합성함수의 미분법

(a) We are asked to find $\frac{dV}{dh}$ (the derivative of $V$ with respect to the independent variable $h$) at the instant $h = 1$. If we first expand $V$ to write $$V = \frac{\pi}{3}\left(12h^2 - h^3\right)$$ we find that $$\frac{dV}{dh} = \frac{\pi}{3} \left(24h - 3h^2\right) = \pi \left(8h-h^2 \right)\,,$$ with units “cubic feet per foot”. Evaluating at $h=1$, we get $$\left. \frac{dV}{dh} \right|_{h=1} = \pi \left(8 \cdot 1 - 1^2 \right) = 7\pi$$ (b) Since $h(t) = \sin(\pi t) + 1$, the instantaneous rate of change of $h$ with respect to $t$ is $h'(t) = \pi \cos(\pi t)$, and therefore $h'(2) = \pi \cos(2\pi) = \pi$ feet per hour. (c) First we want to find $\frac{dV}{dt}$, the derivative of $V$ with respect to the independent variable $t$. Here we can write $V$ as a function of $t$ by writing $V(t) = \frac{\pi}{3}\left(12(h(t))^2 - (h(t))^3\right)$, where $h(t) = \sin(\pi t) + 1$. By the chain rule, it follows that $$\frac{dV}{dt} = \frac{\pi}{3}\left( 12 \cdot 2 h(t) \cdot h'(t) - 3 (h(t))^2 \cdot h'(t) \right)$$ and hence using $h(t) = \sin(\pi t) + 1$ and $h'(t) = \pi \cos(\pi t)$, $$\frac{dV}{dt} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(\pi t) + 1) \cdot \pi \cos(\pi t) - 3 (\sin(\pi t) + 1)^2 \cdot \pi \cos(\pi t) \right]$$ with units “ feet per hour”. Now we want to find $\frac{dV}{dt}$ when $t=2$. Using our most recent work, we have $$\left. \frac{dV}{dt} \right|_{t=2} = \frac{\pi}{3}\left[ 12 \cdot 2 (\sin(2\pi) + 1) \cdot \pi \cos(2\pi) - 3 (\sin(2\pi) + 1)^2 \cdot \pi \cos(2\pi) \right] = 7 \pi^2$$ cubic feet per hour. (d) In (a) we are determining the instantaneous rate at which the volume changes as we increase the height of the water in the tank, while in (c) we are finding the instantaneous rate at which volume changes as we increase time. The first calculation is a direct one, while the second is more indirect, as it involves the composition of $V$ with $h$ as a function of time. In the latter case, both $h$ and $V$ are functions of time, which allows us to think not only of how $V$ is changing as a function of $h$, but also as a function of $t$.