# Min and Max temperature of path in 3D field

## Homework Statement

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

Path is given by r(t)={
$$x(t)=\frac{t^3}{30}+\frac{16t}{9}+7$$
$$y(t)=-\frac{t^3}{120}-\frac{13t^2}{30}+28$$
$$z(t)=\frac{13t^2}{60}+\frac{4t}{3}-14$$

$$t\in \left[0,10\right]$$

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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$$\frac{dT}{dt}=\frac{\partial T}{\partial x}\cdot \frac{\partial x}{\partial t}+\frac{\partial T\:}{\partial \:y}\cdot \:\frac{\partial \:y}{\partial \:t}+\frac{\partial T\:}{\partial \:z}\cdot \:\frac{\partial z}{\partial t}$$

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

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

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.

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HallsofIvy
Homework Helper
I don't understand why you would first differentiate and then "integrate it to get T". You were given T! You seem to have lost track of what you were trying to do.

The reason you differentiated, I hope, was to set the derivative equal to 0 to find the value of t that gives a max or min. Then put whatever value of t you get back into the original equations for x, y, and z.

Hey, thank you for fast response!

That's exactly what I should do, but for some reason there was so much variables that I didn't realize the obvious course of action.
To my surprise, the correct answer happened to be at the limits of t=0 and t=10 for both, so the resoults from the differentation didn't have any affect.

Thank you for clear explanation!

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