Finding the changing rate

Question

When a space shuttle is launched into space, an astronaut's body weight decreases until a state of weightlessness is achieved. The weight $W$ of a $150$-lb astronaut at an altitude of $x$ kilometers above sea level is given by $W=(\frac{6400}{6400+x})^2$. If the space shuttle is moving away from Earth's surface at the rate of $6\frac{km}{sec}$, at what rate is $W$ changing when $x=1000$km?

My Attempt: $$\begin{align} W'(x)=300\frac{6400}{6400+x}\frac{0(6400+x)-1(6400)}{(6400+x)^2}=\frac{300(-6400^2)}{(6400+x)^3} \\\frac{dW}{dt}=\frac{dW}{dx}\frac{dx}{dt}=\frac{300(-6400^2)}{(6400+x)^3}(6)=\frac{1800(-6400^2)}{(6400+x)^3} \end{align}$$ So what I have done is taken $W'(x)$ and made it $W'(t)$, i.e. $W$, instead of being a function of altitude is now a function of time. I think. This is where is gets fuzzy for me.

Question: Is what I am doing making $W$ a function of time? If so, do I just plug in $1000$km into $W'(t)$? The answer I got is ~$-0.1819$.

Any help or guidance is appreciated.:)

Answer 1

It's not entirely clear from the problem statement, but assume the astronaut's weight vs. altitude is $W=W_0\left( {c \over c+x}\right)^2$ where $W_0=150$ and $c=6400$. Then $$ {dW \over dt} = {dW \over dx}{dx \over dt} = vW_0c^2{d \over dx}(c+x)^{-2} = {2vW_0c^2 \over (c+x)^3} $$ For $v=6$ and $x=1000$, ${dW \over dt}=-0.1819 {lb \over sec}$. If $W_o=1$, then the answer is $-0.001213 {lb \over sec}$.

Answer 2

First convert $x$ to into a function of time. $x=6\frac{km}{sec}\cdot t$. now substitute that into the equation to give $W$ in terms of time, $W(t)=(\frac{6400}{6400+6\frac{lb}{sec}\cdot t})^2$ now differentiate that to get $W'(t)=-\frac{61440000}{(3t+3200)^3}$. now find the time when the space shuttle is at $1000\,km$, $\frac{1000km}{6\frac{km}{sec}}=\frac{500}{3}sec$. substitute this value for $t$ in $W'(t)$, $W'(\frac{500}{3})=-\frac{61440000}{(3\cdot\frac{500}{3}+3200)^3}\approx -.001213\frac{lb}{sec}$.