AP Inter 2nd Year Maths Exercise 5f Solutions

Referring to the AP Inter 2nd Year Maths Study Material Chapter 5 Continuity and Differentiability Exercise 5f Solutions makes it easier to understand complex problems.

AP Inter 2nd Year Maths Continuity and Differentiability Solutions Exercise 5f

I. If x and y are connected parametrically by the equations without eliminating the parameter. find \(\frac{\mathrm{dy}}{\mathrm{dx}}\).

Question 1.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = 2at², = at⁴
Solution:
Given that x = 2at² and y = at⁴.
Differentiating both eqns. w.r.t t we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(2at²) = 2a\(\frac{\mathrm{d}}{\mathrm{dt}}\)t² = 2a.2t = 4at
and \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = \(\frac{\mathrm{d}}{\mathrm{dt}}\)(at⁴) = a\(\frac{\mathrm{d}}{\mathrm{dt}}\)t⁴ = a.4t³ = 4at³
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{4 a^3}{4 a t}\) = t².

Question 2.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a cos θ, y = b cos θ
Solution:
Given that x = a cos θ and y = b cos θ.
Differentiating both eqns. w.r.t. θ have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) (a cos θ) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)cos θ = -a sin θ
and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(b cos θ) = b\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)cos θ = – b sin θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{-b \sin \theta}{-a \sin \theta}=\frac{b}{a}\)

Question 3.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = sin t, y = cos 2t
Solution:
Given that x = sin t, y = cos 2t.
Differentiating both eqns. w.r.t t we have
\(\frac{\mathrm{dx}}{\mathrm{dt}}\) = cos t and \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = -sin 2t \(\frac{\mathrm{d}}{\mathrm{dt}}\) (2t) = -2 sin 2t
∴ \(\frac{\mathrm{dy}}{\mathrm{dt}}\) = \(\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=-\frac{-2 \sin 2 \mathrm{t}}{\cos \mathrm{t}}=-2 \frac{2 \sin \mathrm{t} \cos \mathrm{t}}{\cos \mathrm{t}}\) = -4 sin t

Question 4.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = 4t, y = \(\frac{4}{t}\).
Solution:
Given that x = 4t and y = \(\frac{4}{t}\).
Differentiating both eqns. w.r.t. x we have
\(\frac{d x}{d t}=\frac{d}{d t}(4 t)=4 \frac{d}{d t} t=4(1)=4 \text { and } \frac{d y}{d t}=\frac{d}{d t}\left(\frac{4}{t}\right)=\frac{t \frac{d}{d t}(4)-4 \frac{d}{d t} t}{t^2}=\frac{t(0)-4(1)}{t^2}=-\frac{4}{t^2}\)
∴ \(\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\left(-\frac{4}{t^2}\right)}{4}=\frac{-1}{t^2}\)

Question 5.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:
Given that x = cos θ – cos 2θ and y = sin θ – sin 2θ
∴ \(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) (cos θ) – \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) cos 2θ = -sin θ – (-sin 2θ)\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)2θ = -sin θ + (sin 2θ)2 = 2sin 2θ – sin θ
and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = cos θ \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)sin 2θ = cos θ – cos 2θ \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) 2θ = cos θ – cos 2θ(2) = cos θ – 2 cos 2θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{\cos \theta-2 \cos 2 \theta}{2 \sin 2 \theta-\sin \theta}\)

Question 6.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a (θ – sin θ). y = a(1 + cos θ)
Solution:
x = a(θ – sin θ) and y = a(1 + cos θ).
Differentiating both eqns. w.r.t. θ we have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(θ – sin θ) = a[\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)θ – \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\) sin θ] = a(1 – cosθ)
\(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = a\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(1 + cos θ) = a[\(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(1) + \(\frac{\mathrm{d}}{\mathrm{~d} \theta}\)(cos θ)] = a(0 – sin θ) = -a sin θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{-a \sin \theta}{a(1-\cos \theta)}=-\frac{\sin \theta}{1-\cos \theta}=-\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^2 \frac{\theta}{2}}=-\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}=-\cot \frac{\theta}{2} .\)

Question 7.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = \(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\), y = \(\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Solution:
Given that x = \(\frac{\sin ^3 t}{\sqrt{\cos 2 t}}\) and y = \(\frac{\cos ^3 t}{\sqrt{\cos 2 t}}\)
Differentiating both eqns. w.r.t. x, we have

Question 8.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a(cos t + log tan \(\frac{t}{2}\)), y = a sin t
Solution:

Question 9.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a sec θ, y = b tan θ
Solution:
Given that x = a sec θ and y = b tan θ.
Differentiating both eqns. w.r.t. θ we have
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a sec θ tan θ and \(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = b sec² θ
∴ \(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}\)
=\(\frac{b \sec ^2 \theta}{a \sec \theta \tan \theta}=\frac{b \sec \theta}{a \tan \theta}\)
=\(\frac{b \cdot \frac{1}{\cos \theta}}{a \cdot \frac{\sin \theta}{\cos \theta}}=\frac{b}{\cos \theta} \cdot \frac{\cos \theta}{a \sin \theta}=\frac{b}{a \sin \theta}\)
=\(\frac{b}{a} \cos \sec \theta\)

Question 10.
Find \(\frac{\mathrm{dy}}{\mathrm{dx}}\) of x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ).
Solution:
Given that x = a(cos θ + θ sin θ), y = a(sin θ – θ cos θ).
\(\frac{\mathrm{dx}}{\mathrm{~d} \theta}\) = a(-sin θ + θ cos θ + sin θ) = a θ cos θ
\(\frac{\mathrm{dy}}{\mathrm{~d} \theta}\) = a[cos θ – (θ(-sin θ) + cos θ)] = a(cos θ + θ sin θ – cos θ] = a θ sin θ
∴ \(\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a \theta \sin \theta}{a \theta \cos \theta}=\tan \theta\)

Question 11.
If x = \(\sqrt{a^{\sin ^{-1} t}}\), y = \(\sqrt{a^{\cos ^{-1} t}}\) show that \(\frac{d y}{d x}=-\frac{y}{x}\)
Solution: