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anonymous
 one year ago
If z = e ^ (xy ^ 2), x = tcost, and y = tsint
compute dz / dt for t = pi / 2
Thank you have a good day mates!!
anonymous
 one year ago
If z = e ^ (xy ^ 2), x = tcost, and y = tsint compute dz / dt for t = pi / 2 Thank you have a good day mates!!

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4hint: 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
 one year ago
Best ResponseYou've already chosen the best response.4now, please compute the first derivative of both sides with respect to t

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4alternatively, you can use this formula: \[\Large \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay sir will update you after my attempt to answer this question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[z=e^{xy^2},\ x=t \cos t, \ y=t \sin t\\ since\ \frac{ dx }{ dt }= t \sin t+\cos t\\ and\ \frac{ dy }{dt }=t \cos t+\sin t\\ find \ \frac{ dz }{ dt }=\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 1 }{4 }(t^4\sin^2(2t)) \] 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
 one year ago
Best ResponseYou've already chosen the best response.0I think I'm not doing it right. I'm kind of lost right now. Sorryyy

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0They 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 \(\dfrac{dx}{dt}= \dfrac{d}{dt}(t)*cost +t*\dfrac{d}{dt}(cos t)= cos ttsint\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0and 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
 one year ago
Best ResponseYou've already chosen the best response.0is the question really compute dx / dt for t = pi / 2 ? it would make more sense if they were asking for dz/dt

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is dz/dt my bad!!! I'll change it now. Sorrrrry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0^ I thought that is a product rule sir or what? Sorry for my ignorance I'm kinda new in differential calc

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'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
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay. I got it from there! I understand why hmm so what's next about it?

phi
 one year ago
Best ResponseYou've already chosen the best response.0Here is background (if you have time) on the chain rule http://ocw.mit.edu/courses/mathematics/1802multivariablecalculusfall2007/videolectures/lecture11chainrule/
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