18. Related Rates

b. Word Problems

There are many “real” world problems involving related rates. Here are some.

A cone which is \(H=16\,\text{cm}\) high and \(R=4\,\text{cm}\) in radius is filled with water which is leaking out the bottom at \(5\dfrac{\text{cm}^3}{\text{min}}\). How fast is the height of the water decreasing when the water is \(10\,\text{cm}\) deep?
HINTS: Let \(h\) and \(r\) be the height and radius of the water in the cone as functions of time. The volume of a cone is \(V=\dfrac{1}{3}\pi r^2h\).

eg_cone

We start with the volume of a cone: \[ V=\dfrac{1}{3}\pi r^2h \] By similar triangles, the ratio between the height of the water and the radius of the water is: \[ \dfrac{r}{h}=\dfrac{R}{H}=\dfrac{4}{16}=\dfrac{1}{4} \] This means that we can substitute \(r=\dfrac{1}{4}h\) in the volume: \[ V=\dfrac{1}{48}\pi h^3 \] We take the derivative: \[ \dfrac{dV}{dt}=\dfrac{1}{16}\pi h^2\dfrac{dh}{dt} \] The current height is \(10\,\text{cm}\) and the volume is decreasing at \(\dfrac{dV}{dt}=-5\,\dfrac{\text{cm}^3}{\text{min}}\). We plug in and solve for the rate of change of the height: \[ -5=\dfrac{1}{16}\pi 10^2\dfrac{dh}{dt} \] \[ \dfrac{dh}{dt}=-\,\dfrac{80}{100\pi}=-\,\dfrac{.8}{\pi} \] So, the height is decreasing at a rate of \(\dfrac{dh}{dt}=-\,\dfrac{.8}{\pi}\,\dfrac{\text{cm}}{\text{min}}\).

When two resistors with resistances \(R_1\) and \(R_2\) are combined in parallel, the resulting resistance \(R\) satisfies: \[ \dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2} \] Currently, \(R_1=100\,\text{ohms}\) and is increasing at \(\dfrac{dR_1}{dt}=18\,\dfrac{\text{ohms}}{\text{hr}}\rule[-10pt]{0pt}{15pt}\), while \(R_2=200\,\text{ohms}\) and is decreasing at \(\dfrac{dR_2}{dt}=-9\,\dfrac{\text{ohms}}{\text{hr}}\rule[-10pt]{0pt}{15pt}\). Find the current values of \(R\) and \(\dfrac{dR}{dt}\rule[-10pt]{0pt}{15pt}\). Is \(R\) increasing or decreasing?

ex_resistors

This time three variables are related.

\(R\) is increasing at \(\dfrac{dR}{dt}=7\,\dfrac{\text{ohms}}{\text{hr}}\)

This time three variables are related. We first find \(R\): \[\begin{aligned} \dfrac{1}{R}&=\dfrac{1}{R_1}+\dfrac{1}{R_2} =\dfrac{1}{100}+\dfrac{1}{200}=\dfrac{3}{200} \\ R&=\dfrac{200}{3} \end{aligned}\] Next, we differentiate the relation, using the Chain Rule: \[ -\,\dfrac{1}{R^2}\dfrac{dR}{dt} =-\,\dfrac{1}{(R_1)^2}\dfrac{dR_1}{dt}-\dfrac{1}{(R_2)^2}\dfrac{dR_2}{dt} \] We cancel the minuses, plug in numbers and solve for \(\dfrac{dR}{dt}\): \[\begin{aligned} \dfrac{9}{200^2}\dfrac{dR}{dt} &=\dfrac{1}{100^2}(18) +\dfrac{1}{200^2}(-9) \\ \dfrac{dR}{dt} &=\dfrac{200^2}{9}\left(\dfrac{1}{100^2}(18) -\dfrac{1}{200^2}(9)\right) \\ &=8-1=7\,\dfrac{\text{ohms}}{\text{hr}} \end{aligned}\] Since this is positive, \(R\) is increasing.

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