NCERT Solutions for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 5.6

Exercise 5.6

If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find $\mathbf{\frac{dy}{dx}}$.

1. x=2at2 , y=at4

Ans: The given equations are  

x=2at2 …… (1) 

and y=at4 …… (2) 

Then, differentiating both sides of the equation (1) with respect to t gives

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

Also, differentiating both sides of the equation (2) with respect to t gives

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

Now, dividing the equations (4) by (3) gives 

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

Hence $\frac{dy}{dx}=t^2$.

2. x=acosθ, y=bcosθ

Ans: The given equations are  

x=acosθ …… (1) 

and y=bcosθ …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives

$\frac{dx}{d\theta }= \frac{d}{d\theta }= (a\cos \theta ) =a(-\sin \theta ) = -a\sin \theta$.....(3)

Also, differentiating both sides of the equation (1) with respect to θgives

$\frac{dy}{d\theta }= \frac{d}{d\theta }= (b\cos \theta ) =b(-\sin \theta ) = -b\sin \theta$......(4)

Therefore, dividing the equation (4) by (3) gives 

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

Hence, $\frac{dy}{dx} =  \frac{b}{a}$

3. x=sint, y=cos2t

Ans: The given equations are 

x=sint …… (1) 

and y=cos2t …… (2) 

Then, differentiating both sides of the equation (1) with respect to t gives 

$\frac{dx}{dt} = \frac{d}{dt}(\sin\theta ) = \cos\theta ......(3)$

Also, differentiating both sides of the equation (2) with respect to t gives

$\frac{dy}{dt} = \frac{d}{dt}(\cos2\theta ) = \sin 2\theta  \times \frac{d}{dt}(2t)= -2\sin2t......(4)$

Therefore,  by dividing the equation (4) by (3) gives 

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

Hence, $\frac{dy}{dx} =   -4 \sin t$

4. x=4t,  y = $\mathbf{\frac{4}{t}}$

Ans: The given equations are 

x=4t …… (1) 

 y = $\frac{4}{t}$ ……(2)

Now, differentiating both sides of the equation (1) with respect to t gives 

$\frac{dy}{dx} =\frac{d}{dt}(4t)=4......(3)$

Also, differentiating both sides of the equation (2) with respect to t gives 

$\frac{dy}{dt}= \frac{d}{dt}(\frac{4}{t}) = 4 \times \frac{d}{dt}\frac{1}{t}= 4 \times \frac{-1}{t^2}= \frac{-4}{t^2}.......(4)$

Therefore, dividing the equation (4) by (3) gives 

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

Hence, $\frac{dy}{dx} = \frac{-1}{t^2} $

5. x=cosθ-cos2θ, y=sinθ-sin2θ

Ans: The given equations are 

x=cosθ-cos2θ …… (1) 

and y=sinθ-sin2θ …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives 

$\frac{dx}{d\theta } = \frac{d}{d\theta }(\cos \theta - \cos 2\theta ) = \frac{d}{dt}(\cos \theta )- \frac{d}{d\theta }(cos 2\theta )= -\sin \theta (-2 \sin 2\theta ) = 2 \sin 2\theta - \sin\theta ………..(3)$

Also, differentiating both sides of the equation (2) with respect to θ gives 

$\frac{dy}{d\theta } = \frac{d}{d\theta }(\sin \theta - \sin 2\theta ) = \frac{d}{d\theta }(\sin \theta )- \frac{d}{d\theta }(sin 2\theta )= \cos \theta - 2 \cos\theta .......(4)$

Therefore, dividing the equation (4) by (3) gives

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

Hence, $\frac{dy}{dx} =  \frac{\cos\theta - 2 \cos\theta  }{2 \sin 2\theta - \sin\theta }$

6. x=a(θ-sinθ), y=a(1+cosθ)

Ans: The given equations are 

x=a(θ-sinθ) …… (1) 

and y=a(1+cosθ) …… (2) 

Then, differentiating both sides of the equation (1) with respect to θ gives

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

Also, differentiating both sides of the equation (2) with respect to θ gives 

$\frac{dy}{d\theta }= a[\frac{d}{d\theta }(1) - \frac{d}{d\theta }(\cos \theta )] = a[0 + (- \sin \theta )] = -a \sin \theta ........(4)$

