Derivative of projectile parametric y-component?

  • #1
Matt Jacques
81
0
Hi,

I'm looking for the derivative of the projectile parametric y-component?

The y component is:

y = (vi/k)(1-e^-kt)(sin a) + (g/k^2)(1 - kt - e^-kt)

I seem to be doing something wrong and my derivative isnt working out, I just want to check it against the final answer to see where I am going wrong.

Thanks

Matt
 

Answers and Replies

  • #2
Integral
Staff Emeritus
Science Advisor
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I get
dy/dt = Vie-ktsin(a)+g(e-kt-1)/k
 
  • #3
Matt Jacques
81
0
Yep, that's it. Thanks, Integral.

I forgot d/dx [ e^u ] is e^u * du/dx

Ah, the devil is in the details! Now to step two, how to solve it? I logged both sides, used the power rule and everything else I could think of, but it is not working out. Any other suggestions beside logging?

[Edited for spelling mistake]
 
  • #4
Matt Jacques
81
0
No ideas? :(
 
  • #5
HallsofIvy
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Homework Helper
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What do you mean by "solve it"?? Presumably you don't mean "How did you get that answer" because you said "Yep, that's it". What exactly do you want to do with the derivative?
 
  • #6
Matt Jacques
81
0
Ooops, sorry.

Im settng the derivative to zero and solving for t.
 
  • #7
futz
80
0
The first term contains an exp(-kt). Expand out the second term into two seperate terms, one of which will contain the same exp(-kt) factor. Combine terms containing the exponential, move everything else over to the other side and once the exponential is isolated, then take the natural log of both sides.

I got something like

t = (-1/k)*ln[(g/k)*(Vi*sin(a)+g/k)^(-1)]
 
Last edited:
  • #8
Matt Jacques
81
0
Thanks, Futz. I was combining the exponentials wrong.

Here it is solved for t:

http://homepage.mac.com/jjacques2/maxheight.jpg [Broken]
 
Last edited by a moderator:

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