What is the partial derivative of T with respect to time?

[sandy.bridge]

Homework Statement

Hello all,
New to partial derivatives. I was wondering if someone could look over my work and determine if my final step is as far as I can take the proble (ie. that will be my solution). Thanks in advance.

Let the temperature of a 2D domain in polar coordinates (r, \varphi) be given by T=f(r, \varphi, t), where t is time. Suppose a probe follows the straight path, given in Cartesian coordinates by x=X(t), y=Y_0=constant. Using the fact that r^2=x^2+y^2, \varphi=arctan(y/x), find dT/dt.

The Attempt at a Solution

\frac{dr}{dt}=\frac{\delta r}{\delta x}\frac{dx}{dt}+\frac{\delta r}{\delta y}\frac{dy}{dt}=\frac{\delta r}{\delta x}\frac{dx}{dt}

and

\frac{d\varphi}{dt}=\frac{\delta \varphi}{\delta x}\frac{dx}{dt}+\frac{\delta\varphi}{\delta y}\frac{dy}{dt}=\frac{\delta\varphi}{\delta x}\frac{dx}{dt}

thus,

\frac{dT}{dt}=\frac{\delta T}{\delta r}\frac{dr}{dt}+\frac{\delta T}{\delta\varphi}\frac{d\varphi}{dt}+\frac{\delta T}{\delta t}

Next, we have,

\frac{dT}{dt}=\frac{\delta T}{\delta r}\frac{\delta r}{\delta x}\frac{dx}{dt}+\frac{\delta T}{\delta\varphi}\frac{\delta\varphi}{\delta x}\frac{dx}{dt}+\frac{\delta T}{\delta t}
=\frac{\delta T}{\delta r}(x)\frac{d[X(t)]}{dt}-\frac{\delta T}{\delta\varphi}\frac{1}{1+(y/x)^2}\frac{y}{x^2}\frac{d[X(t)]}{dt}+\frac{\delta T}{\delta t}

Is this as far as I can take it with the information given?

 

[sandy.bridge]

also wanted to inform everyone that the lowercase deltas were supposed to be partials

 

[sandy.bridge]

No one?

 

[cepheid]

You should try \partial

Your application of the chain rule seems okay.

I think that partial deriv of r wrt x should be 2x, not x, right?

 

[cepheid]

Actually nvmind my third statement. I forgot about the square root.

 

[sandy.bridge]

Thanks. My real reason for asking was because the answer seems kind of odd. It just feels like I could go furthur with it, but I don't see how.

 

[sandy.bridge]

Now that I have had a second look at the problem I'm going to say that it is 100 percent wrong...

\frac{dT}{dt}=\frac{\partial T}{\partial r}\frac{dr}{dt}+\frac{\partial T}{\partial \varphi}\frac{d\varphi}{dt}+\frac{\partial T}{\partial t}

Hence that couldn't be right since
\frac{dr}{dr}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial t}

and a similar expression for \varphi can be found...

 

[cepheid]

Now that I have had a second look at the problem I'm going to say that it is 100 percent wrong...

Okay, well, this is definitely true:

\frac{dT}{dt}=\frac{\partial T}{\partial r}\frac{dr}{dt}+\frac{\partial T}{\partial \varphi}\frac{d\varphi}{dt}+\frac{\partial T}{\partial t}

That's just how the chain rule works for multivariable functions.

Hence that couldn't be right since
\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial t}

Doesn't the second term above assume that r has some sort of explicit dependence on t that is outside of the implicit dependence on t that comes via its dependence on x? In this case, I don't think that's true. Since there is no explicit dependence on t that is not via some other variable, I think that \partial r / \partial t would be 0.

EDIT: Mathematical way of saying this: r(x,t) = r(x(t)) (no explicit dependence on t).

Another way to look at it: if you posit that r depends on x and on t, then the way to compute \partial r / \partial t is to hold x constant. But for x to be constant, t must be constant as well.

By the way, the itex tags are for LaTeX equations that are inline with regular text. For equations that are not inline, you can just use tex tags instead of itex like so:

\frac{dT}{dt}=\frac{\partial T}{\partial r}\frac{dr}{dt}+\frac{\partial T}{\partial \varphi}\frac{d\varphi}{dt}+\frac{\partial T}{\partial t}

 

[sandy.bridge]

If I understand what you are saying correctly, since dr/dt=0, then essentially dT/dT will simply be the partial derivative of T with respect to the partial derivative of time.

 

[cepheid]

If I understand what you are saying correctly, since dr/dt=0, then essentially dT/dT will simply be the partial derivative of T with respect to the partial derivative of time.

NO, you did not understand me correctly. dr/dt is NOT 0. I was arguing that ∂r/∂t = 0. There's an important difference there. With the latter, the derivative dr/dt reduces to:

\frac{dr}{dt} = \frac{\partial r}{\partial x} \frac{dx}{dt}

So it's just the normal chain rule for composite functions. In other words, we return to what you had in your original post.

 

[sandy.bridge]

Oh wow, thanks. Mind went blank. Thanks a lot!

 

[cepheid]

Oh wow, thanks. Mind went blank. Thanks a lot!

By the way, I also don't think you computed ∂r/∂x correctly. We have:

\frac{\partial r}{\partial x} = \frac{\partial }{\partial x}(x^2 + y^2)^{1/2}

How would you go about differentiating this?