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## Homework Statement

Temperature varies by function T(x,y,z)=3x + 4y + 2z

Path is given by r(t)={

[tex]x(t)=\frac{t^3}{30}+\frac{16t}{9}+7[/tex]

[tex]y(t)=-\frac{t^3}{120}-\frac{13t^2}{30}+28[/tex]

[tex]z(t)=\frac{13t^2}{60}+\frac{4t}{3}-14[/tex]

[tex]t\in \left[0,10\right][/tex]

Question: What is the maximum and minumun temperatures of that path?

## Homework Equations

There's a tip in the assignment to use the chain rule.

## The Attempt at a Solution

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[tex]\frac{dT}{dt}=\frac{\partial T}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial T\:}{\partial \:y}\cdot \:\frac{\partial \:y}{\partial \:t}+\frac{\partial T\:}{\partial \:z}\cdot \:\frac{\partial z}{\partial t}[/tex]

[tex]\frac{dT}{dt}=3\left(\frac{t^2}{10}+\frac{16}{9}\right)+4\left(-\frac{t\left(3t+104\right)}{120}\right)+2\left(\frac{13t}{30}+\frac{4}{3}\right)[/tex]

[tex]\frac{dT}{dt}=\frac{1}{5}\left(t^2-13t+40\right)[/tex]

Then I thought to separate the variables and integrate it to get T, but I end but with constant C.

If I ignore the constant, I end up with wrong answer(min -9/20, max 8)

I cannot think how to continue. I don't know the correct answer either.

Do you think the idea is correct to this point?

Maybe it's something obvious, I might be just baffled with all these derivation steps.

Thank you for reading!

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