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Problems #22 from pg.9 and #5 from pg.24 are graded.
#22,pg.9: A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time.
Solution: The volume of a sphere of radius $r$ is $V=\dfrac{4}{3}\pi r^3$ and the surface area of a sphere of radius $r$ is $S=4\pi r^2$. "Evaporation" is understood as a decrease in the volume of the spherical raindrop and so we express the differential equation in the form
$$(*) \hspace{35pt} \dfrac{dV}{dt} = -\alpha S,$$
where $\alpha>0$ is an "evaporation constant" (which would depend in reality on the environment in which the drop is falling -- an experimentally determined value!). As written, this is not "a differential equation for the volume of the raindrop", because we do not know the precise formula for the surface area of the raindrop (if we did, we wouldn't have to derive an equation to find the volume!). So we must rewrite the right hand side of this equation in terms of the function $V$ instead of the function $S$. To do this we take the formula $V=\dfrac{4}{3}\pi r^3$ and solve for $r$ to get $r = \left( \dfrac{3}{4\pi} V \right)^{\frac{1}{3}}$. Now since $S=4\pi r^2$ we simply substitute the formula for $S$ and the formula for $r$ into $(*)$ to see
$$\dfrac{dV}{dt}=-\alpha S = -\alpha (4 \pi) \left[ \left( \dfrac{3}{4 \pi} V \right)^{\frac{1}{3}} \right]^2=-k V^{\frac{2}{3}},$$
where $k>0$ just combines all the various constants together into one, i.e., $k=\alpha(4\pi) \left(\dfrac{3}{4\pi} \right)^{\frac{2}{3}}$. This is a "differential equation for the volume of the raindrop as a function of time".
#5,pg.24: Determine the order and determine whether or not the equation is a linear or nonlinear equation:
$$\dfrac{d^2 y}{dt^2}+\sin(t+y)=\sin(t).$$
Solution: This is a second order nonlinear equation (there is a $y$ inside of the sine function!).
