Differentiating with multiple variables

[mattb8818]

Homework Statement

Find dw/dt. Check result by substitution and differentiation

w = (x^2 + y^2)^1/2, x = e^2t , y = e^-2t

Homework Equations

The Attempt at a Solution

dx/dw = x/(x^2 + y^2)^1/2 dy/dw = y/(x^2 + y^2)^1/2

Dont really know where to go with it

 

[Mathdope]

One way would be to substitute for x and y, which would result in a function of t alone.

 

[Dick]

One way would be to substitute for x and y, which would result in a function of t alone.

That's the way they suggest to check the answer. They actually do want you to use partial derivatives, mattb8818. Use the chain rule.

 

[tiny-tim]

dx/dw = x/(x^2 + y^2)^1/2 dy/dw = y/(x^2 + y^2)^1/2

Hi matt!

First, it's not dx/dw and dy/dw, it's t'other way up!:

dw/dx = x/(x^2 + y^2)^1/2 dw/dy = y/(x^2 + y^2)^1/2

Does that help?

 

[b0it0i]

since w is a function of x and y
and x and y are functions of t
the chain rule is

dw/dt = dw/dx dx/dt + dw/dy dy/dt

compute everything above and you're set
an easy way to remember the chain rule is to draw a tree diagram

w
x y
t t

 

[mattb8818]

Thanks for the replies

I did it with the chain rule and got

(2xe^2t)/(x^2 + y^2)^1/2 + -(2e^-2t)/(x^2 + y^2)^1/2

The answer is 2(s)^1/2(sinh4t)/(cosh4t)^(1/2)

I don't really understand this solution

 

[tiny-tim]

Hi matt!

I don't know what you did to get 2(s)^1/2(sinh4t)/(cosh4t)^(1/2).

(2xe^2t)/(x^2 + y^2)^1/2 + -(2e^-2t)/(x^2 + y^2)^1/2

Start from that line (which is correct, except you missed out a "y");
then re-write it as:
[(2xe^2t) - (2ye^-2t)]/√(x^2 + y^2)
(this is both to simplify it, and to lessen the risk of making a mistake)
and just substitute for x and y …

so what is the next line?

[size=-2](btw, if you type alt-v, it prints √ )[/size]​

 

[mattb8818]

Ok thanks all

I am glad I atleast did it right. I looked it up and e^x-e^-x/2 = sinh or something like that, but I guess that was my only mistake so I am happy.