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

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?

Hint

This time three variables are related.

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Answer

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

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Solution

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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You should do lots of problems in the exercises.

18. Related Rates

b. Word Problems

Hint

Answer

Solution

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