Relate rates problem

Question

I am looking at this related rates question and since I haven't worked on these types of problems for a while, I just wanted to check if I have done things right.

I simply derived the function giving me

$$Dv=4kr^3dr$$

And then I input $5$ into $r$ and $-0.1$ into $dr$ and solve for $Dv$? I get $-50mm$ per week as the decrease in volume. Have I dont this right or have I missed something?

Answer 1

What is notable about the data given in this question is that the rate of change for the artery radius has units of mm/week. This implies that both the volume and the radius are functions of time themselves. In your answer you have not used that fact. To do this, we can differentiate with respect to time and invoke the chain rule:

$$V = kr^4 \implies \frac{dV}{dt} = 4kr^3 \frac{dr}{dt} $$

If $r=5$ mm and $\frac{dr}{dt} = - 0.1$ mm/week then $\frac{dV}{dt} = 4k(5)^3(-0.1) = -50k \, \frac{\mathrm{mm^4}}{week}$.

You had the right idea, but your notation could use some more clarifying and you forgot to include $k$ in your answer.

Best of luck.

Answer 2

In general, a procedure which helps with all related rates problems is

1. Read the problem. Identify and label the rates that are given, and identify rates that we are searching for.
2. Determine one or more equations that express the basic relationships among the variables.
3. Introduce rates of change by differentiating the equation implicitly with respect to time, $t$.
4. Substitute known values in for the variables and rates after taking the derivative. Solve for the desired quantity and make sure that the units are consistent.

So, we first identify known variables and rates

$$V=\text{Volume of blood flowing through the artery (in $\text{mm}^4$)}$$ $$k=\text{Constant which depends on blood flow}$$ $$r=\text{Radius of the artery}=5\text{mm}$$ $$\frac{dr}{dt}=\text{Rate of change of radius with respect to time}=-0.1\frac{\text{mm}}{\text{week}}$$ $$ t = \text{time is given in weeks}$$

and we also identify the unknown rate that we need to find

$$\frac{dV}{dt}=\text{Rate of change of volume with respect to time}$$

then, the equation relating all variables is given to us as a function of time

$$V=kr^4$$

so we implicitly differentiate this equation with respect to time to form

$$\frac{dV}{dt}=4kr^3\frac{dr}{dt}=4k(5\text{mm})^3 \left(-0.1\frac{\text{mm}}{\text{week}}\right)=-50k\frac{{\text{mm}}^4}{\text{week}}$$

which is reasonable since as the radius decreases, we would expect that the volume would also decrease. As the unit of $V$ is $\text{mm}^4$, it makes sense that the unit of $\frac{dV}{dt}$ is given as $\frac{{\text{mm}}^4}{\text{week}}$ since time is measured in weeks.

I'd pay careful attention to writing down all of the variables at the start of the problem (this will identify the unknown rate which you need to solve when you implicitly differentiate). It is also helpful to write down all of the units after you implicitly differentiate and then simplify.