Class 12 RD Sharma Solutions Chapter 11 Differentiation Exercise 11.7 | Set 1 (original) (raw)

Last Updated : 23 Aug, 2024

**Question 1. Find \frac{dy}{dx} ****, when: x = at** 2 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 = a**eΘ (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 = bsin** 2 θ and y = acos 2 θ

**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 - cos**θ) 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.