Therefore, by dividing the equation (4) by (3) gives 

$\frac{dy}{dx} = \frac{(\frac{dy}{d\theta })}{(\frac{dx}{d\theta })}= \frac{-a \sin \theta }{a(1- \cos\theta )}=\frac{-2 \sin\frac{\theta }{2} \cos \frac{\theta }{2}}{2 \sin ^2\frac{\theta }{2}}= \frac{-\cos\frac{\theta }{2}}{\sin \frac{\theta }{2}}= -\cos\frac{\theta }{2}$

Hence, $\frac{dy}{dx} =  -\cos\frac{\theta }{2}$

7. $-\frac{sin^3 t}{\sqrt{cos 2t}} , y = \frac{cos^3 t}{\sqrt{cos 2t}}$ 

Ans: The given equations are, 

$x = - \frac{\sin^3 t}{\sqrt{\cos 2t}} ......... (1)$

$y = - \frac{\cos^3 t}{\sqrt{\cos 2t}} ......... (2)$

Then, differentiating both sides of the equation (1) with respect to t gives 

$\begin{aligned}&\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\sin ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}\right] \\ \\&=\frac{\sqrt{\cos 2 \mathrm{t}}-\frac{\mathrm{d}}{\mathrm{dt}}\left(\sin ^{3} \mathrm{t}\right)-\sin ^{3} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}} \sqrt{\cos 2 \mathrm{t}}}{\cos 2 \mathrm{t}} \\  \\&=\frac{\sqrt{\cos 2 \mathrm{t}} \times 3 \sin ^{2} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}}(\sin \mathrm{t})-\sin ^{3} \mathrm{t} \times \frac{1}{2 \sqrt{\cos 2 \mathrm{t}}} \times \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t})}{\cos 2 \mathrm{t}} \\ \\&=\frac{3 \sqrt{\cos 2 \mathrm{t}} \times \sin ^{2} \mathrm{t} \cos \mathrm{t}-\frac{\sin ^{3} \mathrm{t}}{2 \sqrt{\cos 2 \mathrm{t}}} \times(-2 \sin 2 \mathrm{t})}{\cos 2 \mathrm{t} \sqrt{\cos 2 \mathrm{t}}}\end{aligned}$

Also, differentiating both sides of the equation (2) with respect to $t$ gives 

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{3 \cos 2 \mathrm{t} \sin ^{2} t \cos t+\sin ^{2} t \sin 2 \mathrm{t}}{\cos 2 \mathrm{t} \sqrt{\cos 2 \mathrm{t}}} . \ldots \ldots$ (3)$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\cos ^{3} \mathrm{t}}{\sqrt{\cos 2 \mathrm{t}}}\right]$

$\begin{aligned}&=\frac{\sqrt{\cos 2 \mathrm{t}} \times \frac{\mathrm{d}}{\mathrm{dt}}\left(\cos ^{3} \mathrm{t}\right)-\cos ^{3} \mathrm{t} \times \frac{\mathrm{d}}{\mathrm{dt}}(\sqrt{\cos 2 \mathrm{t}})}{\cos 2 \mathrm{t}} \\ \\&=\frac{3 \sqrt{\cos 2 \mathrm{t} \cos ^{2} \mathrm{t}(-\sin t)-\cos ^{3} t \times \frac{1}{2(\sqrt{\cos 2 t})}} \times \frac{\mathrm{d}}{\mathrm{dt}}(\cos 2 \mathrm{t})}{\cos 2 \mathrm{t}} \\&\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-3 \cos 2 \mathrm{t} \times \cos ^{2} \mathrm{t} \times \sin t+\cos ^{3} \mathrm{t} \sin 2 \mathrm{t}}{\cos 2 \mathrm{t} \times \sqrt{\cos 2 \mathrm{t}}} \ldots \ldots \text { (4) }\end{aligned}$

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d x}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{-3 \cos 2 t \times \cos ^{2} t \times \sin t+\cos ^{3} t \sin 2 t}{3 \cos 2 t \cos t \sin ^{2} t+\sin ^{3} t \sin 2 t}$

$\begin{aligned}&=\frac{\sin t \cos t\left[-3 \cos 2 t \times \cos t+2 \cos ^{3} t\right]}{\sin t \cos t\left[3 \cos 2 t \sin t+2 \sin ^{3} t\right]}  \\ \\&=\frac{\left[-3\left(2 \cos ^{2} t-1\right) \cos t+2 \cos ^{3} t\right]}{\left[3\left(1-2 \sin ^{3} t\right) \sin t+2 \sin ^{3} t\right]} \quad\left[\begin{array}{l}\cos 2 t=\left(2 \cos ^{2} t-1\right) \\\cos 2 t=\left(1-2 \sin ^{2} t\right)\end{array}\right] \\ \\&=\frac{-4 \cos ^{3} t+3 \cos t}{3 \sin t-4 \sin ^{3} t} \\ \\&=\frac{-\cos 3 t}{\sin 3 t} \\  \\&\text { Hence, } \frac{d y}{d x}=-\cot 3 t .\end{aligned}$

