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

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

I. Find the angle between the curves given below.

Question 1.

x + y + 2 = 0 ; x² + y² – 10y = 0.

Solution:

x + y + 2 = 0 ⇒ x = -(y + 2)

x² + y² – 10y = 0

(y + 2)² + y² – 10y = 0

y² + 4y + 4 + y² – 10y = 0

2y² – 6y + 4 = 0

y² – 3y + 2 = 0

(y + 1) (y – 2) = 0

y = 1 or y – 2

x = – (y + 2)

y = 1 ⇒ x = -(1 + 2) = -3

y = 2 ⇒ x = -(2 + 2) = -4

The points of intersection are P(-3, 1) and Q(-4, 2), equation of the curve is

x² + y² – 10y = 0

Differentiating w.r.to x.

2x + 2y\(\frac{dy}{dx}\) – 10 \(\frac{dy}{dx}\) = 0

2\(\frac{dy}{dx}\)(y – 5) = -2x

\(\frac{dy}{dx}\) = –\(\frac{x}{y-5}\)

f'(x_{1}) = –\(\frac{x}{y-5}\)

Equation of the line is x + y + 2 = 0

1 + \(\frac{dy}{dx}\) = 0 ⇒ \(\frac{dy}{dx}\) = -1

g'(x) = -1

Case (i):

At P(-3, 1), f'(x_{1}) = \(\frac{3}{1-5}=-\frac{3}{4}\), g'(x_{1}) = -1

Case (ii):

Question 2.

y² = 4x, x² + y² = 5.

Solution:

Eliminating y; we get x² + 4x = 5

x² + 4x – 5 = 0

(x – 1) (x + 5) = 0

x – 1 = 0 or x + 5 = 0

x = 1 or -5

Now y² = 4x

x = 1 ⇒ y² = 4

y = ±2

x = -5 ⇒ y is not real.

∴ Points of interstection of P(1, 2) and Q(1, -2) equation of the first curve is y² = 4x

2y.\(\frac{dy}{dx}\) = 4

\(\frac{dy}{dx}\) = \(\frac{4}{2y}\)

f(x) = \(\frac{2}{y}\)

Equation of the second curve is x² + y² = 5 dy

2x + 2y \(\frac{dy}{dx}\) = 0 dx

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

Question 3.

x² + 3y = 3 ; x² – y² + 25 = 0.

Solution:

x² = 3 – 3y; x² – y² + 25 = 0

3 – 3y – y² + 25 = 0

y² + 3y – 28 = 0

(y – 4) (y + 7) = 0

y – 4 = 0 (or) y + 7 = 0

y = 4 or – 7

x² = 3 – 3y

y = 4 ⇒ x² = 3 – 12 = – 9

⇒ x is not real

y = -7 ⇒ x² = 3 + 21 = 24

⇒ x = ± √24 = ± 2√6

Points of intersection are

P(2√6, -7), Q(-2√6, -7)

Equation of the first cur ve is x² + 3y = 3

3y = 3 – x²

3.\(\frac{dy}{dx}\) = -2x

\(\frac{dy}{dx}\) = –\(\frac{2x}{3}\) i.e, f'(x_{1}) = –\(\frac{2x}{3}\)

Equation of the second curve is

x² – y² + 25 = 0

y² = x² + 25

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

Question 4.

x² = 2(y + 1), y = \(\frac{8}{x^{2}+4}\).

Solution:

x² = 2(\(\frac{8}{x^{2}+4}\) + 1) = \(\frac{16+2x^{2}+8}{x^{2}+4}\)

x²(x² + 4) = 2x² + 24

x^{4} + 4x² – 2x² – 24 = 0

x^{4} + 2x² – 24 = 0

(x² + 6) (x² – 4) = 0

x² = -6 or x² = 4

x² = -6 ⇒ x is not real

x² = 4 ⇒ x = ±2

y = \(\frac{8}{x^{2}+4}=\frac{8}{4+4}=\frac{8}{8}\) = 1

∴ Points of intersection are P(2, 1) and Q(-2, 1)

