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Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.7 | Set 3

Last Updated : 20 May, 2021
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Question 21. If x=a\left(\frac{1+t^2}{1-t^2}\right)  and y=\frac{2t}{1-t^2}  , find \frac{dy}{dx}

Solution:

Here,

x=a\left(\frac{1+t^2}{1-t^2}\right)

Differentiate it with respect to t using chain rule,

\frac{dx}{dt}=a\left[\frac{(1+t^2)\frac{d}{dt}(1+t^2)-(1+t^2)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(2t)-(1+t^2)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{2t-2t^2+2t+2t^3}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{4at}{(1-t^2)^2}\ \ \ \ \ ......(1)

And,

y=\frac{2t}{1-t^2}

Differentiate it with respect to t using quotient rule,

\frac{dy}{dt}=a\left[\frac{(1-t^2)\frac{d}{dt}(t)-(t)\frac{d}{dt}(1-t^2)}{(1-t^2)^2}\right]\\ =a\left[\frac{(1-t^2)(1)-(t)(-2t)}{(1-t^2)^2}\right]\\ =a\left[\frac{1-t^2+2t}{(1-t^2)^2}\right]\\ \frac{dy}{dt}=\frac{2(1+t^2)}{(1-t^2)}\ \ \ \ \ ......(2)

Question 22. Find \frac{dy}{dx} , if y = 12(1 – cos t), x = 10(t – sin t), -\frac{\pi}{2}<t<\frac{\pi}{2}

Solution:

It is given that, 

y = 12(1 – cos t),

x = 10(t – sin t)

Therefore,

\frac{dx}{dt}=\frac{d}{dt}[10(t-sin\ t)]\\ =10.\frac{d}{dt}(t-sin\ t)\\ =10(1- cos\ t)

\frac{dy}{dt}=\frac{d}{dt}[12(t-cos\ t)]\\ =12.\frac{d}{dt}(1-cos\ t)\\ =12(0- (-sin\ t)\\ =12sin\ t

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12sin\ t}{10(1-cos\ t)}\\ =\frac{12\times2sin\frac{t}{2}\times cos\frac{t}{2}}{10\times2\ sin^2\frac{t}{2}}\\ =\frac{6}{5}\ cot\ \frac{t}{2}

Question 23. If x = a(θ – sin θ) and y = a(1 – cos θ), find \frac{dy}{dx} , at θ = \frac{\pi}{3}

Solution:

Here,

x = a(θ – sin θ)

and

y = a(1 – cos θ)

Then,

\frac{dx}{dθ}=\frac{d}{dθ}[a(θ-sin\ θ]\\ =a(1-cos\ θ)

\frac{dy}{dθ}=\frac{d}{dθ}[a(1+cos\ θ]\\ =a(-sin\ θ)

Therefore,

\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-asin\ θ}{a(1-cos\ θ)}|_{θ=\frac{\pi}{3}}\\ =-\frac{sin\frac{\pi}{3}}{1-cos\frac{\pi}{3}}\\ =\frac{\frac{\sqrt3}{2}}{1-\frac{1}{2}}=-\sqrt3

Question 24. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that at t = \frac{\pi}{4},\ \frac{dy}{dx}=\frac{b}{a}

Solution:

Consider the given functions,

x = a sin 2t (1 + cos 2t)

and 

y = b cos 2t (1 – cos 2t)

Write again the functions,

x = a sin 2t + \frac{a}{2} sin 4t

Differentiate the above function with respect to t,

\frac{dx}{dt}=2a\ cos\ 2t+2a\ cos\ 4t\ \ \ \ \ ....(1)\\

y = b cos 2t (1 – cos 2t)

y = b cos 2t – b cos2 2t

\frac{dy}{dx}=-2b\ sin\ 2t+2b\ cos\ 2t\ sin\ 2t\\=-2b\ sin\ 2t+b\ sin\ 4t\ \ \ \ ....(2)

From equation (1) and (2)

