Exercise-5.6, Class 12th, Maths, Chapter 5, NCERT

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx.

Question 1

If $x$ and $y$ are connected parametrically by the equations:

$$x=2a{t}^2,y=a{t}^4$$

Find $\frac{dy}{dx}$ without eliminating the parameter.

Solution

Differentiate both equations with respect to parameter $t$:

$$\frac{dx}{dt}=\frac{d}{dt}(2a{t}^2)=4at$$
$$\frac{dy}{dt}=\frac{d}{dt}(a{t}^4)=4a{t}^3$$

Now apply the formula:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
$$\frac{dy}{dx}=\frac{4a{t}^3}{4at}=t^2$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=t^2}}$$

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Question 2

If$$x=a\cos \theta ,y=b\cos \theta$$

find $\frac{dy}{dx}$.

Solution

Differentiate w.r.t. $\theta$:

$$\frac{dx}{d\theta }=-a\sin \theta$$
$$\frac{dy}{d\theta }=-b\sin \theta$$

Apply:

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$
$$\frac{dy}{dx}=\frac{-b\sin \theta }{-a\sin \theta }$$

$$\frac{dy}{dx}=\frac{b}{a}$$

Final Answer

$$\boxed{{\displaystyle \frac{dy}{dx}=\frac{b}{a}}}$$

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Question 3

If$$x=\sin t,y=\cos 2t$$

find $\frac{dy}{dx}$.

Solution

Differentiate both equations with respect to parameter $t$:

$$\frac{dx}{dt}=\cos t$$

$$\frac{dy}{dt}=\frac{d}{dt}(\cos 2t)=-2\sin 2t$$

Apply parametric derivative formula:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
$$\frac{dy}{dx}=\frac{-2\sin 2t}{\cos t}$$

Now use identity:

$$\sin 2t=2\sin t\cos t$$

So:$$\frac{dy}{dx}=\frac{-2(2\sin t\cos t)}{\cos t}$$
$$\frac{dy}{dx}=-4\sin t$$

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Question 4

If$$x=4t,y=\frac{4}{t}$$

find $\frac{dy}{dx}$.

Solution

Differentiate both equations with respect to parameter $t$:

$$\frac{dx}{dt}=4$$
$$\frac{dy}{dt}=\frac{d}{dt}\left(\frac{4}{t}\right)=-\frac{4}{t^2}$$

Now apply:

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
$$\frac{dy}{dx}=\frac{-\frac{4}{t^2}}{4}$$
$$\frac{dy}{dx}=-\frac{1}{t^2}$$

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Question 5

If$$x=\cos \theta -\cos 2\theta ,y=\sin \theta -\sin 2\theta$$

find $\frac{dy}{dx}$.

Solution

Differentiate both equations with respect to $\theta$:

Differentiate $x$

$$\frac{dx}{d\theta }=\frac{d}{d\theta }(\cos \theta )-\frac{d}{d\theta }(\cos 2\theta )$$
$$=-\sin \theta -(-\sin 2\theta )(2)$$
$$=-\sin \theta +2\sin 2\theta$$

Differentiate $y$

$$\frac{dy}{d\theta }=\frac{d}{d\theta }(\sin \theta )-\frac{d}{d\theta }(\sin 2\theta )$$
$$=\cos \theta -(2\cos 2\theta )$$

Apply parametric derivative formula

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$
$$\frac{dy}{dx}=\frac{\cos \theta -2\cos 2\theta }{-\sin \theta +2\sin 2\theta }$$

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Question 6

If$$x=a(\theta -\sin \theta ),y=a(1+\cos \theta )$$

find $\frac{dy}{dx}$.

Solution

Differentiate both $x$ and $y$ with respect to $\theta$:

Differentiate $x$

$$\frac{dx}{d\theta }=a(\frac{d}{d\theta }(\theta )-\frac{d}{d\theta }(\sin \theta ))$$
$$=a(1-\cos \theta )$$

Differentiate $y$

$$\frac{dy}{d\theta }=a\frac{d}{d\theta }(1+\cos \theta )$$
$$=a(0-\sin \theta )$$
$$=-a\sin \theta$$

Apply the formula:

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$
$$\frac{dy}{dx}=\frac{-a\sin \theta }{a(1-\cos \theta )}$$
$$\frac{dy}{dx}=\frac{-\sin \theta }{1-\cos \theta }$$

Use identity:

$$1-\cos \theta =2{\sin }^2\frac{\theta }{2},\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$$
$$\frac{dy}{dx}=\frac{-2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$$
$$=-\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}$$
$$=-\cot \frac{\theta }{2}$$

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Question 7

If

$$x=\frac{{\sin }^{3}t}{\sqrt{\cos 2t}},y=\frac{{\cos }^{3}t}{\sqrt{\cos 2t}}$$

Solution 

Rewrite $x$ and $y$ as:

$$x={\sin }^{3}t(\cos 2t{)}^{-1\mathrm{/}2},y={\cos }^{3}t(\cos 2t{)}^{-1\mathrm{/}2}$$

Divide $y$ by $x$:

$$\frac{y}{x}=\frac{{\cos }^{3}t}{{\sin }^{3}t}={\cot }^{3}t$$

Take cube root:

$${\left(\frac{y}{x}\right)}^{1\mathrm{/}3}=\cot t$$Let:

$$u=\tan t={\left(\frac{x}{y}\right)}^{1\mathrm{/}3}$$Differentiate implicitly:

$${\sec }^{2}t\frac{dt}{dx}=\frac{1}{3}{\left(\frac{x}{y}\right)}^{-2\mathrm{/}3}\cdot \frac{y-x\frac{dy}{dx}}{y^2}$$

$$\frac{dy}{dx}=-\cot 3t$$

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Question 8

If$$x=a(\cos t+\log \tan \frac{t}{2}),y=a\sin t$$

find $\frac{dy}{dx}$.

