Practicing the Intermediate 1st Year Maths 1B Textbook Solutions Inter 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1B Applications of Derivatives Solutions Exercise 10(b)

I.

Question 1.

Find the slope of the tangent to the curve

y = 3x^{4} – 4x at x = 4.

Solution:

Equation of the curve is y = 3x^{4} – 4x

\(\frac{dy}{dx}\) = 12x³ – 4 dx

At x = 4, slope of the tangent = 12 (4)³ – 4

= 12 × 64 – 4

= 768 – 4

= 764

Question 2.

Find the slope of the tangent to the curve

y = \(\frac{x-1}{x-2}\) x ≠ 2 at x = 10.

Solution:

Equation of the curve is

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

= \(\frac{x-2+1}{x-2}\)

= 1 + \(\frac{1}{x-2}\)

\(\frac{dy}{dx}\) = 0 + \(\frac{(-1)}{(x-2)^{2}}=\frac{1}{(x-2)^{2}}\)

At x = 10, slope of the tangent = \(\frac{1}{(10-2)^{2}}\)

= –\(\frac{1}{64}\)

Question 3.

Find the slope of the tangent to the curve y = x³ – x + 1 at the point whose x co-ordinate is 2.

Solution:

Equation of the curve is y = x³ – x + 1

\(\frac{dy}{dx}\) = 3x² – 1

x = 2

Slope of the tangent at (x – 2) is

3(2)² – 1 = 3 x 4 – 1

= 12 – 1 = 11

Question 4.

Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x co-ordinate is 3.

Solution:

Equation of the curve is y = x³ – 3x + 2

\(\frac{dy}{dx}\) = 3x² – 3

At x = 3, slope of the tangent = 3(3)² – 3

= 27 – 3 = 24

Question 5.

Find the slope of the normal to the curve

x = a cos³ θ, y = a sin³ θ at θ = \(\frac{\pi}{4}\).

Solution:

x = a cos³ θ

\(\frac{d x}{d \theta}\) = a(3 cos² θ) (-sin θ)

= -3a cos² θ. sin θ

y = sin³ θ

\(\frac{d y}{d \theta}\) = a (3 sin² θ) cos θ

= 3a sin² θ cos θ

At θ = \(\frac{\pi}{4}\), slope of the tangent = tan \(\frac{\pi}{4}\) = -1

Slope of the normal = – \(\frac{1}{m}\) = 1.

Question 6.

Find the slope of the normal to the curve

x = 1 – a sin θ, y = b cos θ at θ = \(\frac{\pi}{2}\).

Solution:

x = 1 – a sin θ

\(\frac{d x}{d \theta}\) = – a cos θ

y = b cos² θ dy

\(\frac{d y}{d \theta}\) = b(2 cos θ) (- sin θ) = -2b cos θ sin θ

\(\frac{2b}{a}\).sin θ

Slope of the normal = \(\frac{1}{m}=\frac{a}{2b \sin \theta}\)

At θ = \(\frac{\pi}{2}\), slope of the normal = \(\frac{-a}{2 b \sin \frac{\pi}{2}}\)

= \(\frac{-a}{2b.1}\)

= \(\frac{-a}{2b}\)

Question 7.

Find the points at which the tangent to the curve y = x3 – 3×2 – 9x + 7 is parallel to the x-axis.

Solution:

Equation of the curve is y = x³ – 3x² – 9x + 7

\(\frac{dy}{dx}\) = 3x² – 6x – 9 dx

The tangent is parallel to x-axis.

Slope of the tangent = 0

3x² – 6x – 9 = 0

x² – 2x – 3 = 0

(x – 3) (x + 1) = 0

x = 3 or -1

y = x³ – 3x² – 9x + 7

x = 3 ⇒ y = 27 – 27 – 27 + 7 = -20

x = -1, y = -1 – 3 + 9 + 7 = 12

The points required are (3, -20), (-1, 12).

Question 8.

Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2, 0) and (4, 4).

Solution:

Equation of the curve is y = (x – 2)²

\(\frac{dy}{dx}\) = 2(x – 2)

Slope of the chord joining A(2, 0) and B(4, 4)

= \(\frac{4-0}{4-2}=\frac{4}{2}\) = 2.

The tangent is parallel to the chord.

