Respuesta :
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,
[tex]$\frac{dx}{dt} = \cos t, \ \frac{dy}{dt}= - \sin t$[/tex]
[tex]$\therefore \frac{dy}{dx}= \frac{- \sin t}{\cos t}$[/tex]
[tex]$=- \tan t$[/tex]
Taking double derivatives of the above equation:
[tex]$\frac{d^2y}{dx^2}= - \frac{d}{dx}(\tan t) $[/tex]
[tex]$= - \sec^2 t \frac{dt}{dx}$[/tex]
[tex]$= - \sec^2 t \left(\frac{1}{\cos t}\right)$[/tex]
[tex]$= - \sec^3 t$[/tex]
For the concave up, we have
[tex]$\frac{d^2y}{dx^2} > 0$[/tex]
[tex]$\Rightarrow - \sec^3 t > 0$[/tex]
∴ [tex]$t \ \epsilon \left( \frac{\pi }{2}, \pi \right)$[/tex]
For the concave down, we have
[tex]$\frac{d^2y}{dx^2} < 0$[/tex]
[tex]$\Rightarrow - \sec^3 t < 0$[/tex]
[tex]$t \ \epsilon \left( 0,\frac{\pi }{2} \right)$[/tex]
