# Find the slopes of the tangent and the normal to the following curves at the indicted points:

Question:

Find the slopes of the tangent and the normal to the following curves at the indicted points:

(i) $y=\sqrt{x^{3}}$ at $x=4$

(ii) $y=\sqrt{x}$ at $x=9$

(iii) $y=x^{3}-x$ at $x=2$

(iv) $y=2 x^{2}+3 \sin x$ at $x=0$

(v) $x=a(\theta-\sin \theta), y=a(1-\cos \theta)$ at $\theta=-\pi / 2$

(vi) $x=a \cos ^{3} \theta, y=a \sin ^{3} \theta$ at $\theta=\pi / 4$

(vii) $x=a(\theta-\sin \theta), y=a(1-\cos \theta)$ at $\theta=\pi / 2$

(viii) $y=(\sin 2 x+\cot x+2)^{2}$ at $x=\pi / 2$

(ix) $x^{2}+3 y+y^{2}=5$ at $(1,1)$

$(x) x y=6$ at $(1,6)$

Solution:

(i) $y=\sqrt{x^{3}}=x^{\frac{3}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{3}{2} x^{\frac{1}{2}}=\frac{3}{2} \sqrt{x}$

When $x=4, y=\sqrt{x^{3}}=\sqrt{64}=8$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(4,8)}=\frac{3}{2} \sqrt{4}=3$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{(4,8)}}=\frac{-1}{3}$

(ii) $y=\sqrt{x}=x^{\frac{1}{2}}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} x^{\frac{-1}{2}}=\frac{1}{2 \sqrt{x}}$

When $x=9, y=\sqrt{x}=\sqrt{9}=3$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(9,3)}=\frac{1}{2 \sqrt{9}}=\frac{1}{6}$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{(9,3)}}=\frac{-1}{\left(\frac{1}{6}\right)}=-6$

(iii) $y=x^{3}-x$

$\Rightarrow \frac{d y}{d x}=3 x^{2}-1$

When $x=2, y=x^{3}-x=2^{3}-2=6$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(2,6)}=3(2)^{2}-1=11$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{(2,6)}}=\frac{-1}{11}$

(iv) $y=2 x^{2}+3 \sin x$

$\Rightarrow \frac{d y}{d x}=4 x+3 \cos x$

When $x=0, y=2 x^{2}+3 \sin x=2(0)^{2}+3 \sin 0=0$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(0,0)}=4(0)+3 \cos 0=3$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{(0,0)}}=\frac{-1}{3}$

$(v) x=a(\theta-\sin \theta)$

$\Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta)$

$y=a(1+\cos \theta)$

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

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a(-\sin \theta)}{a(1-\cos \theta)}=\frac{-2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=-\cot \frac{\theta}{2}$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}=-\cot \left(\frac{\frac{-\pi}{2}}{2}\right)=-\cot \left(\frac{-\pi}{4}\right)=1$ Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{-\pi}{2}}}=\frac{-1}{1}=-1$

$(v i) x=a \cos ^{3} \theta$

$\Rightarrow \frac{d x}{d \theta}=-3 a \cos ^{2} \theta \sin \theta$

$y=a \sin ^{3} \theta$

$\Rightarrow \frac{d y}{d \theta}=3 a \sin ^{2} \theta \cos \theta$

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 a \sin ^{2} \theta \cos \theta}{-3 a \cos ^{2} \theta \sin \theta}=-\tan \theta$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}=-\tan \frac{\pi}{4}=-1$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{4}}}=\frac{-1}{-1}=1$

$(v i i) x=a(\theta-\sin \theta)$

$y=a(1-\cos \theta)$

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

$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{a(\sin \theta)}{a(1-\cos \theta)}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}=\cot \frac{\theta}{2}$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}=\cot \left(\frac{\frac{\pi}{2}}{2}\right)=\cot \left(\frac{\pi}{4}\right)=1$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{2}}}=\frac{-1}{1}=-1$

(viii) $y=(\sin 2 x+\cot x+2)^{2}$

$\Rightarrow \frac{d y}{d x}=2(\sin 2 x+\cot x+2)\left(2 \cos 2 x-\cos e c^{2} x\right)$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}$

$=2\left[\sin 2\left(\frac{\pi}{2}\right)+\cot \left(\frac{\pi}{2}\right)+2\right]\left[2 \cos 2\left(\frac{\pi}{2}\right)-\operatorname{cosec}^{2}\left(\frac{\pi}{2}\right)\right]$

$=2(0+0+2)(-2-1)$

$=-12$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{x=\frac{\pi}{2}}}=\frac{-1}{-12}=\frac{1}{12}$

$\Rightarrow \frac{d x}{d \theta}=a(1-\cos \theta)$

$(i x) x^{2}+3 y+y^{2}=5$

On differentiating both sides w.r.t. $x$, we get

$2 x+3 \frac{d y}{d x}+2 y \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}(3+2 y)=-2 x$

$\Rightarrow \frac{d y}{d x}=\frac{-2 x}{3+2 y}$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{-2 x}{3+2 y}=\frac{-2}{3+2}=\frac{-2}{5}$

Slope of the normal $=\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,1)}}=\frac{-1}{\left(\frac{-2}{5}\right)}=\frac{5}{2}$

$(x) x y=6$

On differentiating both sides w.r.t. $x$, we get

$x \frac{d y}{d x}+y=0$

$\Rightarrow x \frac{d y}{d x}=-y$

$\Rightarrow \frac{d y}{d x}=\frac{-y}{x}$

Now,

Slope of the tangent $=\left(\frac{d y}{d x}\right)_{(1,6)}=\frac{-y}{x}=\frac{-6}{1}=-6$

Slope of the normal= $\frac{-1}{\left(\frac{d y}{d x}\right)_{(1,6)}}=\frac{-1}{-6}=\frac{1}{6}$