Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 77

$ a.\quad m\displaystyle \frac{dv}{dt}=mg-kv^{2} \quad v=0$ when $ t=0.\quad$ (shown below) $b.\quad\sqrt{\dfrac{mg}{k}}$ $ c.\quad\approx178.89$ ft/sec

$ a.\quad$ $v=\sqrt{\frac{mg}{k}}\tanh(\sqrt{\frac{gk}{m}}\cdot t)$ Use the table of derivatives, and the chain rule do find $\displaystyle \frac{dv}{dt}$ Variables or constants that are not $t$, are constants in the derivation. $\displaystyle \frac{dv}{dt}=\sqrt{\frac{mg}{k}}\cdot{\rm sech}^{2}(\sqrt{\frac{gk}{m}}t)\cdot(\sqrt{\frac{gk}{m}})=g\cdot{\rm sech}^{2}(\sqrt{\frac{gk}{m}}t)$ $m\displaystyle \frac{dv}{dt}=mg\cdot{\rm sech}^{2}(\sqrt{\frac{gk}{m}}t)$ Apply the identity: $\quad \tanh^{2}x=1-{\rm sech}^{2}x$ $m\displaystyle \frac{dv}{dt}=mg\cdot[1- \tanh^{2}x(\sqrt{\frac{gk}{m}}t)]$ $m\displaystyle \frac{dv}{dt}=mg-mg[\tanh^{2}(\sqrt{\frac{gk}{m}}t)]$ The second term is similar to: $v^{2}=\displaystyle \frac{mg}{k}\tanh^{2}(\sqrt{\frac{gk}{m}}t)$ Thus: $m\displaystyle \frac{dv}{dt}=mg-k\cdot\frac{mg}{k}[\tanh^{2}(\sqrt{\frac{gk}{m}}t)]$ $m\displaystyle \frac{dv}{dt}=mg-kv^{2}$ Which we needed to show When $t=0$ $v=\sqrt{\frac{mg}{k}}\tanh(0)$, and, $\displaystyle \tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\Rightarrow\tanh 0=\frac{1-1}{1+1}=0$ So, $v=0$ when $t=0.$ $ b.\quad$ $\displaystyle \lim_{t\rightarrow\infty}v=\lim_{t\rightarrow\infty}\sqrt{\frac{mg}{k}}\tanh(\sqrt{\frac{kg}{m}}t)$ $=\displaystyle \sqrt{\frac{mg}{k}}\lim_{t\rightarrow\infty}\tanh(\sqrt{\frac{kg}{m}}t)$ $\displaystyle \tanh x=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\qquad $ dividing the numerator and denominator with $e^{x}$ gives: $\displaystyle \tanh x=\frac{1-e^{-2x}}{1+e^{-2x}}\qquad ... e^{-2x}\rightarrow 0$ when $ x\rightarrow\infty$ so the above limit exists and equals 1 $\displaystyle \lim_{t\rightarrow\infty}v=\sqrt{\frac{mg}{k}}(1)$ $\displaystyle \lim_{t\rightarrow\infty}v=\sqrt{\frac{mg}{k}}$ $ c.\quad$ Substitute $mg$ with $160$, and $k $ with $0.005$ $\sqrt{\dfrac{160}{0.005}}=\sqrt{\dfrac{160,000}{5}}$ $=\displaystyle \frac{400}{\sqrt{5}}=80\sqrt{5}$ $\approx178.89$ ft/sec