- #1
sandy.bridge
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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 [itex](r, \varphi)[/itex] be given by [itex]T=f(r, \varphi, t)[/itex], where t is time. Suppose a probe follows the straight path, given in Cartesian coordinates by [itex]x=X(t), y=Y_0=constant[/itex]. Using the fact that [itex]r^2=x^2+y^2, \varphi=arctan(y/x)[/itex], find [itex]dT/dt[/itex].
The Attempt at a Solution
[itex]\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}[/itex]
and
[itex]\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}[/itex]
thus,
[itex]\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}[/itex]
Next, we have,
[itex]\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}[/itex]
[itex]=\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}[/itex]
Is this as far as I can take it with the information given?