) is 

y + 6 = \frac{-1}{14}(x+2)

14y + 84 = -x - 2

14y + x + 86 = 0 is the other equation of normal.

Question 22. Find the equations of tangent and normal to the parabola y² = 4ax at the point (at², 2at).

Solution:

Given parabola: y² = 4ax

On differentiating w.r.t. x, we get

2y.\frac{dy}{dx}=4a

\frac{dy}{dx}=\frac{2a}{y}  ;\frac{-dx}{dy}=\frac{-y}{2a}

\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t}

Now, by point slope form, equation of tangent is,

y - y₁ = \frac{dy}{dx}(x-x_1)

y-2at=\frac{1}{t}(x-at^2)

ty = x + at² is the equation of tangent to the parabola y² = 4ax at (at², 2at)

Now \frac{-dx}{dy}=\frac{-2at}{2a}=-t

Now, by point slope form, equation of normal is,

y - y₁ = \frac{-dx}{dy}(x-x_1)

y - 2at = -t(x - at²)

y + xt = 2at + at³ is the equation of normal to the parabola y² = 4ax at (at², 2at)

Question 23. Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.

Solution:

Given curves: x = y² & xy = k

Two curves intersect at right angles f the tangents through their point 

intersection is perpendicular to each other.

Now if tangents are perpendicular their product of their slopes will be equal to -1.

Curve 1: x = y²

1 = 2y.dy/dx

dy/dx = 1/2 y = m₁          -(1)

Curve 2: xy = k

y = k/x

\frac{-k}{x^2}=m_2

Let's find their point of intersection 

x = y² & xy = k

k/y = y² 

y = k^1/3

x = y²      

x = k^2/3

The point is (k^2/3, k^1/3)

m₁ = 1/2k^1/3

For curves to be intersecting each other at right angles,

m₁m₂ = -1

\frac{1}{2k^\frac{1}{3}}.\frac{-1}{k^\frac{1}{3}}=1

2k^\frac{2}{3}=1

8k² = 1  

Hence proved        

Question 24. Find the equations of the tangent and normal to the hyperbola\frac{x^2}{a^2}-\frac{y^2}{b^2}=1  at the point (x_o, y_o).

Solution:

Given curve: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

On differentiating both sides with respect to x,

\frac{2x}{a^2}-\frac{2y}{b^2}.\frac{dy}{dx}=0    

\frac{dy}{dx}=\frac{xb^2}{ya^2}\&\frac{-dx}{dy}=\frac{-ya^2}{xb^2}

Now, \frac{dy}{dx}=\frac{x_ob^2}{y_0a^2}=m_T

Equation of tangent by point slope form is,

y-y_1=m_T(x-x_1)

y-y_0=\frac{x_0b^2}{y_0a^2}(x-x_0)

\frac{yy_0-y_0^2}{b^2}=\frac{xx_0-x_0^2}{a^2}

\frac{yy_0}{b^2}-\frac{xx_0}{a^2}=\frac{y_0^2}{a^2}=-1         -((x₀,y₀) lie on \frac{x^2}{a^2}-\frac{b^2}{b^2}=1 )

\frac{xx_0}{a^2}-\frac{yy_0}{b^2}=1  is the equation of tangent.

Now, \frac{-dx}{dy}=\frac{-y_0a^2}{x_0b^2}=m_N

Equation of normal by point slope form is,

y-y_1=m_N(x-x_1)

y-y_0=\frac{-y_0a^2}{x_0b^2}(x-x_0)

x₀b²y - x₀b²y₀ = -y₀a²x + y₀a²x₀

x₀b²y + y₀a²x = x₀y₀(a² + b²) is the equation of normal

Question 25. Find the equation of the tangent to the curve y = \sqrt{3x-2}  which is parallel to the line 4x - 2y + 5 = 0.

Solution:

Given curve: y=\sqrt{3x-2}

On differentiating w.r.t. x, we get

\frac{dy}{dx}=\frac{1}{2\sqrt{3x-2}}.3

Now, it is given that the tangent is parallel to the line 4x - 2y + 5 = 0, 

so their slopes must be equal.

Slope of 2y = 4x + 5 is 2. 

So,

\frac{dy}{dx}=\frac{3}{2\sqrt{3x-2}}=2

\frac{3}{4}=\sqrt{3x-2}

9/16 = 3x - 2

x₁ = 41/48

Now, y_1=\sqrt{3x_1-2}

y_1=\sqrt{3.\frac{41}{48}-2}=\frac{3}{4}

The point is (41/48, 3/4)

Now by point slope form, the equation of tangent will be 

y - y₁ = m(x - x₁)

y-\frac{3}{4}=2(x-\frac{41}{48})

y-\frac{3}{4}=2x-\frac{41}{24}

24y - 48x + 23 = 0 is the required equation of tangent.

Question 26. The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is 

(A) 3          (B) 1/3          (C)-3          (D) -1/3

Solution:

Given curve: y = 2x² + 3 sin x

On differentiating w.r.t. x, we get

dy/dx = 4x + 3cos x

\frac{-dx}{dy}=\frac{-1}{4x+3\cos x}                  -(Slope of normal)

\frac{-dx}{dy}=\frac{-1}{4(0)+3\cos 0}=\frac{-1}{3}

Hence, the correct option is D.

Question 27. The line y = x + 1 is a tangent to the curve y² = 4x at the point

(A) (1, 2)          (B) (2, 1)          (C) (1, -2)          (D) (-1, 2)

Solution:

Given curve: y² = 4x

On differentiating w.r.t. x, we get

2y\frac{dy}{dx}=4        

\frac{dy}{dx}=\frac{2}{y}

y = x + 1 is tangent, slope is 1, so, 2/y = 2

y₁ = 2,

y₁² = 4x₁ 

x₁ = \frac{2^2}{4}    

x₁ = 1

So, (1, 2) is the point.

Hence, the correct option is A.

Summary

Exercise 6.3 applies the concept of derivatives to solve problems involving rates of change in various real-world scenarios. It covers a wide range of applications, including changes in geometrical figures, motion problems, and related rates. This exercise helps students understand how calculus can be used to model and analyze dynamic situations, bridging the gap between abstract mathematical concepts and practical problem-solving.