Question

If z = e ^ (xy ^ 2), x = tcost, and y = tsint
compute dz / dt for t = pi / 2

Thank you have a good day mates!!

Answer 1

hint:
we can write this:
\[\Large \begin{gathered}
\ln z = {\left( {xy} \right)^2} = {\left\{ {{t^2}\frac{{\sin \left( {2t} \right)}}{2}} \right\}^2} \hfill \\
\hfill \\
\ln z = \frac{{{t^4}{{\left\{ {\sin \left( {2t} \right)} \right\}}^2}}}{4} \hfill \\
\end{gathered} \]

Answer 2

now, please compute the first derivative of both sides with respect to t

Answer 3

alternatively, you can use this formula:
\[\Large \frac{{dz}}{{dt}} = \frac{{\partial z}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial z}}{{\partial y}}\frac{{dy}}{{dt}}\]

Answer 4

Okay sir will update you after my attempt to answer this question

Answer 5

ok!

Answer 6

\[z=e^{xy^2},\ x=t \cos t, \ y=t \sin t\\ since\ \frac{ dx }{ dt }= -t \sin t+\cos t\\ and\ \frac{ dy }{dt }=t \cos t+\sin t\\ find \ \frac{ dz }{ dt }=\]

Answer 7

\[\frac{ 1 }{4 }(t^4\sin^2(2t)) \]
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?

Answer 8

I think I'm not doing it right. I'm kind of lost right now. Sorryyy

Answer 9

Help anyone?

Answer 10

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
\(\dfrac{dx}{dt}= \dfrac{d}{dt}(t)*cost +t*\dfrac{d}{dt}(cos t)= cos t-tsint\)

Answer 11

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

Answer 12

is the question really
compute dx / dt for t = pi / 2
?

it would make more sense if they were asking for dz/dt

Answer 13

I don't know.

Answer 14

It is dz/dt my bad!!! I'll change it now. Sorrrrry

Answer 15

^ I thought that is a product rule sir or what? Sorry for my ignorance I'm kinda new in differential calc

Answer 16

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.

Answer 17

Oh okay. I got it from there! I understand why hmm
so what's next about it?

Answer 18

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/