Solution 

Differentiate both $x$ and $y$ with respect to $t$.

Differentiate $y$

$$y=a\sin t$$

$$\frac{dy}{dt}=a\cos t$$

Differentiate $x$

$$x=a(\cos t+\log \tan \frac{t}{2})$$

$$\frac{dx}{dt}=a(-\sin t+\frac{d}{dt}\log \tan \frac{t}{2})$$

Now recall the identity:

$$\frac{d}{dt}(\log \tan \frac{t}{2})=\frac{1}{\sin t}$$

So:

$$\frac{dx}{dt}=a(-\sin t+\frac{1}{\sin t})$$
$$\frac{dx}{dt}=a\left(\frac{-{\sin }^{2}t+1}{\sin t}\right)$$
$$\frac{dx}{dt}=a\left(\frac{1-{\sin }^{2}t}{\sin t}\right)$$
$$\frac{dx}{dt}=a\left(\frac{{\cos }^{2}t}{\sin t}\right)$$
$$\frac{dx}{dt}=a{\cos }^{2}t\csc t$$

Now apply parametric formula

$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$$$\frac{dy}{dx}=\frac{a\cos t}{a{\cos }^{2}t\csc t}$$Cancel $a$:

$$\frac{dy}{dx}=\frac{\cos t\sin t}{{\cos }^{2}t}$$

$$\frac{dy}{dx}=\sin t\cdot \sec t$$

$$\frac{dy}{dx}=\tan t$$

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Question 9

If

$$x=a\sec \theta ,y=b\tan \theta$$

find $\frac{dy}{dx}$.

Solution

Differentiate both with respect to $\theta$:

Differentiate $x$

$$\frac{dx}{d\theta }=a\sec \theta \tan \theta$$

Differentiate $y$

$$\frac{dy}{d\theta }=b{\sec }^2\theta$$

Apply parametric derivative formula

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$
$$\frac{dy}{dx}=\frac{b{\sec }^2\theta }{a\sec \theta \tan \theta }$$
$$\frac{dy}{dx}=\frac{b\sec \theta }{a\tan \theta }$$
$$\frac{dy}{dx}=\frac{b}{a}\cdot \frac{\sec \theta }{\tan \theta }$$Use identity:

$$\frac{\sec \theta }{\tan \theta }=\csc \theta$$Thus:

$$\frac{dy}{dx}=\frac{b}{a}\csc \mathrm{\theta <hr\; data-start="397"\; data-end="400">}$$

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Question 10

If

$$x=a(\cos \theta +\theta \sin \theta ),y=a(\sin \theta -\theta \cos \theta )$$

find $\frac{dy}{dx}$.

Solution 

Differentiate $x$ and $y$ with respect to $\theta$.

Differentiate $x$

$$x=a(\cos \theta +\theta \sin \theta )$$

Apply product rule to $\theta \sin \theta$:

$$\frac{dx}{d\theta }=a(-\sin \theta +\sin \theta +\theta \cos \theta )$$

$$\frac{dx}{d\theta }=a(\theta \cos \theta )$$

Differentiate $y$

$$y=a(\sin \theta -\theta \cos \theta )$$
$$\frac{dy}{d\theta }=a(\cos \theta -(\cos \theta -\theta \sin \theta ))$$

$$\frac{dy}{d\theta }=a(\theta \sin \theta )$$

Apply parametric derivative formula

$$\frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}$$
$$\frac{dy}{dx}=\frac{a\theta \sin \theta }{a\theta \cos \theta }$$

Cancel $a\theta$:

$$\frac{dy}{dx}=\frac{\sin \theta }{\cos \theta }$$
$$\frac{dy}{dx}=\tan \theta$$

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Question

If$$x=\sqrt{a^{{\sin }^{-1}t}},y=\sqrt{a^{{\cos }^{-1}t}}$$

show that

$$\frac{dy}{dx}=-\frac{y}{x}$$

Solution 

First rewrite expressions using exponent rules.

$$x={\left(a^{{\sin }^{-1}t}\right)}^{1\mathrm{/}2}=a^{\frac{1}{2}{\sin }^{-1}t}$$
$$y={\left(a^{{\cos }^{-1}t}\right)}^{1\mathrm{/}2}=a^{\frac{1}{2}{\cos }^{-1}t}$$

Differentiate w.r.t. $t$ using log differentiation:

Differentiate $x$

$$\log x=\frac{1}{2}({\sin }^{-1}t)\log a$$

Differentiate:

$$\frac{1}{x}\frac{dx}{dt}=\frac{1}{2}\log a\cdot \frac{1}{\sqrt{1-t^2}}$$

$$\frac{dx}{dt}=x\cdot \frac{\log a}{2\sqrt{1-t^2}}$$

Differentiate $y$

$$\log y=\frac{1}{2}({\cos }^{-1}t)\log a$$

Differentiate:

$$\frac{1}{y}\frac{dy}{dt}=\frac{1}{2}\log a\cdot \frac{-1}{\sqrt{1-t^2}}$$

$$\frac{dy}{dt}=-y\cdot \frac{\log a}{2\sqrt{1-t^2}}$$

Now apply parametric derivative formula

$$\frac{dy}{dx}=\frac{dy\mathrm{/}dt}{dx\mathrm{/}dt}$$
$$\frac{dy}{dx}=\frac{-y\cdot \frac{\log a}{2\sqrt{1-t^2}}}{x\cdot \frac{\log a}{2\sqrt{1-t^2}}}$$

Cancel common factors:

$$\frac{dy}{dx}=\frac{-y}{x}$$