# Thread: Total Differential

1. ## Total Differential

Struggling with this textbook question:

"The equations

x+2y+3z+4u=-3

x2+y2+z2+u2=10

x3+y3+ z3+u3=0

define u as a function of y if x and z are eliminated. Find du/dy when x=1, y=-1, z=2, u=-2"

The problem I am having is why they give three equations? There must be a reason and I am missing it.

Is anyone able to explain just what the question is asking for?

Thanks
Laura

2. ## Re: Total Differential

Hey lauramorrison93.

Basically since you have three equations and four unknowns, you will be able to reduce your system to one un-known (or one as a function of the other).

Its like the situation in linear algebra where you reduce a matrix to get the simplest solution but in your case you have non-linear functions of the variables.

An example is say

x + y = 1
2x + y = 2

We let y = 2 - 2x and sub into 1 giving

x + 2 - 2x = 1 which gives solution x = 1

Its exactly the same thing except you have non linear terms.

3. ## Re: Total Differential

Is the question definitely asking for this?

I attempted to do this and I can eliminate x but not z because I get z
and z terms in the equation where x is eliminated.

Also, why do they give values for x, y, z and u? What do I use them for?

4. ## Re: Total Differential

It is asking for you to do the same thing except you will get one function in terms of another variable.

Think about the example where you have two equations in three unknowns. Something like this:

2x + y + z = 3
3x + 2y - 5z = 4

If you solved that system you would get a lot of solutions (in terms of one free variable).

The same thing will happen in your system except that it is non-linear and that you have three equations in four knowns.

So instead of having something like x = ay + b you will have u as a function of y where u(y) = some function of y.

The ideas are the same whether they are linear or non-linear: you are cancelling out variables by substituting in one equation to another.