2(x – 2) = 2

x – 2 = 1

x = 3

y = (x – 2)² = (3 – 2)² = 1

The required point is P(3, 1).

Question 9.

Find the point on the curve

y = x³ – 11x + 5 at which the tangent is y = x – 11.

Solution:

Equation of the curve is y = x³ – 11x + 5

\(\frac{dy}{dx}\) = 3x² – 11

The tangent is y = x – 11

Slope of the tangent = 3x² – 11 = 1

3x² = 12

x² = 4

x = ±2

y = x – 11

x = 2 ⇒ y = 2 – 11 = -9

The points on the curve is P(2, -9).

Question 10.

Find the equations of all lines having slope 0 which are tangents to the curve y = \(\frac{1}{x^{2}-2x+3}\).

Solution:

Equation of the curve is y = \(\frac{1}{x^{2}-2x+3}\)

Given slope of the tangent = 0

Equation the point is P(1, \(\frac{1}{2}\))

Slope of the tangent = 0

Equation of the required tangent is

y – \(\frac{1}{2}\) = 0(x – 1)

⇒ 2y – 1 = 0

II.

Question 1.

Find the equations of tangent and normal to the following curves at the points indicated against.

i) y = x^{4} – 6x³ + 13x² – 10x + 5 at (0, 5).

Solution:

\(\frac{dy}{dx}\) = 4x³ – 18x² + 26x – 10

At x = 0,

Slope of the tangent = 0 – 0 + 0 -10 = -10

Equation of the tangent is y – 5 = -10(x – 0)

= -10x

10x + y – 5 = 0

Slope of the normal = – \(\frac{1}{m}=\frac{1}{10}\)

Equation of the normal is y – 5 = \(\frac{1}{10}\) (x – 0)

10y – 50 = x ⇒ x – 10y + 50 = 0

ii) y = x³ at (1, 1).

Solution:

\(\frac{dy}{dx}\) = 3x²

At (1, 1), slope of the tangent = 3 (1)² = 3

Equation of the tangent at P(1, 1) is

y – 1 = 3(x – 1)

= 3x – 3

3x – y – 2 = 0

Slope of the normal = – \(\frac{1}{m}=-\frac{1}{3}\)

Equation of the normal is y – 1 = \(-\frac{1}{3}\)(x – 1)

3y – 3 = -x + 1

x + 3y – 4 = 0

iii) y = x² at (0, 0).

Solution:

Equation of the curve is y = x²

\(\frac{dy}{dx}\) = 2x

At P(0, 0), slope of the tangent = 2.0 = 0

Equation of the tangent is y – 0 = 0 (x – 0)

⇒ y = 0

The normal is perpendicular to the tangent.

Equation of the normal is x = k.

The normal passes through (0, 0) ⇒ k = 0

Equation of the normal is x = 0.

iv) x = cos t, y = sin t at t = \(\frac{\pi}{4}\).

Solution:

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

Equation of the tangent is

Slope of the normal = –\(\frac{1}{m}=\frac{-1}{-1}\) = 1

Equation of the normal is y \(\frac{1}{\sqrt{2}}\) = x – \(\frac{1}{\sqrt{2}}\)

i.e., x – y = 0

v) y = x² – 4x + 2 at (4, 2).

Solution:

Equation of the curve is y = x² – 4x + 2

\(\frac{dy}{dx}\) = 2x – 4

At P(4, 2), slope of the tangent =2.4 – 4

= 8 – 4 = 4

Equation of the tangent at P is

y – 2 = 4(x – 4)

= 4x – 16

4x – y – 14 = 0

Slope of the normal = –\(\frac{1}{m}=-\frac{1}{4}\)

Equation of the normal at P is

y – 2 = \(-\frac{1}{4}\) (x – 4)

⇒ 4y – 8 = -x + 4

⇒ x + 4y – 12 = 0

vi) y = \(-\frac{1}{1+x^{2}}\) at (0, 1)

Solution:

Equation of the curve is y = \(-\frac{1}{1+x^{2}}\)

\(\frac{dy}{dx}\) = \(-\frac{1}{(1+x^{2})^{2}}\)

At (0, 1), x = 0, slope of the tangent = 0

Equation of the tangent at P(0, 1) is

y – 1 = 0(x – 0)

y = 1

The normal is perpendicular to the tangent.

