Question

The tangent to the curve x = e^t cost, y = e^t sin t at t = $\frac{\mathrm{\pi }}{4}$ makes with x-axis an angle
(a) 0 (b) $\frac{\mathrm{\pi }}{4}\left(\mathrm{c}\right)\frac{\mathrm{\pi }}{3}\left(\mathrm{d}\right)\frac{\mathrm{\pi }}{2}$

Answer 1

The given curve is x = e^t cost, y = e^t sint.

x = e^t cost

Differentiating both sides with respect to t, we get

$\frac{dx}{dt}=e^t\times \left(-\sin t\right)+\cos t\times e^t$

$\Rightarrow \frac{dx}{dt}=e^t\left(\cos t-\sin t\right)$

y = e^t sint

Differentiating both sides with respect to t, we get

$\frac{dy}{dt}=e^t\times \cos t+\sin t\times e^t$

$\Rightarrow \frac{dy}{dt}=e^t\left(\cos t+\sin t\right)$

Now,

Slope of tangent to the given curve $=\frac{dy}{dx}$$=\frac{{\displaystyle \frac{dy}{dt}}}{{\displaystyle \frac{dx}{dt}}}$$=\frac{e^t\left(\cos t+\sin t\right)}{e^t\left(\cos t-\sin t\right)}$$=\frac{\cos t+\sin t}{\cos t-\sin t}$

∴ Slope of tangent to the given curve at $t=\frac{\mathrm{\pi }}{4}$ $={\left(\frac{dy}{dx}\right)}_{t=\frac{\mathrm{\pi }}{4}}$ $=\frac{\cos {\displaystyle \frac{\mathrm{\pi }}{4}}+\sin {\displaystyle \frac{\mathrm{\pi }}{4}}}{\cos {\displaystyle \frac{\mathrm{\pi }}{4}}-\sin {\displaystyle \frac{\mathrm{\pi }}{4}}}$$=\frac{{\displaystyle \frac{1}{\sqrt{2}}}+{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}-{\displaystyle \frac{1}{\sqrt{2}}}}$$=\frac{\sqrt{2}}{0}=\infty$

Let the angle made by the tangent to the given curve at t = $\frac{\mathrm{\pi }}{4}$ with the x-axis be θ.

∴ tanθ = Slope of tangent to the given curve at $t=\frac{\mathrm{\pi }}{4}$ = $\infty$

$\Rightarrow \tan \theta =\tan \frac{\mathrm{\pi }}{2}$

$\Rightarrow \theta =\frac{\mathrm{\pi }}{2}$

Thus, the tangent to the curve x = e^t cost, y = e^t sint at t = $\frac{\mathrm{\pi }}{4}$ makes with x-axis an angle $\frac{\mathrm{\pi }}{2}$.

Hence, the correct answer is option (d).