Finding turning points in a curve.

[Stani]

Hi, sorry if some wordings are not right, english is not my native language.

so i am given the following curve
$x(t)=\frac{4cos(2t)}{e^t},y(t)=\frac{2sin(2t)}{e^t},$
and i need find the turning points. which i understand comes with
$\frac{dy}{dx}=0$, which means $dy = 0$.

after differentiating i get $y'=2\frac{2cos(2t)-sin(2t)}{e^t}$, so $2cos(2t)-sin(2t)$must be 0. which isn't obvious to me how to solve. with $y''$ it gets even worse, as i need to distinguish between maximum and minimum.

am i missing something here?

also, i have already seeked help on another community and i got one answer that states turning points comes with both $\frac{dy}{dx}=0$ or $\frac{dx}{dy}=0$, which seems wrong to me. can someone clarify?

many thanks
stan

 

[Dr.Peterson]

so i am given the following curve
$x(t)=\frac{4cos(2t)}{e^t},y(t)=\frac{2sin(2t)}{e^t},$
and i need find the turning points. which i understand comes with
$\frac{dy}{dx}=0$, which means $dy = 0$.

after differentiating i get $y'=2\frac{2cos(2t)-sin(2t)}{e^t}$, so $2cos(2t)-sin(2t)$must be 0. which isn't obvious to me how to solve. with $y''$ it gets even worse, as i need to distinguish between maximum and minimum.

am i missing something here?

also, i have already seeked help on another community and i got one answer that states turning points comes with both $\frac{dy}{dx}=0$ or $\frac{dx}{dy}=0$, which seems wrong to me. can someone clarify?

many thanks
stan

First, what definition have you been given for "turning point"? Does it include dx/dy = 0? What would the curve look like at such a point?

Second, to solve $2\cos(2t)-\sin(2t)=0$, you might try dividing by $\cos(2t)$.