Equation of the normal can be taken at x = 10.

The normal passes through P(0, 1) ⇒ 0 = k

Equation of the normal at P is x = 0.

Question 2.

Find the equations of tangent and normal to the curve xy = 10 at (2, 5).

Solution:

Equation of the curve is xy = 10.

y = \(\frac{10}{x}\); \(\frac{dy}{dx}=\frac{10}{x^{2}}\)

At P(2, 5), f'(x_{1}) = –\(\frac{10}{4}=-\frac{5}{2}\)

Equation of the tangent is

y – y_{1} = f'(x_{1}) (x – x_{1})

y – 5 = – \(\frac{5}{2}\) (x – 2)

2y – 10 = -5x + 10

5x + 2y – 20 = 0

Equation of the normal is

y – y_{1} = \(\frac{1}{f'(x_{1})}\)(x – x_{1})

y – 5 = \(\frac{5}{2}\) (x – 2)

5y – 25 = 2x – 4

i.e., 2x – 5y + 21 = 0.

Question 3.

Find the equations of tangent and normal to the curve y = x³ + 4x² at (-1, 3).

Solution:

Equation of the curve is y = x³ + 4x²

\(\frac{dy}{dx}\) = 3x² + 8x

At P(-1, 3),

Slope of the tangent

= 3(-1)² + 8(-1)

= 3 – 8 = -5

Equation of the tangent at P(-1, 3) is

y – y_{1} = f'(x_{1}) (x – x_{1})

y – 3 = -5(x + 1) = -5x – 5

5x + y + 2 = 0

Equation of the nonnal at P is

y – y_{1} = –\(\frac{1}{f'(x_{1})}\) (x – x_{1})

y – 3 = \(\frac{1}{5}\) (x + 1)

5y – 15 = x + 1

x – 5y + 16 = 0

Question 4.

If the slope of the tangent to the curve x² – 2xy + 4y = 0 at a point on it is –\(\frac{3}{2}\), then find the equations of tangent and normal at that point.

Solution:

Equation of the curve is

x² – 2xy + 4y = 0 ………… (1)

Differentiating w.r.to x

2x – 2y = -3x + 6; 5x – 2y = 6

2y = 5x – 6 ……. (2)

P(x, y) is a point on (1)

x² – x(5x – 6) + 2(5x – 6) = 0

x² – 5x² + 6x + 10x – 12 = 0

-4x² + 16x – 12 = 0

-4(x² – 4x + 3) = 0

x² + 4x + 3 = 0

(x – 1) (x – 3) = 0

x – 1 = 0 or x – 3 = 0

∴ x = 1 or x = 3

Case (i): x = 1

Substituting in (1)

1 – 2y + 4y = 0

2y = -1 ⇒ y = –\(\frac{1}{2}\)

The required point is P(1, –\(\frac{1}{2}\))

Equation of the tangent is

y + \(\frac{1}{2}\) = –\(\frac{3}{2}\)(x – 1)

\(\frac{2y+1}{2}=\frac{-3(x-1)}{2}\)

2y + 1 = -3x + 3

3x + 2y – 2 = 0

Equation of the normal isy + \(\frac{1}{2}=\frac{2}{3}\)(x – 1)

\(\frac{2y+1}{2}=\frac{2}{3}\) (x – 1)

6y + 3 = 4x – 4

4x – 6y – 7 = 0

Case (ii) : x = 3

Substituting in (1), 9 – 6y + 4y = 0

2y = 9 ⇒ y = \(\frac{9}{2}\)

∴ The required point is (3, \(\frac{9}{2}\))

Equation of the tangent is

y – \(\frac{9}{2}=-\frac{3}{2}\) (x – 3)

\(\frac{2y-9}{2}=\frac{-3(x-3)}{2}\)

2y – 9 = -3x + 9

3x + 2y- 18 = 0

Equation of the normal is y – \(\frac{9}{2}=\frac{2}{3}\) (x – 3)

\(\frac{2y-9}{2}=\frac{2(x-3)}{3}\)

6y – 27 = 4x – 12

i.e., 4x – 6y + 15 = 0.

Question 5.

