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.

Last edited:

HallsofIvy