For the parametric curve whose equation is $x=4\cosθ$ and $y=4\sinθ$, what would be the concavity of curve at $θ = π/4$?
What I'm actually confused is that I found the slope to be $-1$ which is correct. But when for finding the concavity I take the double derivative of the given function it gives the answer $\tan π/4$ which is equal to $1$ so it would be concave up.
But when I checked in the solution manual, their answer is $-\frac{1}{\sqrt 2}$ and they're calling it concave up which makes no sense to me. Could anyone explain if I'm doing it the right way or I should follow the manual?
Your derivative part is correct. $$\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=-\cot\theta$$
I think you would have messed up with second derivative, which is as follows:$$\begin{align}\frac{d^2y}{dx^2}=\frac{d(\frac{dy}{dx})}{dx}&=\frac{\frac{d(\frac{dy}{dx})}{d\theta}}{\frac{dx}{d\theta}}\\ &=\frac{\text{cosec}^2\theta}{-4\sin\theta}\\ &=-\frac{1}{4\sin^3\theta}\end{align}$$