Equation of the first curve is x² = 2(y + 1)

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

f'(x_{1}) = x_{1}

Equation of the second curve is y = \(\frac{8}{x^{2}+4}\)

∴ The given curves cut orthogonally

i.e., θ = \(\frac{\pi}{2}\)

At Q (-2, -1),f'(x_{1}) = -2, g'(x_{1}) = \(\frac{32}{64}=\frac{1}{2}\)

f'(x_{1}) g'(x_{1}) = -2 × \(\frac{1}{2}\) = -1

∴ The given curves cut orthogonally

⇒ θ = \(\frac{\pi}{2}\)

Question 5.

2y² – 9x = 0, 3x² + 4y = 0 (in the 4^{th} quadrant).

Solution:

2y² – 9x = 0 ⇒ 9x = 2y²

x = \(\frac{2}{9}\)y²

3x² + 4y = 0

⇒ 3.\(\frac{4}{8}\) y^{4} + 4y = 0

\(\frac{4y^{2}+108y}{27}\) = 0

4y(y³ + 27) = 0

y = 0 or y³ = -27 ⇒ y = -3

9x = 2y² 2 × 9 ⇒ x = 2

Point of intersection (in 4^{th} quadrant) is P(2, -3)

Equation of the first curve is 2y² = 9x

4y\(\frac{dy}{dx}\) = 9 ⇒ \(\frac{dy}{dx}=\frac{9}{4y}\)

f'(x_{1}) = \(\frac{9}{4y}\)

At P(2, -3), f'(x_{1}) = \(\frac{9}{-12}=-\frac{3}{4}\)

Equation of the second curve is

3x² + 4y = 0

4y = -3x²

4.\(\frac{dy}{dx}\) = -6x

Question 6.

y² = 8x, 4x² + y = 32

Solution:

4x² + 8x = 32 ⇒ x² + 2x = 8

x² + 2x – 8 = 0

(x – 2) (x + 4) = 0

x = 2 or -4

y² = 8x

x = -4 ⇒ y² is not real

x = 2 ⇒ y² = 16 ⇒ y = ±4

Point of intersection are P(2, 4), Q(2, -4)

Equation of the first curve is y² = 8x

2y.\(\frac{dy}{dx}\) = 8 ⇒ \(\frac{dy}{dx}=\frac{8}{2y}=\frac{4}{y}\)

f'(x_{1}) = \(\frac{4}{y}\)

Equation of the second curve is

4x² + y² = 32

8x + 2y.\(\frac{dy}{dx}\) = 0

Question 7.

x²y = 4, y(x² + 4) = 8.

Solution:

x²y = 4 ⇒ x = \(\frac{4}{y}\)

y(x² + 4) = 8

y(\(\frac{4}{y}\) + 4y) = 8

y\(\frac{(4+4y)}{y}\) = 8

4y – 4 ⇒ y = 1

x² = 4 ⇒ x = ±2

Points of intersection are P(2, 1), Q(-2, 1)

x²y = 4 ⇒ y = \(\frac{4}{x^{2}}\)

\(\frac{dy}{dx}\) = –\(\frac{8}{x^{3}}\) ⇒ f'(x_{1}) = –\(\frac{8}{x^{3}}\)

y(x² + 4) = 8 ⇒ y = \(\frac{8}{x^{2}+4}\)

Question 8.

Show that the curves 6x² – 5x + 2y = 0 and 4x² + 8y² = 3 touch each other at (\(\frac{1}{2}\), \(\frac{1}{2}\))

Solution:

Equation of the first curve is

6x² – 5x + 2y = 0

2y = 5x – 6x²

2.\(\frac{dy}{dx}\) = 5 – 12x

\(\frac{dy}{dx}=\frac{5-12x}{2}\)

Equation of the second curve is 4x² + 8y² = 3

∴ f'(x_{1}) = g'(x_{1})

The given curves touch each other at P(\(\frac{1}{2}\), \(\frac{1}{2}\)).