Question 1. Find \frac{dy}{dx} , when: x = at2 and y = 2at
Solution:
Given that x = at2, y = 2at
So,
\frac{dx}{dt}=\frac{d}{dt}(at^2)=2at\\ \frac{dy}{dt}=\frac{d}{dt}(2at)=2a Therefore,
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2a}{2at}=\frac{1}{t}
Question 2. Find \frac{dy}{dx} , when: x = a(θ + sinθ) and y = a(1 - cosθ)
Solution:
Here,
x = a(θ + sinθ)
Differentiating it with respect to θ,
\frac{dx}{dθ}=a(1+cosθ)\ \ \ ..........(1) and,
y = a(1 - cosθ)
Differentiate it with respect to θ,
\frac{dy}{dθ}=a(θ+sinθ)\\ \frac{dy}{dθ}=asinθ\ \ \ \ .....(2) Using equation (1) and (2),
\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}\\ =\frac{asinθ}{a(1-cosθ)}\\ =\frac{\frac{2sinθ}{2}\frac{cosθ}{2}}{\frac{2sin^2θ}{2}},\ \ \ \ \ \ \ \left\{Since,\ 1-cosθ=\frac{2sin2θ}{2},\frac{2sinθ}{2}\frac{cosθ}{2}=sinθ\right\}\\ =\frac{dy}{dx}=\frac{tanθ}{2}
Question 3. Find \frac{dy}{dx} , when: x = acosθ and y = bsinθ
Solution:
Then x = acosθ and y = bsinθ
Then,
\frac{dx}{dθ}=\frac{d}{dθ}(acosθ)=-asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(bsinθ)=bcosθ Therefore,
\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{bcosθ}{-asinθ}=-\frac{b}{a}cotθ
Question 4. Find \frac{dy}{dx} , when: x = aeΘ (sinθ -cosθ), y = aeΘ (sinθ +cosθ)
Solution:
Here,
x = aeΘ (sinθ - cosθ)
Differentiating it with respect to θ,
\frac{dx}{dθ}=a[e^θ\frac{d}{dθ}(sinθ-cosθ)+(sinθ-cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ+sinθ)+(sinθ-cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θsinθ]\ \ \ \ \ \ .......(1) And,
y = aeΘ(sinθ+cosθ)
Differentiating it with respect to θ
\frac{dy}{dθ}=a[e^θ\frac{d}{dθ}(sinθ+cosθ)+(sinθ+cosθ)\frac{d}{dθ}(e^θ)]\\ =a[e^θ(cosθ-sinθ)+(sinθ+cosθ)e^θ]\\ \frac{dx}{dθ}=a[2e^θcosθ]\ \ \ \ \ \ .......(2) Dividing equation (2) by equation (1)
\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(2e^θ cosθ)}{a(2e^θ sinθ)}\\ \frac{dy}{dx}=cotθ
Question 5. Find \frac{dy}{dx} , when: x = bsin2θ and y = acos2θ
Solution:
Here,
x = bsin2θ and y = acos2θ
Then,
\frac{d}{dθ}=\frac{d}{dθ}(bsin^2θ)=2bsinθ cosθ\\ \frac{dy}{dθ}=\frac{d}{dθ}(acos^2θ)=-2acosθ sinθ \\ \frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{-2acosθ sinθ}{2bsinθ cosθ}=-\frac{a}{b}\\
Question 6. Find \frac{dy}{dx} , when: x = a(1 - cosθ) and y = a(θ +sinθ) at θ =\frac{\pi}{2}
Solution:
Here,
x = a(1 - cosθ) and y = a(θ + sinθ)
Then,
\frac{dx}{dθ}=\frac{d}{dθ}[a(1-cosθ)]=asinθ\\ \frac{dy}{dθ}=\frac{d}{dθ}[a(θ +sinθ)]=a(1+cosθ) Therefore,
\frac{dy}{dx}=\frac{\frac{dy}{dθ}}{\frac{dx}{dθ}}=\frac{a(1+cosθ)}{a(sinθ)}|_{θ =\frac{x}{2}}\\ =\frac{a(1+0)}{a}=1
Question 7. Find \frac{dy}{dx} , when: x=\frac{e^t+e^{-t}}{2} andy=\frac{e^t-e^{-t}}{2}
Solution:
Here,
x=\frac{e^t+e^{-t}}{2} Differentiate it with respect to t,
\frac{dx}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)+\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t+e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t-e^{-t})=y\ \ \ ......(1) and,
y=\frac{e^t-e^{-t}}{2} Differentiating it with respect to t,
\frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x \frac{dy}{dt}=\frac{1}{2}\left[\frac{d}{dt}(e^t)-\frac{d}{dt}(e^{-t})\right]\\ =\frac{1}{2}\left[e^t-e^{-t}\frac{d}{dt}(e^{-t})\right]\\ \frac{dx}{dt}=\frac{1}{2}(e^t+e^{-t})=x\ \ \ \ .....(2) Dividing equation (2) and (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{x}{y}\\ \frac{dy}{dt}=\frac{x}{y}
