Chain rule for several variables

Question

Suppose that the temperature $T$ in a certain liquid varies with depth $z$ and time $t$ according to the formula $T=e^{-t} \cdot z$. Find the rate $of$ change of temperature with respect to time at a point that is moving through the liquid so that at time $t$ its depth is $f(t)$. What is this rate if $f(t)=e^t$? What is happening in this case?

I am not really sure how to set up the dependence chart for this problem, so I am not sure how to go about it.

Answer 1

In this case, the depth $z(t)$ is equal to the function $f(t) = e^t$. So, when you compose $T(f(t), t)$, you find $T(f(t),t) = e^{-t}e^t = 1$. When you differentiate, there is really no more need of the chain rule, since $T(f(t),t)$ turns out to be constant. Hence the rate of change of temperature of the body is $0$.

In general, if the body has an arbitrary $z(t) = f(t)$ governing its depth at a time $t$, then simply apply the derivative to this function that is now a function of a single variable: $$ T(f(t), t) = e^{-t}f(t) \\ D_t T = -e^{-t}f(t) + e^{-t}f'(t). $$

Answer 2

$z$ is a function of $t$ and $T$ is a function of $t, z$

$T = e^{-t}z\\ z = e^{t}$

$\frac {dT}{dt} = (-e^{-t})z + (e^-t)\frac {dz}{dt}$ product rule.

$\frac {dz}{dt} = e^t$

Substitute:

$\frac {dT}{dt} = (-e^{-t})e^t + (e^{-t})(e^t) = 0$