If the slope of the tangent to the curve y = x log x at a point on it is \(\frac{3}{2}\), then find the equations of tangent and normal at that point.

Solution:

Equation of the curve is y = x log x

\(\frac{dy}{dx}\) = x. \(\frac{1}{x}\) + log x.1 = 1 + log x.

Given 1 + log x = \(\frac{3}{2}\)

log_{e} x =\(\frac{1}{2}\) ⇒ x = e^{½} = √e

y = √e . log .√e = \(\frac{\sqrt{e}}{2}\)

The required point is P (√e, \(\frac{\sqrt{e}}{2}\))

Equation of the tangent is y \(\frac{\sqrt{e}}{2}=\frac{3}{2}\)(x – √e)

\(\frac{2 y-\sqrt{e}}{2}=\frac{3(x-\sqrt{e})}{2}\)

2y – √e = 3x – 3 √e

3x – 2y – 2√e = 0

Equation of the normal is

y – y_{1} = – \(\frac{1}{f'(x_{1})}\)(x – x_{1})

y – \(\frac{\sqrt{e}}{2}=-\frac{2}{3}\)(x – √e)

\(\frac{2 y-\sqrt{e}}{2}=-\frac{2}{3}\)(x – √e)

6y – 3√e = -4x + 4√e

i.e., 4x + 6y – 7√e =0

Question 6.

Find the tangent and normal to the curve y = 2e^{-x/3} at the point where the curve meets the Y-axis.

Solution:

Equation of the curve is y = 2e^{-x/3}

Equation of Y-axis is x = 0

y = 2.e° = 2.1 = 2

Required point is P(0, 2)

\(\frac{dy}{dx}\) = 2(-\(\frac{1}{3}\)) . e^{-x/3}

When x = 0, slope of the tangent = –\(\frac{2}{3}\) .e° = \(\frac{-2}{3}\)

Equation of the tangent at P is

y – y_{1} = f'(x_{1}) (x – x_{1})

y – 2 = –\(\frac{2}{3}\) (x – 0)

3y – 6 = -2x

2x + 3y – 6 = 0

Equation of the normal is

y – y_{1} = –\(\frac{1}{f'(x_{1})}\) (x – x_{1}

y – 2 = \(\frac{3}{2}\) (x – 0)

2y – 4 = 3x; 3x – 2y + 4 = 0

III.

Question 1.

Show that the tangent at P(x_{1}, y_{1}) on the curve √x + √y = √a is yy_{1}^{-½} + xx_{1}^{-½} = a^{½}.

Solution:

Equation of the curve is √x + √y = √a

Differentiating w.r.to x

Slope of the tangent at P(x_{1} y_{1}) = –\(\frac{\left(y_{1}\right)^{1 / 2}}{\left(x_{1}\right)^{1 / 2}}\)

Equation of the tangent at P is

= x_{1}^{½} + y_{1}^{½}

x. x_{1}^{-½} + y. y_{1}^{-½} = a^{½}

(P is a point on the curve)

Equation of the tangent at P is

y. y_{1}^{-½} + x. x_{1}^{-½} = a^{½}

Question 2.

At what points on the curve x² – y² = 2, the slopes of the tangents are equal to 2?

Solution:

Equation of the curve is x² – y² = 2 ………. (1)

Differentiating w.r.to x

2x – 2y.\(\frac{dy}{dx}\) = 0 ⇒ \(\frac{dy}{dx}=\frac{x}{y}\)

Slope of the tangent = \(\frac{dy}{dx}\) = 2

∴ \(\frac{x}{y}\) = 2 ⇒ x = 2y

Substituting in (1), 4y² – y² = 2

3y² = 2

y ² = \(\frac{2}{3}\) ⇒ y = ± \(\sqrt{\frac{2}{3}}\)

x = 2y = ± 2 \(\sqrt{\frac{2}{3}}\)

∴ The required points are

Question 3.

Show that the curves x² + y² = 2 and 3x² + y² = 4x have a common tangent at the point (1, 1).

Solution:

Equation of the first curve is x² + y² = 2

Differentiating w.r.to x

At P (1, 1) slope of the tangent = \(\frac{-1}{1}\) = -1

Equation of the second curve is 3x² + y² = 4x.

