# Math Help - Partial derivative. Completely stuck

1. ## Partial derivative. Completely stuck

Evaluate $\frac{dw}{dt}$ at t=2 for $w(x,y,z)=e^{xyz^{2}}$; $x=t$, $y=t$, $z=\frac{1}{t}$

From what I gather, I'm looking for:

$(\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt})e^{xyz^{2}}$

Is this even correct? If so, I've tried to go further and gotten:

$[(yz^{2})(yz^{2})+(xz^{2})(xz^{2})+(2xyz)(\frac{-2xy}{t})]e^{xyz^{2}}$

and then:

$[(y^{2}z^{4})+(x^{2}z^{4})-(\frac{4x^{2}y^{2}}{t})]e^{xyz^{2}}$

then I substitute :

$[(t^{2}(\frac{1}{t})^{4})+(t^{2}(\frac{1}{t})^{4})-(\frac{4t^{2}t^{2}}{t})]e^{(t)(t)(\frac{1}{t})^{2})}$

Then set t=2

$[\frac{4}{16}+\frac{4}{16}-(\frac{64}{2})]e^{1}=\frac{-63}{2}e$

I know that isn't the correct answer, so what am I doing wrong? Thanks.

2. Is the result by any chance 0?

3. I'm not sure. The question is for an exam review and the professor hasn't put the answers up yet. He was supposed to on Saturday night. Anyway, there are 4 answer choices.

a. e
b. 1
c. 0
d. -e

I can see how you would come to the answer 0 if, at the beginning of the problem, you substituted all of the variables for their corresponding value of t. Then you'd have:

$\frac{d}{dt}e^{(t)(t)(\frac{1}{t^{2}}}}=\frac{d}{d t}e^0=0$

But I wanna know how to do it without substituting the t values in.

4. Originally Posted by downthesun01
I'm not sure. The question is for an exam review and the professor hasn't put the answers up yet. He was supposed to on Saturday night. Anyway, there are 4 answer choices.

a. e
b. 1
c. 0
d. -e

I can see how you would come to the answer 0 if, at the beginning of the problem, you substituted all of the variables for their corresponding value of t. Then you'd have:

$\frac{d}{dt}e^{(t)(t)(\frac{1}{t^{2}}}}=\frac{d}{d t}e^0=0$

But I wanna know how to do it without substituting the t values in.
Ok you said:
$(\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt})e^{xyz^{2}}$

$yz^2e^{xyz^2}\cdot 1 + xz^2e^{xyz^2}\cdot 1+2xyze^{xyz^2}\cdot (-\frac{1}{t^2})=$
$=t\cdot \frac{1}{t^2}e^1+t\cdot \frac{1}{t^2}e^1-2t\cdot t\cdot \frac{1}{t}e^1\cdot\frac{1}{t^2}$
$=\frac{e}{t}+\frac{e}{t}-2\frac{e}{t}=0$

5. Originally Posted by downthesun01
Evaluate $\frac{dw}{dt}$ at t=2 for $w(x,y,z)=e^{xyz^{2}}$; $x=t$, $y=t$, $z=\frac{1}{t}$

From what I gather, I'm looking for:

$(\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt})e^{xyz^{2}}$

Is this even correct?

No, it's not. You should not have the " $e^{xyz^2}$" multiplied at the end. You should just have the general formula for the chain rule, $\frac{\partial w}{\partial x}\frac{dx}{dt}+\frac{\partial w}{\partial y}\frac{dy}{dt}+\frac{\partial w}{\partial z}\frac{dz}{dt}$
and take $w= e^{xyz^2}$.

$\frac{\partial w}{partial x}= yz^2 e^{xyz^2}$ and $\frac{dx}{dt}= 1$

$\frac{\partial w}{partial y}= xz^2 e^{xyz^2}$ and $\frac{dy}{dt}= 1$

$\frac{\partial w}{partial z}= 2xyz e^{xyz^2}$ and $\frac{dx}{dt}= -\frac{1}{t^2}$

So $\frac{dw}{dt}= yz^2e^{xyz^2}+ xz^2e^{xyz^2}- \frac{2xyz e^{xyz^2}}{t^2}$

At t= 2, x= t= 2, y= t= 2, $z= \frac{1}{2}$ so that is
$\frac{dw}{dt}(3)= 2(1/4)e^{2(2)/4}+ 2(1/4)e^{2(2)/4}- \frac{2(2)(2)(1/4)e^{2(2)(1/4)}}{4}$
$\frac{\frac{1}{2}e+ \frac{1}{2}e- \frac{2e}{4}= \frac{1}{2}e$

[/quote] If so, I've tried to go further and gotten:

$[(yz^{2})(yz^{2})+(xz^{2})(xz^{2})+(2xyz)(\frac{-2xy}{t})]e^{xyz^{2}}$[/quote]
Because of the additional $e^{xyz^2}$, you are getting each derivative of w twice.

and then:

$[(y^{2}z^{4})+(x^{2}z^{4})-(\frac{4x^{2}y^{2}}{t})]e^{xyz^{2}}$

then I substitute :

$[(t^{2}(\frac{1}{t})^{4})+(t^{2}(\frac{1}{t})^{4})-(\frac{4t^{2}t^{2}}{t})]e^{(t)(t)(\frac{1}{t})^{2})}$

Then set t=2

$[\frac{4}{16}+\frac{4}{16}-(\frac{64}{2})]e^{1}=\frac{-63}{2}e$

I know that isn't the correct answer, so what am I doing wrong? Thanks.

6. Thanks. I just reworked the problem and got the answer that you posted.