Differentiating with multiple variables (1 Viewer)

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1. The problem statement, all variables and given/known data
Find dw/dt. Check result by substitution and differentiation

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


2. Relevant equations



3. 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
 
Last edited:
One way would be to substitute for x and y, which would result in a function of t alone.
 

Dick

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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

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dx/dw = x/(x^2 + y^2)^1/2 dy/dw = y/(x^2 + y^2)^1/2
Hi matt! :smile:

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? :smile:
 
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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
 
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 dont really understand this solution
 

tiny-tim

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Hi matt!

I don't know what you did to get 2(s)^1/2(sinh4t)/(cosh4t)^(1/2). :confused:
(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? :smile:
[size=-2](btw, if you type alt-v, it prints √ )[/size]​
 
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 Im happy.
 

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