Question

The tangent to the curve given by,$x=e^t.\cos t,y=e^t.\sin t\mathrm{at}t=\frac{\pi }{4},$ makes an angle of with x-axis.

• A. \N
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

$\frac{\mathrm{dx}}{\mathrm{dt}}=e^t\cdot \cos t-e^t\cdot \sin t\frac{\mathrm{dy}}{\mathrm{dt}}=e^t\cdot \cos t+e^t\cdot \sin t\mathrm{Now},\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{e^t\cdot \cos t+e^t\cdot \sin t}{e^t\cdot \cos t-e^t\cdot \sin t}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\cos t+\sin t}{\cos t-\sin t}{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{t=\frac{\pi }{4}}=\frac{\sqrt{2}}{0}\therefore \mathrm{The}\mathrm{angle}\mathrm{that}\mathrm{the}\mathrm{tangent}\mathrm{makes}\mathrm{with}x-\mathrm{axis}\mathrm{is}\frac{\pi }{2}.$