anonymous
  • anonymous
If z = e ^ (xy ^ 2), x = tcost, and y = tsint compute dz / dt for t = pi / 2 Thank you have a good day mates!!
Mathematics
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SOLVED
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
hint: we can write this: \[\Large \begin{gathered} \ln z = {\left( {xy} \right)^2} = {\left\{ {{t^2}\frac{{\sin \left( {2t} \right)}}{2}} \right\}^2} \hfill \\ \hfill \\ \ln z = \frac{{{t^4}{{\left\{ {\sin \left( {2t} \right)} \right\}}^2}}}{4} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
now, please compute the first derivative of both sides with respect to t
Michele_Laino
  • Michele_Laino
alternatively, you can use this formula: \[\Large \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\]

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anonymous
  • anonymous
Okay sir will update you after my attempt to answer this question
Michele_Laino
  • Michele_Laino
ok!
anonymous
  • anonymous
\[z=e^{xy^2},\ x=t \cos t, \ y=t \sin t\\ since\ \frac{ dx }{ dt }= -t \sin t+\cos t\\ and\ \frac{ dy }{dt }=t \cos t+\sin t\\ find \ \frac{ dz }{ dt }=\]
anonymous
  • anonymous
\[\frac{ 1 }{4 }(t^4\sin^2(2t)) \] Apply product rule? \[\frac{1}{4}\left(\frac{d}{dt}\left(t^4\right)\sin ^2\left(2t\right)+\frac{d}{dt}\left(\sin ^2\left(2t\right)\right)t^4\right)\] Am I doing it right?
anonymous
  • anonymous
I think I'm not doing it right. I'm kind of lost right now. Sorryyy
anonymous
  • anonymous
Help anyone?
Loser66
  • Loser66
They give you x = t cost is it not that if you find dx/dt, you just take derivative of this?? Apply product rule, you get \(\dfrac{dx}{dt}= \dfrac{d}{dt}(t)*cost +t*\dfrac{d}{dt}(cos t)= cos t-tsint\)
Loser66
  • Loser66
and then for t = pi/2 , it is = -pi/2?? I don't know why @Michele_Laino mess around with z while the question is just dx/dt
phi
  • phi
is the question really compute dx / dt for t = pi / 2 ? it would make more sense if they were asking for dz/dt
Loser66
  • Loser66
I don't know.
anonymous
  • anonymous
It is dz/dt my bad!!! I'll change it now. Sorrrrry
anonymous
  • anonymous
^ I thought that is a product rule sir or what? Sorry for my ignorance I'm kinda new in differential calc
anonymous
  • anonymous
I'm afraid I'm hard finding it. I thought my product rule answer is going in the right way, I'm sorry I just got lost from there.
anonymous
  • anonymous
Oh okay. I got it from there! I understand why hmm so what's next about it?
phi
  • phi
Here is background (if you have time) on the chain rule http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-11-chain-rule/

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