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Simple differentiation question

  • Thread starter DeanBH
  • Start date
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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?
 

Answers and Replies

Dick
Science Advisor
Homework Helper
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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?
 
454
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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?
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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