Question

The motion of a spring that is subject to dampening (such as a car's shock absorber)is often modeled by the product of an exponential function and a sine or cosine function. Suppose the equation of motion for a point on such a spring is


s(t)=3∗e−2tsin(3ðt)
where t is given in seconds.

a. Find the velocity of the point after t seconds.
I found the derivative for a to be e^(-2t)(9picos(3pit)-6sin(3pit))
for part b I have to find t when the velocity=0. I keep getting .2122 on my ti 89 but I know that's not the right answer. Can someone at least show me by hand how to solve?

Answer 1

s(t) = 3e^-2t sin 3πt
s'(t) = e^-2t (9π cos 3πt - 6sin 3πt)

so far, so good

when s'(t) = 0, we have

3π cos 3πt - 2sin 3πt = 0
t = 0.1445

If you visit
http://rechneronline.de/function-graphs/

and enter

exp(-2x) * sin(3*pi*x)

for your function, you can see that's where it changes direction. You can also plot the derivative, to verify that.

wolframalpha.com gives the solution as

0.21221(nπ + 0.68085)

Better see whether your TI calculator is also tacking on some extra factors.