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
[imath]x(t)=\frac{4cos(2t)}{e^t},y(t)=\frac{2sin(2t)}{e^t},[/imath]
and i need find the turning points. which i understand comes with
[imath]\frac{dy}{dx}=0[/imath], which means [imath]dy = 0[/imath].

after differentiating i get [imath]y'=2\frac{2cos(2t)-sin(2t)}{e^t}[/imath], so [imath]2cos(2t)-sin(2t)[/imath]must be 0. which isn't obvious to me how to solve. with [imath]y''[/imath] 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 [imath]\frac{dy}{dx}=0[/imath] or [imath]\frac{dx}{dy}=0[/imath], which seems wrong to me. can someone clarify?

many thanks
stan

 

[Dr.Peterson]

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

after differentiating i get [imath]y'=2\frac{2cos(2t)-sin(2t)}{e^t}[/imath], so [imath]2cos(2t)-sin(2t)[/imath]must be 0. which isn't obvious to me how to solve. with [imath]y''[/imath] 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 [imath]\frac{dy}{dx}=0[/imath] or [imath]\frac{dx}{dy}=0[/imath], 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 [imath]2\cos(2t)-\sin(2t)=0[/imath], you might try dividing by [imath]\cos(2t)[/imath].