# The position of a particle moving in a straight line is given by s(t) = (e^(-t))(cos(5t)) for t>0, where t is in seconds. If the particle changes direction at time T seconds, then T must satisfy the equation:

a. tan(5T) = 4.8
b. cos(5T) = 0
c. e^(-T) = [cos(5t)+5sin(5T)] / [-5sin(5T)+25cos(5T)]
d. tan(5T) = 2.4
e. e^(-T) = [5sin(5T)-25cos(5T)] / [cos(5T)+5sin(5T)]

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1. Just to give an update to this problem, the correct answer is actually d. This is because this problem is asking for the relative maxima of the SPEED function, not of the position function. Therefore, you would need to find the second derivative of the position function, e^-t*(10sin(5t)-24cos(5t)) and set that equal to zero, which is where 10sin(5t)-24cos(5t)=0. From it, 10sin(5t)=24cos(5t) comes out, which simplifies to sin(5t)/cos(5t)=24/10 or tan(5t)=2.4

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2. The x-coordinate is steadily decreasing. See the graph at
https://www.wolframalpha.com/input/?i=plot+x%3De%5E%28-t%29%2Cy%3Dcos%285t%29
so if the particle changes direction, we must have dy/dt = 0
that is -5sin(5t) = 0
so, what do you think?

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3. Would the answer be B (where it changes direction)?

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4. I can not make any of the answer choices work
I agree that clearly sin 5T = 0
that means 5 T = 0 or pi or 2 pi or onward
The cosine of 5 T is 1 or -1 there and the tangent is zero

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5. Ah. I see I have misread the problem. Looking at the choices, I figure you must have been sloppy typing your fraction, and meant
s(t) = e^(-t)/cos(5t)
now we want ds/dt = 0
ds/dt = e^(-t)(5tan5t - 1)/cos5t
that means that tan5t = 1/5

But again, that is not one of the choices.

How about s(t) = e^(-t)cos(5t)?
Nope. In that case,
ds/dt = -e^-t (5sin5t+cos5t)
and we need tan5t = -1/5

So, what is the real equation for s(t)?

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6. Oh, I guess it it the product, not x and y components
s = e^-t cos 5 t
ds/dt = - 5e^-t sin 5 t -e^-t cos 5 t = 0 for our change of direction
+5 sin 5 t = -cos 5 t
tan 5 t = -1/5
still no luck

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7. agggghhhh

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