Differentiating with multiple variables

In summary, the homework statement is to find dw/dt. Check result by substitution and differentiation.
  • #1
mattb8818
6
0

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
 
Last edited:
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  • #2
One way would be to substitute for x and y, which would result in a function of t alone.
 
  • #3
Mathdope said:
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.
 
  • #4
mattb8818 said:
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:
 
  • #5
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
 
  • #6
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
 
  • #7
Hi matt!

I don't know what you did to get 2(s)^1/2(sinh4t)/(cosh4t)^(1/2). :confused:
mattb8818 said:
(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]​
 
  • #8
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.
 

1. What is the purpose of differentiating with multiple variables?

Differentiating with multiple variables allows us to analyze the rate of change of a function with respect to more than one independent variable. This is useful in many areas of science, such as physics and economics, where multiple variables are often involved in complex systems.

2. How is differentiating with multiple variables different from single variable differentiation?

In single variable differentiation, we are finding the rate of change of a function with respect to one independent variable. In multiple variable differentiation, we are finding the rate of change of a function with respect to two or more independent variables. This requires the use of partial derivatives and can be more complex than single variable differentiation.

3. What is the process for differentiating with multiple variables?

The process for differentiating with multiple variables involves finding the partial derivatives of the function with respect to each independent variable, and then combining them using the appropriate rules of differentiation. This can be done using the product rule, quotient rule, and chain rule, as well as the basic rules of differentiation.

4. Why is it important to consider multiple variables when differentiating?

In many real-world situations, there are multiple factors that can affect a particular outcome. By considering multiple variables, we can better understand the relationship between these factors and the overall behavior of the system. This can help us make more accurate predictions and inform decision-making.

5. Are there any applications of differentiating with multiple variables?

Yes, there are many applications of differentiating with multiple variables in various fields such as engineering, economics, and physics. For example, in physics, differentiating with respect to time can help us analyze the velocity and acceleration of an object in motion. In economics, differentiating with respect to multiple variables can help us understand the relationship between supply and demand in a market.

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