The rate of change in the volume of icicle, given the rates of change of its radius and length

Solution 1

Don't worry, these questions seem hard until you get an intuition for what all the numbers represent!

To start, lets look at the formula for the volume of an icicle, since that is what the question is asking about:

$$V=\frac{1}{3} \cdot \pi \cdot r^2 \cdot h\quad \mathrm{cm}^3$$

...but we know that $r$ and $h$ are varying with respect to time, so these are actually functions of a time variable $t$:

$$V(t)=\frac{1}{3} \cdot \pi \cdot r(t)^2 \cdot h(t) \quad \mathrm{cm}^3$$

The question, mathematically, is asking you to find $\frac{d V}{d t}$ at $t=0$. So we know that we have to differentiate $V(t)$ with respect to $t$, and evaluate the result at 0.

Before we can do that though, we have to find out what $V(t)$ is, and to do that we need to know what $r(t)$ and $h(t)$ are. So now let's find $r(t)$ and $h(t)$.

We're given information about how $r(t)$ and $h(t)$ are changing with respect to $t$:

$$\frac{d r}{d t}=-0.2\frac{\mathrm{cm}}{\mathrm{hr}}$$

$$\frac{d h}{d t}=0.8\frac{\mathrm{cm}}{\mathrm{hr}}$$

Notice that if we integrate the above two forms, the fundamental theorem of calculus says that we get the functions:

$$r(t)=\int\frac{dr}{dt}dt=\int -0.2\ dt=-0.2t+C_1$$

$$h(t)=\int\frac{dh}{dt}dt =\int0.8\ dt=0.8t+C_2$$

Now all that's left is to use our initial conditions to find $C_1$ and $C_2$. We're given:

$$r(0)=4\ \mathrm{cm}$$ $$h(0)=20\ \mathrm{cm}$$

Now we can use this information to complete the formulas we found for $r(t)$ and $h(t)$:

$$r(0)=-0.2\cdot 0 + C_1=4 \Rightarrow C_1 = 4$$ $$r(0)=0.8\cdot 0 + C_2=20 \Rightarrow C_2 = 20$$

and finally with this we can substitute in to complete our formulas for $r(t)$ and $h(t)$:

$$r(t)=4-0.2t$$ $$h(t)=20+0.8t$$

Now let's substitute these back into $V(t)$:

$$V(t)=\frac{1}{3}\cdot \pi \cdot (4-0.2t)^2\cdot(20+0.8t)$$

Now all we have to do is differentiate $V(t)$ and evaluate it at $t=0$.

Solution 2

I guess another way of approaching the problem is to write the total derivative given that $V=V(r,h)$ as: $$ dV = \frac{\partial V}{\partial h}dh + \frac{\partial V}{\partial r}dr.$$ Then we can divide by $dt$ to see how the volume changes with both the radius and height changing simultaneously, namely: $$ \frac{dV}{dt} = \frac{\partial V}{\partial h}\frac{dh}{dt} + \frac{\partial V}{\partial r}\frac{dr}{dt}.$$ The problem statement gives $\frac{dh}{dt}=+0.8 \frac{cm}{hr}$ and $\frac{dr}{dt}=-0.2 \frac{cm}{hr}$. Compute $\frac{\partial V}{\partial h}$ and $\frac{\partial V}{\partial r}$ and insert in the following expression for how the volume changes with time. Note, the sign of $\frac{dV}{dt}$ tells you if the volume is increasing or decreasing. $$ \frac{dV}{dt}=\frac{1}{3}\pi r^{2}(+0.8)+\frac{2}{3}\pi r h (-0.2) $$ $$ \frac{dV}{dt}=\frac{1}{3}\pi (4)^{2}(+0.8)+\frac{2}{3}\pi (4) (20) (-0.2) $$ $$ \frac{dV}{dt}=-20.11 \frac{cm^{3}}{hr}.$$ You were correct!

The volume is decreasing at about 20 mls per hour. Hope this helps some.

Cheers,

Paul Safier

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Updated on August 04, 2020