Finding Turning Points of y=e^{-t}sin(2t)

[mlazos]

Homework Statement

I have this function
$$y=e^{-t}sin(2t)$$

Homework Equations

$$\frac{dy}{dt}=e^{-t}(-sin(2t)+2cos(2t))$$
and I know that we can find from this the turning points. Then we substitute to the second derivative to know if they are positive or negative and thus make them min or max. The substitution is not easy though
$$\frac{d^2y}{dt^2}=-4 e^{-t} Cos(2t)-3e^{-t}Sin(2t)$$

The Attempt at a Solution

I try to find the turning points but honestly I am little lost since the second derivative is little complicated. So any easy way to understand which one is min and which is maximum?
I know i have to find the stationary points from the first derivative and to substitute to the second but this substitution is little complicated. Any ideas are welcome.

 

[bob1182006]

well to find the turning points you find where the first derivative = 0 right? where would that be? when will e^-t =0? or when will -sin2t+2cos2t=0?

also your notation is wrong, you took the derivative w.r.t. time on the RHS but on the LHS you have dy/dx should be dy/dt.

 

[mlazos]

i didnt notice that, sorry, yes i mean dt not dx

 

[mlazos]

I know that the exponential is never zero i can fine the turning points its just the substitution to the second derivative that creates problem.
i know that $$t=\frac{arctan2}{2}+k\pi$$ but the substitution is the problem. I know how to solve it, I am not sure about the substitution to the second derivative.

 

[bob1182006]

well what would be some numbers from -sin2t+2cos2t=0?
yes you could find EVERY point which is that formula you have but try finding 1-2 first and then 2 more and see if it's repeating since sine/cosine oscillate they might have only a local max/min at t=(arctan2)/2+kpi

 

[mlazos]

Thats a good idea, thanks. I did what you said and i got an idea of the function. even k gives second derivative negative so is min and odd k gives positive second derivative so is maximum (local of course). Thank for the help