Chapter 5: Continuity and Differentiability

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

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, $\frac{dy}{dx} =  \frac{b}{a}$

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, $\frac{dy}{dx} =   -4 \sin t$

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} $

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

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

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

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

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

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

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

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