next up previous
Next: About this document ...

Math 112 Solutions for 4.2, p. 329

2.
(b) $ y^{4}=\frac{8}{3}x^{3/2}$. (c) $ y^{3}=3x^{2}+1$. (d) $ y^{2}=6(x+\frac{1}{3}x^{3})-2$. (f) $ \tan y=\frac{1}{2}x^{2}+1$.
3.
(a) about 1986 or 1987. (b) about 1983.
4.
$ \dfrac{dP}{dt}=kP^{2/3}$; $ P=(t+10)^{3}$; $ t=0\Rightarrow P=1000$
5.
$ \dfrac{dC}{dt}=\dfrac{k}{C}$; $ C^{2}=\frac{1}{2}t$; $ C=\frac{1}{2}\Rightarrow t=\frac{1}{2}$ hr; $ C=\frac{1}{3}\Rightarrow
t=\frac{2}{9}$, so it takes $ 2-\frac{2}{9}=\frac{16}{9}$ hr to completely charge.
8.
$ \dfrac{dh}{dt}=\dfrac{k}{\sqrt h}$; $ h^{3/2}=\dfrac{5\sqrt{10}}{2} t$; $ t=8\Rightarrow h=10\cdot 2^{2/3}$.
9.
10,506
12.
resistive force $ =\frac{3}{2}v$; $ F=ma=120-\frac{1}{2}v-\frac{3}{2}v\Rightarrow
a=\dfrac{dv}{dt}=\dfrac{g}{400}(60-v)$; $ v=60-Ae^{-gt/400}$; maximum $ v$ is 60.
13.
$ a=g-\frac{1}{7}v\Rightarrow v=7g+Ae^{-t/7}\Rightarrow v\to 7g$ as $ t\to\infty$.
14.
$ \dfrac{dV}{dt}=576\pi \dfrac{dh}{dt}=-\pi\sqrt{2g}h^{1/2}$; $ h=\left(4\sqrt 3-\dfrac{\sqrt g}{576\sqrt 2}t\right)^{2}$; $ h=0\Rightarrow
t=288$ sec.
15.
(a) Since $ g$ is inversely proportional to the square of the distance $ x$ from the center of the earth, the acceleration on the object is $ a=-g=-\frac{k}{x^{2}}$.

(b) $ a=v\frac{dv}{dx}=-\frac{k}{x^{2}}\Rightarrow v^{2}=\frac{2k}{x}+C$; $ x=x_{0}$ and $ v=v_{0} \Rightarrow C=v_{0}^{2}-\frac{2k}{x_{0}}$, and the result follows.

(c) $ v>0\Rightarrow x\to\infty$ as $ t\to\infty$, so that $ v^{2}\to v_{0}^{2}-\frac{2k}{x_{0}}>0\Rightarrow
v_{0}>\sqrt{\frac{2k}{x_{0}}}$.

(d)$ k=95040$ and $ v_{0}>6.93$ mi/sec.

21.
(a) i (b) iii (c) v (d) ii (e) iv





Jason Grout 2003-03-19