Question

Determine the open t-intervals on which the curve is concave downward or concave upward. (Enter your answer using interval notation.) x = sin t, y = cos t, 0 < t < π

Answer 1

Solution :

We have been given a parametric curve :

x = sin t , y = cos t , 0 < t < π

In order to determine concavity of the given parametric curve, we need to evaluate its second derivative first.

Therefore,

$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$

$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$

        $=- \tan t$

Taking double derivatives of the above equation:

$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t)$

      $= - \sec^2 t \frac{dt}{dx}$

     $= - \sec^2 t \left(\frac{1}{\cos t}\right)$

    $= - \sec^3 t$

For the concave up, we have

$\frac{d^2y}{dx^2} > 0$

$\Rightarrow - \sec^3 t > 0$

∴ $t \ \epsilon \left( \frac{\pi }{2}, \pi \right)$

For the concave down, we have

$\frac{d^2y}{dx^2} < 0$

$\Rightarrow - \sec^3 t < 0$

$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$