Rates Question (1 Viewer)

hfreemannnnnnn · Oct 10, 2021

Hi, the question is:
Consider an open inverted conical container of height 3m and base radius of 1m. Water is evaporating at a rate proportional to the water surface (which is exposed to air). If it takes one hour to reduce the height of the water by half, how long will it take to empty the container, assuming it is initially full?

A) 1.5 hours
B) 2 hours
C) 2.5 hours
D) 3 hours

I've tried fiddling around with volume formulas and surface area formulas and their derivatives but can't actually answer the question. Thanks for any help

Answer is B - 2 hours

notme123 · Oct 10, 2021

Can someone check if somethings wrong here:
$\frac{dV}{dt}=-kA = -k\pi{r}^2$
$V=\frac{4}{3}\pi{r}^3$
$\frac{dV}{dr}=4\pi{r}^2$
$\pi{r}^2 = \frac{\frac{dV}{dr}}{4}$
$\therefore \frac{dV}{dt}= -k\frac{\frac{dV}{dr}}{4}$
$\frac{dr}{dt} = -\frac{k}{4}$
$r=-\frac{k}{4}t + C$
when t=0, r=3
$C=1$
From cone: $\frac{r}{h} = \frac{1}{3}$
$h=3r$
When h=1.5, r=1/2, t =1
$\frac{1}{2}=-\frac{k}{4} + 1$
$k=2$
$\therefore r=-\frac{1}{2}t + 1$
r=0: t=2 ?????

CM_Tutor · Oct 10, 2021

notme123 said:
Can someone check if somethings wrong here:
$\frac{dV}{dt}=-kA = -k\pi{r}^2$
$V=\frac{4}{3}\pi{r}^3$

The shape is a cone, not a sphere.

$V = \cfrac{1}{3}\pi r^2 h$

We know, using similar triangles, that

$\cfrac{r}{h} = \cfrac{1}{3} \qquad \implies \qquad r = \cfrac{h}{3}$

And thus

$V = \cfrac{1}{3} \times \pi \times \left(\cfrac{h}{3}\right)^2 \times h = \cfrac{\pi h^3}{27} \qquad \implies \qquad \cfrac{dV}{dh} = \cfrac{\pi h^2}{9}$

Applying the Chain Rule and using the correct formula given above that $\cfrac{dV}{dt} = -k\pi r^2$ :

$\cfrac{dh}{dt} = \cfrac{dh}{dV} \times \cfrac{dV}{dt} = \cfrac{9}{\pi h^2} \times -k\pi \left(\cfrac{h}{3}\right)^2 = -k$

$\begin{align*} \text{After integrating:} \qquad h &= -kt + H \qquad \text{where $H$ m is the height at time $t = 0$ h} \\ \text{At $t = 1$ h, $h = \cfrac{H}{2}$ m:} \qquad \cfrac{H}{2} &= -k + H \\ k &= \cfrac{H}{2} \\ \text{We seek $t$ when $h = 0$ m:} \qquad 0 &= -\cfrac{H}{2}t + H \\ t &= \cfrac{2}{H} \times H = 2\ \text{h} \end{align*}$

notme123 · Oct 10, 2021

CM_Tutor said:
The shape is a cone, not a sphere.

$V = \cfrac{1}{3}\pi r^2 h$

We know, using similar triangles, that

$\cfrac{r}{h} = \cfrac{1}{3} \qquad \implies \qquad r = \cfrac{h}{3}$

And thus

$V = \cfrac{1}{3} \times \pi \times \left(\cfrac{h}{3}\right)^2 \times h = \cfrac{\pi h^3}{27} \qquad \implies \qquad \cfrac{dV}{dh} = \cfrac{\pi h^2}{9}$

Applying the Chain Rule and using the correct formula given above that $\cfrac{dV}{dt} = -k\pi r^2$ :

$\cfrac{dh}{dt} = \cfrac{dh}{dV} \times \cfrac{dV}{dt} = \cfrac{9}{\pi h^2} \times -k\pi \left(\cfrac{h}{3}\right)^2 = -k$

$\begin{align*} \text{After integrating:} \qquad h &= -kt + H \qquad \text{where $H$ m is the height at time $t = 0$ h} \\ \text{At $t = 1$ h, $h = \cfrac{H}{2}$ m:} \qquad \cfrac{H}{2} &= -k + H \\ k &= \cfrac{H}{2} \\ \text{We seek $t$ when $h = 0$ m:} \qquad 0 &= -\cfrac{H}{2}t + H \\ t &= \cfrac{2}{H} \times H = 2\ \text{h} \end{align*}$

omg what a move i made there

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