THIS IS MY EFFORT:

I just wanted to know if there are mistakes in my work...

#1

http://i104.photobucket.com/albums/m...emike1/pd1.jpg

#2

http://i104.photobucket.com/albums/m...emike1/pd2.jpg

THANKS FOR THOSE WHO WILL HELP ^_^

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- Oct 7th 2010, 01:44 AMcutiemike1Chain Rule & Substitution in Partial Derivatives
THIS IS MY EFFORT:

I just wanted to know if there are mistakes in my work...

#1

http://i104.photobucket.com/albums/m...emike1/pd1.jpg

#2

http://i104.photobucket.com/albums/m...emike1/pd2.jpg

THANKS FOR THOSE WHO WILL HELP ^_^ - Oct 7th 2010, 04:26 AMHallsofIvy
Those are correct. However, I notice that in using chain rule for the second problem, you have

$\displaystyle \frac{\partial u}{\partial t}= 2e^{\frac{y}{x}}sec^2 t$

while using direct substitution, you get

$\displaystyle \frac{\partial u}{\partial t}= 2e^{2tan t}sec^2 t$

Those are, of course, the same because $\displaystyle x= 2r cos t$ and $\displaystyle y= 4r cos t$ so that $\displaystyle \frac{y}{x}= \frac{4r cos t}{2r cos t}= 2 tan t$

but the second, with only the variable t is the better expression. - Oct 7th 2010, 04:39 AMcutiemike1
oh yeah!! since it is

*with respect to t*..thanks a lot for that one...

however, i am just wondering in my number 1 answer...try to take a look in Chain Rule the, are the derivatives under it are correct??**y=re^-s** - Oct 7th 2010, 04:40 AMcutiemike1
oh yeah!! since it is

*with respect to t*..thanks a lot for that one...

however, i am just wondering in my number 1 answer...try to take a look in Chain Rule ung y=re^-s, are the derivatives under it are correct??