http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Wed, 09 Jul 2014 21:30:23 -0500http://calc3.askbot.com/question/72/chain-rule/So I've just now been working on 14.4 the chain rule (late, I know). And I'm having some issues with 3 and 4. I understand the chain rule and how it works, and got 1 and 2 with no issues. My issue I'm running into now has to do with it being with respect to s and t now instead of x and y. So the problem states to use the chain rule to solve for $\frac {\partial z} { \partial s}$ and $\frac{\partial z}{\partial t}$ $$z = x^2y \quad x=\sin(st) \quad y= t^2+s^2$$ So I guess here is where I get confused. Do you replace x and y with their respective equations, then take the partial derivatives from there? Or is it some sort of combination of the original z equation plus the derivitavies of x and y separrately. The book gets the answer..$$ 2xyt\cos(st) + 2x2s, \quad 2xys\cos(st) +2x2t$$ So I'm lost, any help would be greatly appreciated! Thanks!Wed, 09 Jul 2014 20:54:50 -0500http://calc3.askbot.com/question/72/chain-rule/http://calc3.askbot.com/question/72/chain-rule/?answer=74#post-id-74It helped me to look at page 365, example 14.4.2. It shows the following: $$\frac{\partial f}{\partial x} = f_x\frac{\partial x}{\partial x} + f_y\frac{\partial y}{\partial x} + f_z\frac{\partial z}{\partial x}$$ So starting with problem 3, I began with the following: $$\frac{\partial z}{\partial s} = z_x \frac{\partial x}{\partial s} + z_y \frac{\partial y}{\partial s}$$ Then I calculated $z_x$, $z_y$, $ \frac{\partial x}{\partial s}$ and $\frac{\partial y}{\partial s}$ as follows: $$z_x = 2yx$$$$z_y = x^2$$ $$ \frac{\partial x}{\partial s} = tcos(st)$$$$\frac{\partial y}{\partial s} = 2s$$ Plugging those values in gives us $$\frac{\partial z}{\partial s} = 2xytcos(st) + 2sx^2$$ Which is what the answer is! Hope this helps in solving the $\frac{\partial z}{\partial t}$Wed, 09 Jul 2014 21:30:23 -0500http://calc3.askbot.com/question/72/chain-rule/?answer=74#post-id-74http://calc3.askbot.com/question/72/chain-rule/?answer=73#post-id-73-Comparing these answers back to the ones in the book, instead of $2s2x$ i have calculated $2sx^2$. I am not sure if I am wrong or if the book is wrong in this instance, but i calculated the partial derivative on a calculator and i got the answer i have provided- The way the problem is state, I believe you simply swap out $x$ and $y$ for their $x(s,t)$ and $y(x,t)$ counterparts thus generating the equation: $$ z = x^2y $$ Substituting here for $x$ and $y$: $$ z = sin^2(st)(t^2 + s^2) $$ Now it should be a simple matter of taking the partial derivatives (with quite a bit of chain rule in here) $$ \frac {\partial z} {\partial t} = 2tsin^2(st) + (t^2 + s^2)(2ssin(st)cost(st)) $$ And then with respect to s $$ \frac {\partial z} {\partial s} = 2ssin^2(st) + (t^2 + s^2)(2tsin(st)cos(st)) $$ When we simplify this back to $x$ and $y$ terms we get the equations: $$ \frac {\partial z} {\partial s} = 2sx^2 + 2txycos(st) $$ and $$ \frac {\partial z} {\partial t} = 2tx^2 + 2sxycos(st) $$ The trick to this question is recognizing that $x$ and $y$ are their own functions of 's' and 't' and plugging this into the equation given to create the new function $z(s, t)$. Hope this helpsWed, 09 Jul 2014 21:13:34 -0500http://calc3.askbot.com/question/72/chain-rule/?answer=73#post-id-73