Let the curve be represented by $x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}), y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t})$. Then normal at any point ' $t$ ' of the curve is at a distance of __________ units from the origin.

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Aicharya · Accepted Answer

$\begin{array}{ll} & x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}) \ 	herefore \quad & \frac{\mathrm{d} x}{\mathrm{dt}}=2(-\sin \mathrm{t}+\sin \mathrm{t}+\mathrm{t} \cos \mathrm{t})=2 \mathrm{t} \cos \mathrm{t} \ & y=2(\sin \mathrm{t}-\mathrm{t} \cos \mathrm{t}) \ 	herefore \quad & \frac{\mathrm{d} y}{\mathrm{dt}}=2(\cos \mathrm{t}-\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t})=2 \mathrm{t} \sin \mathrm{t} \ 	herefore \quad & \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\frac{\mathrm{d} y}{\mathrm{dt}}}{\frac{\mathrm{d} x}{\mathrm{dt}}}=\frac{2 \mathrm{t} \sin \mathrm{t}}{2 \mathrm{t} \cos \mathrm{t}}=\mathrm{tan} \mathrm{t} \ & 	ext { Slope of normal }=-\frac{1}{\frac{\mathrm{d} y}{\mathrm{~d} x}}=-\frac{1}{	an \mathrm{t}}=-\frac{\cos \mathrm{t}}{\sin \mathrm{t}}\end{array}$ $	herefore \quad$ Equation of the normal is $\begin{aligned} & y-2(\sin t-t \cos t)=-\frac{\cos t}{\sin t}[x-2(\cos t+t \sin t)] \ & \Rightarrow y \sin \mathrm{t}-2 \sin ^2 \mathrm{t}+2 \mathrm{t} \sin \mathrm{t} \cos \mathrm{t} \ & =-x \cos t+2 \cos ^2 t+2 t \sin t \cos t \ & \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2\left(\sin ^2 \mathrm{t}+\cos ^2 \mathrm{t}ight) \ & \Rightarrow x \cos \mathrm{t}+y \sin \mathrm{t}=2 \ & 	herefore \quad 	ext { Distance from origin }=\left|\frac{-2}{\sqrt{\cos ^2 \mathrm{t}+\sin ^2 \mathrm{t}}}ight|=2 	ext { units } \ & \end{aligned}$

Let the curve be represented by $x=2(\cos \mathrm{t}+\mathrm{t} \sin \mathrm{t}), y=2(\sin…

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