Finding partial derivative with 4 unknowns in 4 equations

[Monsterman222]

I'm trying to figure out Ch 4, Sec. 7, Q 25.c of Mathematical Methods in the Physical Sciences, 3rd Ed. It's not homework I'm working on since I'm not in school.

Assume that f\left(x, y, z\right) = 0
If x, y and z are each functions of t, show that
\left(\frac{\partial y}{\partial z}\right)_{x} = \left(\frac{\partial y}{\partial t} \right)_{x} / \left(\frac{\partial z}{\partial t}\right)_{x}.

This doesn't make any sense to me. We have 4 unknowns in 4 equations, so NO independent variables. And what does it mean to say "the partial derivative of y with respect to t holding x constant", when x is a function of t only (and therefore cannot be held constant if t is changing)?

 

[Fredrik]

The function f:\mathbb R^3\rightarrow\mathbb R is used to define a relationship between three variables: x,y,z. Variables aren't functions. They are just symbols that represent numbers. The equation f(x,y,z)=0 tells you that no matter what values have been assigned to x and z, the value of y is such that f(x,y,z)=0.

Depending on the properties of f, there may be several such values, or no such value at all. If f is such that there's exactly one such value for each (x,z), then the equality f(x,y,z)=0 implicitly defines the function g:\mathbb R^2\rightarrow\mathbb R that takes any pair of real numbers (p,q) to the unique real number r that satisfies f(p,r,q)=0. This r is then denoted by g(p,q).

The problem also involves a curve in the set S={(x,y,z)|f(x,y,z)=0}. (When they tell you that x,y,z are functions of t, this means that what they have in mind is a curve in \mathbb R^3. Since they also specified the condition f(x,y,z)=0, it's clear that what they have in mind is a curve in S). Let's denote this curve by C:\mathbb R\rightarrow S and let's write C(t)=(C_1(t),C_2(t),C_3(t)), where each C_i is a function from \mathbb R into \mathbb R.

The book is causing some confusion by using each of the symbols x,y and z for three different things. For example, y is sometimes a variable, sometimes the function g, and sometimes the function C₂.

When you're asked to compute (\partial y/\partial t)_x, what they want you to compute is (g\circ C_3)'(t), which by the chain rule is equal to g'(C_3(t))C_3'(t). So \left(\frac{\partial y}{\partial z}\right)_{\!\!x} =g'(C_3(t))=\frac{(g\circ C_3)'(t)}{C_3'(t)}=\left(\frac{\partial y}{\partial t}\right)_{\!\!x} \bigg/\left(\frac{\partial z}{\partial t}\right)_{\!\!x}. Note that we will usually not provide complete solutions even if you say that it's not homework. Every textbook-style question will be treated as homework. I'm making an exception in this case because I just don't see any way to help you at all without giving you a complete solution.

 

[Monsterman222]

Hi Fredrik, thanks a lot for the help! I'll make sure to post questions from textbooks in the homework section next time.

I have a follow up question.

I'm a bit confused by: \left(g\circ C_{3}\right)'(t). Since g is a function of two variables, does this mean: g' \left(C_{1} \left(t \right), C_{3} \left(t \right)\right) ?

If so, wouldn't that be: \frac{\partial g}{\partial C_{1}} \frac{\partial C_{1}}{\partial t} + \frac{\partial g}{\partial C_{3}} \frac{\partial C_{3}}{\partial t}.

If that is correct so far, then you seem to have assumed that \frac{\partial C_{1}}{\partial t}=0 , but I don't see how that's justified.


Thanks again!

 

[Fredrik]

You're absolutely right. I made a major blunder there. I'm going to think it through and post an update.

OK, I don't have time to work it out completely. This is what I've got so far.

Let's define three functions g_1,g_2,g_3:\mathbb R^2\rightarrow\mathbb R instead of just one. g_1 is the function that takes (y,z) to the x that satisfies f(x,y,z)=0. The other two are defined similarly.

We still interpret the claim that x,y,z are "functions of t" as saying that we're dealing with a curve C:\mathbb R\rightarrow S, where S=\{(x,y,z)\in\mathbb R^3|\,f(x,y,z)=0\}. This indirectly defines three more functions C_1,C_2,C_3, satisfying C(t)=(C_1(t),C_2(t),C_3(t)).

Now note that C_1(t)=g_1(C_2(t),C_3(t)), and similarly for C_2,C_3. So we get C_1'(t)=\frac{\partial g_1}{\partial C_2}C_2'(t)+\frac{\partial g_1}{\partial C_3}C_3'(t), and two similar equations.

I interpret the identity we're trying to prove as \frac{\partial g_2}{\partial C_3}=C_2'(t)C_3'(t). Perhaps you can get that result from the system of equations above? If that doesn't work, you should look at the text again and see if there's anything that indicates that my interpretation of what we're supposed to prove is wrong.

 

[Monsterman222]

Yes, I think you're right in interpreting the identity that way. I think the author must have made a mistake.

Given that t is an independent variable, then x(t) (that is, C₁(t)) is not an independent variable, so when the author writes \left(\frac{\partial y}{\partial t}\right)_{x}\;\;, it makes no sense. You can't hold x(t) constant.

 

[Fredrik]

What if the x subscript is her way of saying that the curve C is such that C_1'(t)=0 when the value of t is such that C(t) is the point where we want to evaluate the partial derivatives? Then things are starting to make sense again. C_2'(t)=\frac{\partial g_2}{\partial C_1}\underbrace{C_1'(t)}_{=0}+\frac{\partial g_2}{\partial C_3}C_3'(t)=\frac{\partial g_2}{\partial C_3}C_3'(t) \frac{C_2'(t)}{C_3'(t)}=\frac{\partial g_2}{\partial C_3}

 

[Monsterman222]

Possibly, but I'm still thinking the book has an error. The formula we would be deriving would only work for specific curves. It would not work for any curve which does not have any spot where C₁'(t) = 0.

 

[Fredrik]

But for every point p in the set S, there's a curve C in S that goes through p and has constant C₁. The x subscript could be defined to mean precisely that the C appearing in my calculations is such a curve. If that definition appears in the book, she hasn't made a mistake. (If it doesn't, she has).