if $z=xe^{xy}$ and $x=3st$ and $y=t^2-s$ find $\large \frac{\partial z}{\partial s}$ at $(3,0)$

Question

jhannybean · Answer

I'm just unsure where to plug in the point $(3,0)$

jhannybean · Answer

@hartnn @iambatman

xapproachesinfinity · Answer

this asking to find slope at the point s=3!
what's the problem here?

jhannybean · Answer

$\large \frac{\partial z}{\partial s}= \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} +\frac{\partial z}{\partial y} \frac{\partial y}{\partial s}$ 

$\large\frac{\partial z}{\partial s}=(e^{xy}+xye^{xy})(3t) + (x^2e^{xy})(-1)$

from here... do I plug in the point (3,0) into x and y? or what do I do after this?

xapproachesinfinity · Answer

did you find the partial derivative

jhannybean · Answer

Yep :)

xapproachesinfinity · Answer

well (3,0) is related to s
so you can't plug it in for x and y the question is  asking the derivative of z at that point
when s=3
I don't see any s in your partial derivative

amistre64 · Answer

x=3st   and   y=t^2−s

use the point to determine s and t if needed

3 = 3st
0 = t^2-s

zepdrix · Answer

Yah you want to think of z as a function of x and y.
x=3, y=0.
I can certainly see the confusion though :) Since z is a function of s and t, not directly though.

jhannybean · Answer

Ah that way... so 

3 = 3st                        0 = t^2 - s
st = 1                          0 = (1/s)^2 - s 
t = 1/s ?                      0 = 1/(s^2) - s
                                   0  = s^-2 - s
                                 
something along this path? :\

zepdrix · Answer

Hmmm that seems like a longer way to go about it...$$\Large\rm 3=3st$$$$\Large\rm 0=t^2-s\qquad\implies\qquad s=t^2$$$$\Large\rm 3=3(t^2)t\qquad \implies\qquad t^3=1$$Should work out nicer that way, yes? :o

amistre64 · Answer

yeah, or s=t=1 seems to work

jhannybean · Answer

I was embarrassed to say I forgot how to factor out the s in my method <.< and yeah, the alternate method does work better.

jhannybean · Answer

Ohh, I just remembered...

1/s^2 - s = 0

(1-s^3)/s^2 = 0

1-s^3 = 0 

s^3 = 1

s = 1

Thank you guys so much! Haha.

amistre64 · Answer

yw

zepdrix · Answer

Ah there's the factoring c: hehe

jhannybean · Answer

It's not even the calculus that kills me in these problems, it's mostly the factoring...

xapproachesinfinity · Answer

I totally miss understood this!
thanks to this guys i recaptured what's happening!

xapproachesinfinity · Answer

these*