## 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!!

1. Michele_Laino

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

2. Michele_Laino

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

3. Michele_Laino

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

4. anonymous

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

5. Michele_Laino

ok!

6. anonymous

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

7. 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?

8. anonymous

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

9. anonymous

Help anyone?

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

11. 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

12. phi

is the question really compute dx / dt for t = pi / 2 ? it would make more sense if they were asking for dz/dt

13. Loser66

I don't know.

14. anonymous

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

15. anonymous

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

16. 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.

17. anonymous

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

18. 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/