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Homework Help: Simple differentiation question

  1. Apr 24, 2008 #1
    helicopter moves vertically. Height above start point = y

    height at t seconds =
    y= (1/4)t^4 -26t^2 + 96t t = between 0 and 4
    differentiate to find velocity = t^3 + 52t + 96
    acceleration = 3t^2-52

    thats part 1 of question done, easy.

    Verify that y has a stationary value when t = 2 determine whether it is min/max value.

    i can put this into 3t^2-52t+96 and find out that it is.

    but how is it min/max

    nevermind : i put it into acceleration and find out if acceleration is more or less than the value i got for velocity. If it's less i know that the thing will never accelerate again after this given time. so it would be a maximum. Where as if it was more than the value i got for velocity i know that the object would be able to accelerate again, which is minimum.. right?
  2. jcsd
  3. Apr 24, 2008 #2


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    No. For one thing you've got some scrambled formulas in your post, but assuming they are just typos, then if you did everything correctly you should have found v=0 at t=2. That's what a 'stationary' point means. Now check acceleration at t=2. What sign is it? What does that have to do with it being a min or a max?
  4. Apr 24, 2008 #3
    At a stationary point the velocity is 0, so you can just compare the acceleration with 0.
    if the acceleration is negative, you know there is a local maximum.
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