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I have an older Calculus book that I'm using to prepare for Calculus II. I'm just trying to brush up on some things. I find I often get thrown off by word problems.

"An object increases in height x above the surface of the earth, its weight w decreases because the pull of gravity on the object decreases. Suppose that a rocket weighs 80/(x + 4000)^2 million pounds at an altitude of x miles and is rising at the constant rate of 100 mi/min. Find the rate of change of weight with respect to time at any instant."

I'm given the answer dw/dt = -16,000/(x + 4000)^3, but I'm not sure how to arrive at this answer.

I suspect this has something to do with the Chain Rule. I know the derivative of (x + 4000)^2 = 2(x + 4000), gravity is s = 16t^2 and that the acceleration of the rocket is a=100. The velocity of the rocket would be v(t)=100t and the distance would be d(t) = 50t^2. Unfortunately I am not able to put everything together to solve this equation.

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We have $w = \displaystyle \frac{80}{(x+4000)^2}$, so

$\displaystyle\frac{dw}{dt} = -\frac{160}{(x + 4000)^3}\frac{dx}{dt}$.

We are given that the rocket changes in height at a constant rate of $100$ miles per pinute. So thus $\frac{dx}{dt} = 100$.

Then $\displaystyle\frac{dw}{dt} = -\frac{160}{(x + 4000)^3}\frac{dx}{dt} = -\frac{16,000}{(x + 4000)^3}$, as desired.

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    David, thanks for the quick response. This makes sense. I was able to use the quotient rule to get -160/(x+4000)^3. – 2017-01-01
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    Charles, you don't even need to use the quotient rule there, but simply the chain rule. You have $-160(x+4000)^{-2}$ and can simply apply the chain rule. Of course, quotient rule works too, but is more cumbersome. Glad I could help! – 2017-01-01
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    So it seems that the initial equation should be dw/dx = -160/(x+4000)^3? You put dw/dt. And using the Chain Rule we'd have dw/dt = dx/dt * dw/dx? – 2017-01-01
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    Right, the Chain Rule from the start would have been much easier. Thanks for pointing that out. – 2017-01-01
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    No, what I have is right. What I wrote was $\frac{dw}{dt} = \frac{dw}{dx} \frac{dx}{dt}$, where I actually wrote out $\frac{dw}{dx}$. I should've been more clear. – 2017-01-01