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 4tDifferentiate 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
Summary
Exercise 11.7 Set 3 in RD Sharma's Class 12 Mathematics textbook further expands on the application of derivatives to solve complex rate of change problems. This set introduces more advanced scenarios, often involving multiple interrelated variables or sophisticated real-world applications. Students are challenged to apply higher-order differentiation techniques, interpret complex physical situations, and analyze rates of change in multivariable systems.