8. $\mathbf{x=a\left(\operatorname{cost}+\log \tan \frac{t}{2}\right), y=a \sin t}$

Ans: The given equations are

$x=a\left(\cos t+\log \tan \frac{t}{2}\right) \ldots \ldots$ (1)

and $y=a \sin t \quad \ldots \ldots(2)$

Then, differentiating both sides of the equation (1) with respect to $t$ gives $\frac{d x}{d t}=a \times\left[\frac{d}{d \theta}(\operatorname{cost})+\frac{d}{d \theta}\left(\log \tan \frac{t}{2}\right)\right]$

$\begin{aligned}&=\mathrm{a}\left[-\operatorname{sint}+\frac{1}{\tan\frac{\mathrm{t}}{2}} \times \frac{\mathrm{d}}{\mathrm{dt}}\left(\tan \frac{\mathrm{t}}{2}\right)\right] \\&=\mathrm{a}\left[-\sin \mathrm{t}+\cot \frac{\mathrm{t}}{2} \times \sec ^{2} \frac{\mathrm{t}}{2}\times\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{t}}{2}\right)\right]\\&=\mathrm{a}\left[-\operatorname{sint}+\frac{\cos \frac{\mathrm{t}}{2}}{\sin \frac{\mathrm{t}}{2}} \times \frac{1}{\cos ^{2} \frac{\mathrm{t}}{2}} \times \frac{1}{2}\right)\end{aligned}$

$\begin{aligned}&=a\left(-\sin t+\frac{1}{2 \sin \frac{t}{2} \cos \frac{t}{2}}\right) \\&=a\left(-\sin t+\frac{1}{\sin t}\right) \\&=a\left(\frac{-\sin ^{2} t+1}{\sin t}\right) \\&\text { Therefore, } \frac{d x}{d t}=a \frac{\cos ^{2} t}{\sin t}\end{aligned}$

Also, differentiating both sides of the equation (2) with respect to $t$ gives $\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{a} \frac{\mathrm{d}}{\mathrm{dt}}(\sin t)=\mathrm{acost} \ldots \ldots(4)$

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}=\frac{a \cos t}{\left(a \frac{\cos ^{2} t}{\sin t}\right)}=\frac{\sin t}{\cos t}=\operatorname{tant}$

Hence, $\frac{d y}{d x}=\tan t$.

9. $\mathbf{x=a s e c \theta, y=b \tan \theta}$

Ans: The given equations are

$\mathrm{x}=\operatorname{asec} \quad \ldots \ldots$ (1)

and $y=\tan \theta \quad \ldots \ldots$ (2)

Then, differentiating both sides of the equation (1) with respect to $\theta$ gives $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a} \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\sec \theta)=\operatorname{asec} \theta \tan \theta \ldots \ldots$ (3)

Also, differentiating both sides of the equation (2) with respect to $\theta$ gives $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{b} \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\tan \theta)=\mathrm{b} \sec ^{2} \theta \ldots \ldots$ (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{b \sec ^{2} \theta}{a \sec \theta \tan \theta}=\frac{b}{a} \sec \theta \tan \theta=-\frac{b \cos \theta}{a \cos \theta \sin \theta}=\frac{b}{a} \times \frac{1}{\sin \theta}=\frac{b}{a} \cos$

Hence, $\frac{d y}{d x}=\frac{b}{a} \operatorname{cosec} \theta$.

10. $\mathbf{x=a(\cos \theta+\theta \sin \theta), y=a(\sin \theta-\theta \cos \theta)}$

Ans: The given equations are

$\mathrm{x}=\mathrm{a}(\cos \theta+\theta \sin \theta) \quad \ldots \ldots$ (1)

and $\mathrm{y}=\mathrm{a}(\sin \theta-\theta \cos \theta) \ldots \ldots$ (2)

Then, differentiating both sides of the equation (1) with respect to $\theta$ gives $\frac{\mathrm{dx}}{\mathrm{d} \theta}=\mathrm{a}\left[\frac{\mathrm{d}}{\mathrm{d} \theta} \cos \theta+\frac{\mathrm{d}}{\mathrm{d} \theta}(\theta \sin \theta)\right]=\mathrm{a}\left[-\sin \theta+\theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)+\sin \theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\theta)\right]$ $=\mathrm{a}[-\sin \theta+\theta \cos \theta+\sin \theta]$.

