Question

A curve is represented parametrically by equation $x=e^t\cos t,y=e^{t}sint$ where ${}^{\prime }{t}^{\prime }$ is parameter. If $\tan \alpha$ be the slope of any tangent to the curve then value of 't' can be

• A. $\alpha +\frac{3\pi }{4}$
• B. $\alpha -\frac{\pi }{4}$
• C. $\frac{\pi }{4}+\alpha$
• D. $\frac{3\pi }{4}-\alpha$

Answer 1

The correct option is B $\alpha -\frac{\pi }{4}$

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

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

$\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$=\frac{e^t(\cos t+\sin t)}{e^t(\cos t-\sin t)}$

$=\frac{\cos t+\sin t}{\cos t-\sin t}$ by dividing both numerators and denominators by $\cos t$

$=\frac{1+\frac{\sin t}{\cos t}}{1-\frac{\sin t}{\cos t}}$

$=\frac{1+\tan t}{1-\tan t}$

$\Rightarrow \tan \alpha =\frac{\tan \frac{\pi }{4}+\tan t}{1-\tan \frac{\pi }{4}\tan t}$ since $\tan \frac{\pi }{4}=1$

$\Rightarrow \tan \alpha =\tan (\frac{\pi }{4}+t)$

$\therefore \alpha =\frac{\pi }{4}+t$

$\therefore t=\alpha -\frac{\pi }{4}$