AP Calculus BC 7.6 General DE Solutions Overview

Review general differential-equation solutions by separating variables, integrating both sides and retaining the constant.

[Maximum number: 4]

A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height h of the water in the barrel with respect to time t is modeled by $\frac{d h}{d t}=-\frac{1}{10} \sqrt{h}$ , where h is measured in feet and t is measured in seconds. (The volume V of a cylinder with radius r and height h is $V=\pi r^{2} h$ .)

At time t=0 seconds, the height of the water is 5 feet. Use separation of variables to find an expression for h in terms of t.

$\frac{d h}{\sqrt{h}}=-\frac{1}{10} d t$ $\int \frac{d h}{\sqrt{h}}=\int-\frac{1}{10} d t$ $2 \sqrt{h}=-\frac{1}{10} t+C$ $2 \sqrt{5}=-\frac{1}{10} \cdot 0+C \Rightarrow C=2 \sqrt{5}$ $2 \sqrt{h}=-\frac{1}{10} t+2 \sqrt{5}$ $h(t)=\left(-\frac{1}{20} t+\sqrt{5}\right)^{2}$ : $\left\{\begin{array}{l}1: \frac{d V}{d t}=\pi \frac{d h}{d t} \\ 1: \text { answer with units }\end{array}\right.$ $3:\left\{\begin{array}{l}1: \frac{d}{d h}\left(-\frac{1}{10} \sqrt{h}\right)=-\frac{1}{20 \sqrt{h}} \\ 1: \frac{d^{2} h}{d t^{2}}=-\frac{1}{20 \sqrt{h}} \cdot \frac{d h}{d t} \\ 1: \text { answer with explanation }\end{array}\right.$ $4:\left\{\begin{array}{l}1: \text { separation of variables } \\ 1: \text { antiderivatives } \\ 1: \text { constant of integration } \\ \quad \text { and uses initial condition } \\ 1: h(t)\end{array}\right.$

Note: 0/4 if no separation of variables
Note: max 2/4 [1-1-0-0] if no constant of integration

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Covered syllabus points

7.6 Finding General Solutions Using Separation of Variables question 1