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2. calculus

7 answers

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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   answered by Dan

   Mar 7, 2022

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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   oobleck

   Jun 3, 2021

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

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   answered by lovejoy

   Jun 3, 2021

5. 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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   answered by Anonymous

   Jun 3, 2021

6. 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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   oobleck

   Jun 3, 2021

7. 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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   answered by Anonymous

   Jun 3, 2021

8. agggghhhh

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   answered by Anonymous

   Jun 3, 2021

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