Problem
We have function which defines temperature for every point in $\mathbb{R^3}$
$$ T(x,y,z)=3x+3y+3z $$
Now we move along curve which is given in parametric form.
$$ r(t)=\begin{bmatrix} x(t)= \frac{t^3}{30}+\frac{14t}{5}+7 \\ y(t)= \frac{-t^3}{90}-\frac{32t^2}{45}+28 \\ z(t)= \frac{8t^2}{45}+\frac{7t}{5}-14 \end{bmatrix} $$
when $t\in[0,10]$
$x,y,z$ are unit vectors.
$$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\z \end{bmatrix} $$
Define the coldest and the hottest point along the curve.
Attempt to solve
Now we can create new function which defines temperature for each point that are on the curve. Simply by joining these two functions together. $g(t)$ is our joint function $$ g(t)=T(x(t),y(t),z(t))=3\cdot (\frac{t^3}{30}+\frac{14t}{5}+7)+3\cdot (\frac{-t^3}{90}-\frac{32t^2}{45}+28) +3 \cdot (\frac{8t^2}{45}+\frac{7t}{5}-14) $$ Now since we want to find the coldest and the hottest point on the curve we want to find local minima and local maxima. We can try to computer when $g'(t)=0$
According to chain rule:
$$ \frac{d}{dx}f(g(x))=f'(g(x))g'(x) $$
$$ g'(t)=T1(x(t),y(t),z(t))\cdot x'(t)+T2(x(t),y(t),z(t))\cdot y'(t) + T3(x(t),y(t),z(t))\cdot z'(t) $$
$(T1,T2,T3)$ are partial derivatives
$$T1=\frac{\delta}{\delta x}(3x+3y+3z) \\ T2=\frac{\delta}{\delta y}(3x+3y+3z) \\ T3=\frac{\delta}{\delta z}(3x+3y+3z)$$
Partial derivatives are $(T1,T2,T3)=(3,3,3)$
Now we can can compute tangent vector from r(t):
$$r'(t)=\begin{bmatrix} x'(t)= \frac{d}{dt}(\frac{t^3}{30}+\frac{14t}{5}+7)=\frac{t^2}{10}+\frac{14}{15} \\ y'(t)= \frac{d}{dt} (\frac{-t^3}{90}-\frac{32t^2}{45}+28)=-\frac{t^2}{30}-\frac{64}{45} \\ z'(t)= \frac{d}{dt} (\frac{8t^2}{45}+\frac{7t}{5}-14)=\frac{14t}{45}+\frac{7}{5} \end{bmatrix}$$
Now combining this information we get:
$$ g'(t)=3\cdot (\frac{t^3}{10}+\frac{14}{5})+3\cdot (-\frac{t^2}{30}-\frac{64}{45})+3 \cdot (\frac{14t}{45}+\frac{7}{5}) $$
$$ g'(t)=\frac{3t^3}{10}-\frac{t^2}{10}+\frac{42t}{45}-\frac{1}{15} $$
if we try to solve when $g'(t)=0$
$$\frac{3t^3}{10}-\frac{t^2}{10}+\frac{42t}{45}-\frac{1}{15}=0 $$
we get two solutions:
$$ t_1=-\frac{7}{3}-\frac{2\sqrt{13}}{3} $$ $$ t_2=-\frac{7}{3}+\frac{2\sqrt{13}}{3} $$
plugging in these solutions to $g(t)$:
$ g(-\frac{7}{3}-\frac{2\sqrt{13}}{3})\approx -39.67626227$
$ g(-\frac{7}{3}+\frac{2\sqrt{13}}{3})\approx 63.8787314$
It would seem that we have found local minima and local maxima. I am unable to find accurate temperature for these so approximations have to do.
Now these result are very likely false. I should be able to find accurate result for both of these temperatures so there has to be mistake but since this becomes very complicated fast it's hard to find errors. If someone knows alternative way to approach this problem or if someone can somehow spot the mistake that would be highly appreciated.
This is not actually multivariable calculus, because $T(r(t)):\mathbb R\longrightarrow \mathbb R$, that is, $T\circ r$ is a real function of one variable. Your usual single-variable calculus methods can deal with it just fine.
Notice that you already have a formula for $T\circ r$; more specifically, you found that it is a third degree polynomial in $t$. Rather than apply the chain rule to find $g'$, why not differentiate the polynomial you found with respect to $t$?