Question

[CALCULUS III—DIFFERENTIATION] Given z(x, y), express ∂z/∂t if given x(s, t) and y(s, t); figure inside

Answer 1

\[z=x\ sin(y/x)\]\[x=e^{s+t},\ y=e^{s-t}\]
express ∂z/∂t using the chain rule.

Answer 2

http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx

Answer 3

ok, this doesn't look too bad.

Answer 4

hmmm...can't we just find dz/dx and dx/dt without needing dy/dt?

Answer 5

Partials, that is.

Answer 6

I am confused, I understand it when it is dz/dt but I feel ∂z/∂t requires different consideration.

Answer 7

Partial derivative chain rule. \[
\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} +\frac{\partial z}{\partial y} \frac{\partial y}{\partial t}
\]

Answer 8

I recall the ∂z/∂x and ∂z/∂y from earlier.

So I need to find:

∂x/∂t and ∂y/∂t

Answer 9

lemme rephrase that. yes....you need all partials...my mistake.

Answer 10

Does ∂x/∂t = dx/dt in this situation?

Answer 11

It's a partial derivative in this case because x also varies with t.

Answer 12

With s, sorry

Answer 13

A wolfram inquiry provides that dx/dt = ∂x/∂t.

I am not saying this applies to all cases, but when referring to e^(s+t) ∂t = dt

Answer 14

and ∂t for y is -e^(s-t)

Answer 15

It's a partial derivative. That means you treat \(s\) as a constant and differentiate with respect to \(t\).

Answer 16

I understand partial derivatives, that is how I was able to evaluate it and make the observation dt = ∂t. Just verifying the evaluation with yall

Answer 17

I am going to apply the evaluations to @wio 's form for ∂z/∂t

Answer 18

\[∂z/∂t=\frac{-y\ cos(y/x)+x\ sin(y/x)}{x}e^{s+t}+-e^{s-t}cos(y/x)\]

Can anybody proficient with wolfram verify this via a wolfram inquiry? @wio

Answer 19

I really need a new laptop...it's acting funny...and slow. But I just had it cleaned and debugged. Too many programs makig it run too slow.

Answer 20

It's too complex for wolfram.

Answer 21

http://www.wolframalpha.com/input/?i=d%2Fdt+e%5E%28s%2Bt%29+sin%28e%5E%28s%2Bt%29%2Fe%5E%28s-t%29%29

Answer 22

Check each of the partials independently. If you multiply and add without extra simplyfying there shouldn't be error.

Answer 23

I got the result on paper that you have there, stamp. But that doesn't mean much...lol

Answer 24

I think I am going to steal my wife's laptop..it's much faster.

Answer 25

Here's what I got. I believe it is the same thing as yours....can't tell because my laptop has major issues.