Therefore, $\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=\mathrm{a} \theta \cos \theta$(3)

Also, differentiating both sides of the equation (2) with respect to $\theta$ gives $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}\left[\frac{\mathrm{d}}{\mathrm{d} \theta}(\sin \theta)-\frac{\mathrm{d}}{\mathrm{d} \theta}(\theta \cos \theta)\right]=\mathrm{a}\left[\cos \theta-\left\{\theta \frac{\mathrm{d}}{\mathrm{d} \theta}(\cos \theta)+\cos \theta \times \frac{\mathrm{d}}{\mathrm{d} \theta}(\theta)\right\}\right]$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a}[\cos \theta+\theta \sin \theta-\cos \theta]$

Therefore, $\frac{\mathrm{dy}}{\mathrm{d} \theta}=\mathrm{a} \theta \sin \theta \quad \ldots \ldots$ (4)

Thus, dividing the equation (4) by the equation (3) gives

$\frac{d y}{d x}=\frac{\left(\frac{d y}{d \theta}\right)}{\left(\frac{d x}{d \theta}\right)}=\frac{a \theta \sin \theta}{a \theta \sin \theta}=\tan \theta \text {. }$

Hence, $\frac{d y}{d x}=\tan \theta$.

11. If $\mathbf{x=\sqrt{a^{\sin ^{-1} t}}, y=\sqrt{a^{\cos ^{-1} t}}}$.Show that $\mathbf{\frac{d y}{d x}=-\frac{y}{x}}$

Ans: The given parametric equations are $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$.

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

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

$\Rightarrow \mathrm{x}=\mathrm{a}^{\frac{1}{2} \sin ^{-1 \mathrm{t}}}$ and $\mathrm{y}=\mathrm{a}^{\frac{1}{2} \cos ^{-1} \mathrm{t}}$

Therefore, first consider $x=a^{\frac{1}{2} \sin ^{-1} t}$.

Take logarithms on both sides of the equation.

Then, we have

$\log x=\frac{1}{2} \sin ^{-1}$ tloga .

Then, differentiating both sides of the equation with respect to $t$ gives

$\begin{aligned}&\frac{1}{x} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\sin ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{x}{2} \log a \times \frac{1}{\sqrt{1-t^{2}}} \\&\text { Therefore, } \frac{d x}{d t}=\frac{x \log a}{2 \sqrt{1-t^{2}}} . \ldots \ldots \text { (1) }\end{aligned}$

Again, consider the equation $y=a^{\frac{1}{2} \cos ^{-1} t}$.

Take logarithm both sides of the equation.

Then, we have

$\log y=\frac{1}{2} \cos ^{-1} \text { tloga }$

Differentiating both sides of the equation with respect to $\mathrm{t}$ gives $\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right)$

$\begin{aligned}&\frac{1}{y} \times \frac{d x}{d t}=\frac{1}{2} \log a \times \frac{d}{d t}\left(\cos ^{-1} t\right) \\&\Rightarrow \frac{d x}{d t}=\frac{y \log a}{2} \times\left(\frac{1}{\sqrt{1-t^{2}}}\right)\end{aligned}$

Therefore, $\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{-\mathrm{yloga}}{2 \sqrt{1-\mathrm{t}^{2}}}$.

$\frac{dy}{dx} = \frac{(\frac{dy}{dt})}{(\frac{dx}{dt})} = \frac{(\frac{-y \log a}{2\sqrt{1- t^2}})}{(\frac{x \log }{2 \sqrt{1 - t^2}})} = \frac{y}{x}$

 Hence, $\frac{dy}{dx} = \frac{y}{x}$

Conclusion

In conclusion, Ex 5.6 Class 12 Maths Chapter 5 is essential for mastering the derivatives of functions expressed in parametric forms. This exercise builds a strong foundation in applying the chain rule to relate the derivatives of parametric equations. It's important to focus on understanding how to convert parametric equations into standard forms to effectively find their derivatives. By practising these solutions, you will enhance your problem-solving skills and be well-prepared for CBSE board exams. Download the NCERT Solutions for Class 12 Ex 5.6 Maths and practice regularly to score well in your exams.

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