Question

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

(i) $y=\sqrt{x^3}\mathrm{at}x=4$
(ii) $y=\sqrt{x}\mathrm{at}x=9$
(iii) y = x³ − x at x = 2
(iv) y = 2x² + 3 sin x at x = 0
(v) x = a (θ − sin θ), y = a(1 − cos θ) at θ = −π/2
(vi) x = a cos³ θ, y = a sin³ θ at θ = π/4
(vii) x = a (θ − sin θ), y = a(1 − cos θ) at θ = π/2
(viii) y = (sin 2x + cot x + 2)² at x = π/2
(ix) x² + 3y + y² = 5 at (1, 1)
(x) xy = 6 at (1, 6)

Answer 1

$$\begin{gathered} \left(i\right)y=\sqrt{x^3}=x^{\frac{3}{2}} \\ \Rightarrow \frac{dy}{dx}=\frac{3}{2}{x}^{\frac{1}{2}}=\frac{3}{2}\sqrt{x} \\ \text{When}\mathit{\text{x}}\text{=4,}y=\sqrt{x^3}=\sqrt{64}=8 \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(4,8\right)}\text{=}\frac{3}{2}\sqrt{4}=3 \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(4,8\right)}}\text{=}\frac{-1}{3} \end{gathered}$$

$$\begin{gathered} \left(ii\right)y=\sqrt{x}=x^{\frac{1}{2}} \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2}{x}^{\frac{-1}{2}}=\frac{1}{2\sqrt{x}} \\ \text{When}\mathit{\text{x}}\text{=9,}y=\sqrt{x}=\sqrt{9}=3 \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(9,3\right)}\text{=}\frac{1}{2\sqrt{9}}\text{=}\frac{1}{6} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(9,3\right)}}\text{=}\frac{-1}{\left({\displaystyle \frac{1}{6}}\right)}\text{=-6} \end{gathered}$$

$$\begin{gathered} \left(iii\right)y=x^3-x \\ \Rightarrow \frac{dy}{dx}=3x^2-1 \\ \text{When}\mathit{\text{x}}\text{=2,}y\text{=}{x}^3-x=2^3-2=6 \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(2,6\right)}\text{=3}{\left(2\right)}^2\text{-1=11} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(2,6\right)}}\text{=}\frac{-1}{11} \end{gathered}$$

$$\begin{gathered} \left(iv\right)y=2x^2+3\sin x \\ \Rightarrow \frac{dy}{dx}=4x+3\cos x \\ \text{When}\mathit{\text{x}}\text{=0,}y\text{=}2x^2+3\sin x=2{\left(0\right)}^2+3\sin 0=0 \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}\text{=4}\left(0\right)\text{+ 3 cos 0=3} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(0,0\right)}}\text{=}\frac{-1}{3} \end{gathered}$$


$$\begin{gathered} \left(v\right)x=a\left(\theta -\sin \theta \right) \\ \Rightarrow \frac{dx}{d\theta }=a\left(1-\cos \theta \right) \\ y=a\left(1+\cos \theta \right) \\ \Rightarrow \frac{dy}{d\theta }=a\left(-\sin \theta \right) \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\left(-\sin \theta \right)}{a\left(1-\cos \theta \right)}=\frac{-2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}{2{\sin }^2{\displaystyle \frac{\theta }{2}}}=-cot\frac{\theta }{2} \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{-\mathrm{\pi }}{2}}\text{=-cot}\left(\frac{\frac{-\mathrm{\pi }}{2}}{2}\right)\text{=-cot}\left(\frac{-\mathrm{\pi }}{4}\right)\text{=1} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\theta =\frac{-\mathrm{\pi }}{2}}}\text{=}\frac{-1}{1}\text{=-1} \end{gathered}$$

$$\begin{gathered} \left(vi\right)x=a{\cos }^3\theta \\ \Rightarrow \frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \\ y=a{\sin }^3\theta \\ \Rightarrow \frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}\text{=-tan}\frac{\mathrm{\pi }}{4}\text{=-1} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{4}}}\text{=}\frac{-1}{-1}\text{=1} \end{gathered}$$


$$\begin{gathered} \left(vii\right)x=a\left(\theta -\sin \theta \right) \\ \Rightarrow \frac{dx}{d\theta }=a\left(1-\cos \theta \right) \\ y=a\left(1-\cos \theta \right) \\ \Rightarrow \frac{dy}{d\theta }=a\left(\sin \theta \right) \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{a\left(\sin \theta \right)}{a\left(1-\cos \theta \right)}=\frac{2\sin {\displaystyle \frac{\theta }{2}}\cos {\displaystyle \frac{\theta }{2}}}{2{\sin }^2{\displaystyle \frac{\theta }{2}}}=\cot \frac{\theta }{2} \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{2}}\text{=cot}\left(\frac{\frac{\mathrm{\pi }}{2}}{2}\right)\text{=cot}\left(\frac{\mathrm{\pi }}{4}\right)\text{=1} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\theta =\frac{\mathrm{\pi }}{2}}}\text{=}\frac{-1}{1}\text{=-1} \end{gathered}$$

$$\begin{gathered} \left(viii\right)y={\left(\sin 2x+\cot x+2\right)}^2 \\ \Rightarrow \frac{dy}{dx}=2\left(\sin 2x+\cot x+2\right)\left(2\cos 2x-\cos e{c}^{2}x\right) \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{2}} \\ \text{=}2\left[\sin 2\left(\frac{\mathrm{\pi }}{2}\right)+\cot \left(\frac{\mathrm{\pi }}{2}\right)+2\right]\left[2\cos 2\left(\frac{\mathrm{\pi }}{2}\right)-{\mathrm{cosec}}^2\left(\frac{\mathrm{\pi }}{2}\right)\right] \\ =2\left(0+0+2\right)\left(-2-1\right) \\ =-12 \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{2}}}\text{=}\frac{-1}{-12}\text{=}\frac{1}{12} \end{gathered}$$

$$\begin{gathered} \left(ix\right)x^2+3y+y^2=5 \\ \text{On differentiating both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ 2x+3\frac{dy}{dx}+2y\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}\left(3+2y\right)=-2x \\ \Rightarrow \frac{dy}{dx}=\frac{-2x}{3+2y} \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}\text{=}\frac{-2x}{3+2y}\text{=}\frac{-2}{3+2}\text{=}\frac{-2}{5} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(1,1\right)}}\text{=}\frac{-1}{\left(\frac{-2}{5}\right)}\text{=}\frac{5}{2} \end{gathered}$$

$$\begin{gathered} \left(x\right)xy=6 \\ \text{On differentiating both sides w.r.t.}x,\mathrm{we}\mathrm{get} \\ x\frac{dy}{dx}+y=0 \\ \Rightarrow x\frac{dy}{dx}=-y \\ \Rightarrow \frac{dy}{dx}=\frac{-y}{x} \\ \mathrm{Now}, \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{\left(1,6\right)}\text{=}\frac{-y}{x}\text{=}\frac{-6}{1}\text{=-6} \\ \text{Slope of the normal=}\frac{-1}{{\left(\frac{dy}{dx}\right)}_{\left(1,6\right)}}\text{=}\frac{-1}{-6}\text{=}\frac{1}{6} \end{gathered}$$