The slope of the normal to the curve $x=\sqrt{t}$ and $y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}$ at $\mathrm{t}=4$ is

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{1+\frac{1}{2 \mathrm{t}^{\frac{3}{2}}}}{\frac{1}{2 \sqrt{\mathrm{t}}}}=\frac{2 \mathrm{t}^{\frac{3}{2}}+1}{\mathrm{t}} \ 	herefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{\mathrm{t}=4}= & \frac{2(4)^{\frac{3}{2}}+1}{4}=\frac{16+1}{4}=\frac{17}{4} \end{aligned}$ $	herefore \quad$ Slope of normal at $\mathrm{t}=4$ is $-\frac{1}{\left(\frac{\mathrm{d} y}{\mathrm{~d} x}ight)_{t=4}}=-\frac{4}{17}$

The slope of the normal to the curve $x=\sqrt{t}$ and $y=\mathrm{t}-\frac{1}{\sqrt{\mathrm{t}}}$ at…

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