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-2b\ sin\ 2t+b\ sin\ 4t}{2a\ cos\ 2t+2a\ cos\ 4t}\\ \therefore\frac{dy}{dx}|_{\frac{\pi}{4}}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}|_{t=\frac{\pi}{4}}=\frac{-2b}{-2a}=\frac{b}{a}

Question 25. If x = cos t (3 – 2cos2t) and y = sin t (3 – 2 sin2t), find the value of \frac{dy}{dx}  at t = \frac{\pi}{4}

Solution:

Here, the given function:

x = cos t (3 – 2cos2t)

x = cos t – 2cos3t

\frac{dx}{dt}=-3sin\ t+6cos^2t\ \ \ \ ......(1)

y = sin t (3 – 2 sin2t)

y = 3cos t – 2sin3t

\frac{dy}{dt}=3cos\ t-6sin^2tcos\ t\ \ \ \ .....(2)\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ =\frac{3cos\ t-6sin^tcos\ t}{-3sin\ t+6cos^2t\ sin\ t}\\ =\frac{3cos\ t(1-2sin^2t)}{3sin\ t(2cos^2t-1)}\\ =cot\ t\frac{(1-2(1-cos^2t))}{(2cos^2t-1)}\\ =cot\ t\\ \frac{dy}{dx}|_{\frac{\pi}{4}}=cot\frac{\pi}{4}=1

Question 26. If x=\frac{1+log\ t}{t^2} , y=\frac{3+2log\ t}{t}  find \frac{dy}{dx}

Solution:

Here,

x=\frac{1+log\ t}{t^2}

 and

y=\frac{3+2log\ t}{t}

\frac{dx}{dt}=\frac{t^2\left(\frac{1}{t}\right)-(1+log\ t)(2t)}{t^4}\\ =\frac{t-2t-2tlog\ t}{t^4}\\ =\frac{-2log\ t-1}{t^3}

\frac{dy}{dt}=\frac{t\left(\frac{2}{t}\right)-(-3+2log\ t)(1)}{t^2}\\ =\frac{2-3-2tlog\ t}{t^2}\\ =\frac{-2log\ t-1}{t^2}

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{-2log\ t-1}{t^2}}{\frac{-2log\ t-1}{t^3}}=t

Question 27. If x = 3sin t – sin3t, y = 3cos t – cos3t, find \frac{dy}{dx}\ at\ t=\frac{\pi}{3}

Solution:

x = 3sin t – sin3t

and,

y = 3cos t – cos3t

\frac{dx}{dt}=3cos\ t-3cos3t\\ \frac{dy}{dt}=-3sin\ t+3sin3t\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-3sin\ t+3sin3t}{3cos\ t-3cos3t}

When, t=\frac{\pi}{3}

\frac{dy}{dx}=\frac{-3sin(\frac{\pi}{3})+3sin(\pi)}{3cos(\frac{\pi}{3})-3cos(\pi)}=\frac{-3\times\frac{\sqrt3}{2}+0}{3\times\frac{1}{2}-3(-1)}=-\frac{1}{\sqrt3}

Question 28. If sin\ x=\frac{2t}{1+t^2} , tan\ y=\frac{2t}{1-t^2}  find \frac{dy}{dx}

Solution:

sin\ x=\frac{2t}{1+t^2}

and,

tan\ y=\frac{2t}{1-t^2}

\Rightarrow x=sin^{-1}\left(\frac{2t}{1+t^2}\right)

and 

\Rightarrow y=tan^{-1}\left(\frac{2t}{1-t^2}\right)

\frac{dx}{dt}=\frac{1}{\sqrt{1-\left(\frac{2t}{1+t^2}\right)^2}}\times\frac{2(1+t^2)-(2t)(2t)}{(1+t^2)^2}\\ \frac{dx}{dt}=\frac{2}{1+t^2}\\ \frac{dy}{dt}=\frac{1}{\left(\frac{2t}{1-t^2}\right)2+1}\times\frac{2(1-t^2)-(2t)(-2t)}{(1-t^2)^2}\\ \frac{dy}{dt}=\frac{2}{1+t^2}

\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{\frac{2}{(1+t^2)}}{\frac{2}{(1+t^2)}}=1



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