Differentiating w.r.to x, 6x + 2y.\(\frac{dy}{dx}\) = 4

2y.\(\frac{dy}{dx}\) = 4 – 6x

\(\frac{dy}{dx}=\frac{4-6x}{2y}=\frac{2-3x}{y}\)

At P( 1, 1) slope of the tangent = \(\frac{2-3}{1}\) = –\(\frac{1}{1}\) = -1

The slope of the tangents to both the curves at P( 1, 1) are same and pass through the same point (1, 1)

∴ The given curves have a common tangent at P (1, 1)

Question 4.

At a point (x_{1}, y_{1}) on the curve x³ + y³ = 3axy, show that the tang;ent is

(x_{1}² – ay_{1}) x+ (y_{1}² – ax_{1})y = ax_{1}y_{1}

Solution:

Equation of the curve is x³ + y³ = 3axy

Differentiating w. r. to x

Slope of the tangent P(x_{1}, y_{1}) = –\(\frac{\left(x_{1}^{2}-a y_{1}\right)}{\left(y_{1}^{2}-a x_{1}\right)}\)

Equation of the tangent at P(x_{1}, y_{1}) is

y(y – y_{1}) = –\(\frac{\left(x_{1}^{2}-a y_{1}\right)}{\left(y_{1}^{2}-a x_{1}\right)}\)(x – x_{1})

y(y_{1}² – ax_{1}) – y_{1}(y_{1}² – ax_{1}) = – x(x_{1}² – ay_{1}) + x_{1}(x_{1}² – ay_{1})

x_{1}(x_{1}² – ay_{1}) + y_{1}(y_{1}² – ax_{1})

= x_{1}(x_{1}² – ay_{1}) + y_{1}(y_{1}² – ax_{1})

= x_{1}³ – ax_{1}y_{1} + y_{1}³ – ax_{1}y_{1}

= x_{1}³ + y_{1}³ – 2ax_{1}y_{1}

3ax_{1}y_{1} – 2ax_{1}y_{1} (P is a point on the curve)

= ax_{1}y_{1}

Question 5.

Show that the tangent at the point P (2, -2) on the curve y (1 – x) = x makes intercepts of equal length on the co-ordinate axes and the normal at P passes through the origin.

Solution:

Equation of the curve is

y (1 – x) = x

y = \(\frac{x}{1-x}\)

Differentiating w.r. to x

Equation of the tangent at P is

y + 2 = +(x – 2) = x – 2; x – y = 4

\(\frac{x}{4}-\frac{y}{4}\) ⇒ \(\frac{x}{4}-\frac{y}{(-4)}\) = 1

∴ a = 4, b = – 4

∴ The tangent makes equal intercepts on the co-ordinate axes but they are in opposite in sign. Equation of the normal at P is

y – y_{1} = \(\frac{1}{f'(x_{1})}\) (x – x_{1})

y + 2 = -(x – 2)= -x + 2

x + y = 0

There is no constant term in the equation.

∴ The normal at P(2, -2) passes through the origin.

Question 6.

If the tangent at any point on the curve x^{2/3} + y^{2/3} = a^{2/3} intersects the coordinate axes in A and B then show that length AB is a constant.

Solution:

Equation of the curve is

x^{2/3} + y^{2/3} = a^{2/3}

Differentiating w.r. to x

Equation of the tangent at P (x_{1}, y_{1}) is

AB = a = constant.

Question 7.

If the tangent at any point P on the curve x^{m} y^{n} = a^{m+n} (mn ≠ 0) meets the co-ordinate axes in A, B, then show that AP: PB is a constant.

Solution:

Equation of the curve is x^{m}.y^{n} = a^{m+n}

Differentiating w.r. to x

Slope of the tangent at P(x_{1}, y_{1}) = –\(\frac{my_{1}}{nx_{1}}\)

Equation of the tangent at P is

Co-ordinates of A are [\(\frac{m+n}{m}\).x_{1}, o] and B are [0, \(\frac{m+n}{m}\).y_{1}]

Let P divide AB in the ratio k : l

Co-ordinates of P are

Dividing (1) by (2) \(\frac{l}{k}=\frac{m}{n}\) ⇒ \(\frac{k}{l}=\frac{n}{m}\)

∴ P divides AB in the ratio n : m

i.e., AP : PB = n : m = constant.