Answer:
Approximately [tex]0.0435\; \text{in} \cdot \text{s}^{-1}[/tex]. (Given that [tex]V = \pi\, r\, h^{2} - (1/3)\, h^{3}[/tex].)
Step-by-step explanation:
Let [tex]t[/tex] denote time. Both the volume of water in the bowl, [tex]V[/tex], and the height of water in the bowl, [tex]h[/tex], are dependent on [tex]t\![/tex] (as if the [tex]V\![/tex] and [tex]h[/tex] are functions of [tex]\! t[/tex].)
Implicitly differentiate both sides of the equation [tex]V = \pi\, r\, h^{2} - (1/3)\, h^{3}[/tex] with respect to [tex]\! t[/tex]:
[tex]\displaystyle \frac{d}{d t}[V] = \frac{d}{d t}\left[\pi\, r\, h^{2} - (1/3)\, h^{3}\right][/tex].
[tex]\displaystyle \frac{d V}{d t} = 2\, \pi\, r\, h\, \frac{d h}{d t} + (1/3)\, (3\, h^{2})\, \frac{d h}{d t}[/tex].
Both [tex]h[/tex] (current height of water in the bowl) and [tex](d V) / (d t)[/tex] (rate at which water is poured into the bowl) are given: [tex]h = 2.4\; \text{in}[/tex] and [tex](d V) / (d t) = 5\; \text{in}^{3}\cdot \text{s}^{-1}[/tex]. The question is asking for [tex](d h) / (d t)[/tex], the rate at which the water level in the bowl is rising.
Thus, rearrange the equation to find an expression for [tex](d h) / (d t)[/tex] in terms of [tex]h[/tex], [tex]r[/tex], and [tex](d V) / (d t)[/tex]. Substitute in the values [tex]h = 2.4\; \text{in}[/tex], [tex](d V) / (d t) = 5\; \text{in}^{3}\cdot \text{s}^{-1}[/tex], and [tex]r= 8\; \text{in}[/tex]; solve for [tex](d h) / (d t)\![/tex], the rate at which the water level is rising:
[tex]\displaystyle \frac{d V}{d t} = \left(2\, \pi\, r\, h\, + (1/3)\, (3\, h^{2})\right)\, \frac{d h}{d t}[/tex].
[tex]\begin{aligned} \frac{d h}{d t} &= \frac{\displaystyle \frac{d V}{d t}}{2\, \pi\, r\, h\, + (1/3)\, (3\, h^{2})} \\ &= \frac{\displaystyle \frac{d V}{d t}}{2\, \pi\, r\, h\, + h^{2}} \\ &= \frac{5\; \text{in}^{3} \cdot \text{s}^{-1}}{2\, \pi \, (8\; \text{in})\, (2.4\; \text{in}) + (2.4\; \text{in})^{2}} \\ &\approx 0.0435\; \text{in} \cdot \text{s}^{-1} \end{aligned}[/tex].