Question 8. Find \frac{dy}{dx} , when: x=\frac{3at}{1+t^2} andy=\frac{3at^2}{1+t^2}
Solution:
Here,
x=\frac{3at}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dx}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at)-3at\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(3a)-3at(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{3a+3at^2-6at^2}{(1+t^2)^2}\right]\\ =\left[\frac{3a-3at^2}{(1-t^2)^2}\right]\\ \frac{dx}{dt}=\frac{3a(1-t^2)}{(1+t^2)^2}\ \ \ \ ....(1) and,
y=\frac{3at^2}{1+t^2} Differentiating it with respect to t using quotient rule,
\frac{dy}{dt}=\left[\frac{(1+t^2)\frac{d}{dt}(3at^2)-3at^2\frac{d}{dt}(1+t^2)}{(1+t^2)^2}\right]\\ =\left[\frac{(1+t^2)(6at)-(3at^2)(2t)}{(1+t^2)^2}\right]\\ =\left[\frac{6at+6at^3-6at^3}{(1+t^2)^2}\right]\\ \frac{dy}{dt}=\frac{6at}{(1+t^2)^2}\ \ \ \ ....(2) Dividing equation (2) by (1)
\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{6at}{(1+t^2)^2}\times\frac{(1+t^2)^2}{3a(1-t^2)}\\ \frac{dy}{dt}=\frac{2t}{1-t^2}
Question 9. If x and y are connected parametrically by the equation, without eliminating the parameter, find\frac{dy}{dx} when: x = a(cosθ +θsinθ), y = a(sinθ -θcosθ)
Solution:
The given equations are
x = a(cosθ +θ sinθ) and y = a(sinθ -θcosθ)
Then,
\frac{dx}{dθ}=a\left[\frac{d}{dθ}cosθ +\frac{d}{dθ}(θ sinθ)\right]\\ =a\left[-sinθ +θ \frac{d}{dθ}(sinθ)+sinθ \frac{d}{dθ}(θ)\right] = a[-sinθ + θcosθ + sinθ] = aθcosθ
\frac{dy}{dθ}=a\left[\frac{d}{dθ}sinθ +\frac{d}{dθ}(θ cosθ)\right]\\ =a\left[cosθ -\{θ \frac{d}{dθ}(cosθ)+cosθ \frac{d}{dθ}(θ)\}\right] = a[cosθ +θsinθ -cosθ]
= aθsinθ
Therefore,
\frac{dy}{dx}=\frac{\left(\frac{dy}{dθ}\right)}{\left(\frac{dx}{dθ}\right)}=\frac{aθsinθ}{aθ cosθ}=tanθ
Question 10. Find \frac{dy}{dx} , when: x=e^θ \left(θ +\frac{1}{θ}\right) andy=e^{-θ} \left(θ -\frac{1}{θ}\right)
Solution:
Here,
x=e^θ \left(θ +\frac{1}{θ}\right) Differentiating it with respect to θ using product rule,
\frac{dx}{dθ}=e^θ \frac{d}{dθ}\left(θ +\frac{1}{θ}\right)+\left(θ +\frac{1}{θ}\right)\frac{d}{dθ}(e^θ)\\ =e^θ \left(1-\frac{1}{θ^2}\right)+\left(\frac{θ ^2+1}{θ}\right)e^θ \\ =e^θ \left(\frac{θ ^2-1+θ ^3+θ}{θ ^2}\right)\\ \frac{dx}{dθ}=\frac{e^θ (θ ^3+θ ^2+θ -1)}{θ ^2}\ \ \ .....(1) and,
y=e^θ \left(θ -\frac{1}{θ}\right) Differentiating it with respect to θ using product rule and chain rule
\frac{dy}{dθ}=e^{-θ} \frac{d}{dθ}\left(θ -\frac{1}{θ}\right)+\left(θ -\frac{1}{θ}\right)\frac{d}{dθ}(e^{-θ})\\ =e^{-θ} \left(1+\frac{1}{θ^2}\right)+\left(θ -\frac{1}{θ}\right)e^{-θ} \\ =e^{-θ} \left(1+\frac{1}{θ ^2}-θ +\frac{1}{θ}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{θ ^2+1 -θ^3 +θ}{θ ^2}\right)\\ \frac{dy}{dθ}=e^{-θ}\left(\frac{-θ ^3+θ ^2+θ +1}{θ ^2}\right)\ \ \ \ ......(2)
Summary
Exercise 11.7 Set 1 in RD Sharma's Class 12 Mathematics textbook focuses on the application of derivatives to solve problems related to rates of change. This set covers problems involving instantaneous rates, average rates, and related rates in various contexts. Students are required to apply differentiation techniques to analyze how quantities change with respect to time or other variables, often in real-world scenarios involving motion, growth, or other dynamic processes.