Find the points on the curve where the tangent is horizontal or vertical: $x= e^{\sin(t)}$ ;$ y= e^{\cos(t)}$

Question

What I understand is that where the t-value of $\frac{dy}{dt}$ and $\frac{dx}{dt}$ is equal to $0$, you can use that $t$ to determine the $x$ and $y$ coordinates of the point. What I got for my vertical tangent was $(e,1)$ and horizontal $(1,e)$ using a t-value of $t=1$ for vertical and $t=0$ for horizontal. This is wrong though.

Answer 1

For horizontal tangents you want $$ \frac {dy}{dx}=0 $$and for vertical tangent lines you want $$ \frac {dx}{dy}=0$$

Using chain rule we get $$ \frac {dx}{dy} =\frac {dx/dt}{dy/dt} \\ \frac {dy}{dx} =\frac {dy/dt}{dx/dt}$$

Thus for horizontal tangent line we need $$\frac {dy}{dt}=0$$

Thus $$ -\sin te^{\cos t} =0 $$ which implies $t=0$ or $t=\pi$ The points are therefore, $(1,e)$ and $(1, 1/e)$

Similarly we find the vertical tangency happening at $(e,1)$ and $(1/e , 1).$