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.


Paul Safier


Related videos on Youtube

Author by


Updated on August 04, 2020


  • Lillyym
    Lillyym almost 3 years

    I was studying for my midterm and I couldn't figure this one out. I don't know where to begin and this is really hard.

    A cone shaped icicle is dripping from the roof. The radius of the icicle is decreasing at a rate of .2 cm per hour, while the length is increasing at a rate of .8cm per hour. If the icicle is currently 4cm in radius and 20 cm long, is its volume increasing or decreasing, and at what rate?

    I tried it and I got decreasing at -20.106 but Im pretty sure thats wrong. If you know how to solve it please help! Thank you so much!

    • Admin
      Admin over 9 years
      Can you add how you got your answer? This way, people can suggest improvements or find errors, if there are any.
    • Lillyym
      Lillyym over 9 years
      V= (1/3)pir^2h dr/dt= -.2 dh/dt=.8 dv/dt= (1/3)pi(2rh(dr/dt)+ r^2(dh/dt) dv/dt= (1/3)pi((2times4times20times-.2)+(16times.8)) dv/dt= -20.10619298cm/hour therefore decreasing at rate of -20.106
    • Ross Millikan
      Ross Millikan over 9 years
      Why are you sure it's wrong? It does need some units. The volume change is $cm^3/hour$
    • Lillyym
      Lillyym over 9 years
      I was pretty sure I was wrong because I'm extremely bad at rates of change problems haha
    • Ross Millikan
      Ross Millikan over 9 years
      This one is fine.
    • Baby Dragon
      Baby Dragon over 9 years
      A small tip: when a question like this comes up, try to answer the sub-question "is the volume increasing or decreasing".
    • Christian Chapman
      Christian Chapman over 9 years
      Another way to solve the problem: did you notice that $$\frac{dV}{dt}=\frac{1}{3}\pi\left[2r\cdot \frac{dr}{dt}\cdot h + r^2\cdot \frac{dh}{dt}\right]$$?
  • André Nicolas
    André Nicolas over 9 years
    In principle we cannot "find" $r(t)$ and $h(t)$, since we are only given their rates of change at a certain instant. We could assume these rates of change are unchanging, and that would give the right numerical answer. But differentiating directly works fine with no assumptions.
  • Christian Chapman
    Christian Chapman over 9 years
    That is true, it is also a bit less legwork. See the comment I put on the question.