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November 22, 2014

November 22, 2014

Posted by **Anonymous** on Thursday, March 29, 2012 at 6:20pm.

cos(5T) = 0

5T = arctan(-1/5)

5e^(-t) sin(5t) = 0

tan(5T) = -1/5

cos(5T) = 5

I know that a change in direction will be marked by a change from positive to negative or vice versa, but I don't understand the equations the question gives me. Could someone please talk me through this process to find the right answer?

- Calculus (Related Rates) -
**Steve**, Friday, March 30, 2012 at 10:29amyou are correct, as far as you go. When it changes from pos to neg, it will be zero.

Note that the particle changes direction, not position. So, its velocity changes sign. The velocity is given by the derivative.

s = e^-t cos5t

s' = e^-t (-cos5t - 5sin5t)

so, since e^-t is always positive, we need

-cos5t - 5sin5t = 0

tan 5t = -1/5

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