Question

Related Rates (Please help!)
If an object weighs 100kg on earth, then it weighs 100(1+1/4000r)^-2 kg when it is r km above the earth. At what rate is the weight decrease when object is 800km above the earth if its altitude increase at 15km/sec.
Thanks!

Answer 1

I'm having a problem with the wording of this question... kilograms are a unit of mass, not weight, so are independent of gravitational influence. But beyond that, is it

\[W = \frac{w}{\left(1+\frac{1}{4000r}\right)^2}\]
Because that equation also doesn't make a lot of sense, since at r=infinity, the "weight" that it gives would be the same...

Answer 2

with w=100 for this problem

Answer 3

the equation is supposed to be like this :
\[100(1+\frac{ 1 }{ 4000 }r)^{-2}\]

Answer 4

OH! That at least gives us something useful, regardless of the wording ^_^
So then it's a bit nicer. Can you just take the derivative wrt time of
\[W = 100kg\left(1+\frac{1}{4000}r\right)^{-2}\]

Answer 5

I'm not sure how to? Do I use the product rule for this?

Answer 6

Yeah. the form can be analogous to if we had
\[y=100(1+ax)^{-2}\]
So take the derivative of the stuff on the inside of the parenthesis and drop the power
\[\frac{dy}{dt} = \frac{dx}{dt}\frac{-2}{4000}\left(100(1+ax)^{-3}\right)\]

Answer 7

ops, sorry, got the example mixed up with the problem, 1 sec

Answer 8

\[\frac{dy}{dt} = -2a\frac{dx}{dt}(1+ax)^{-3}\]

Answer 9

grrr, there's a 100 in there, too. Last time's the charm

\[\frac{dy}{dt} = -2a\frac{dx}{dt}(100(1+ax)^{-3}\]

Answer 10

Make sense, or go through it step by step?

Answer 11

Okay, I think I got it.

Answer 12

Sorry for the wait...

Answer 13

No worries ^_^ What'd ya get for the answer?

Answer 14

\[dw/dt = 100(2(1+\frac{ 1 }{ 4000 }(dr/dt))^{-3}\]
I'm not sure if this is correct?

Answer 15

The thing that's in the parenthesis never changes, but the power it's raised to is correct - the only thing you need to change is to have the dr/dt in the parenthesis just be an r, and then multiply the whole thing by the dr/dt.

Answer 16

Like if you had
\[y = (3+ax)^-2\]
\[u = (3+ax)\]
\[y = (u)^-2\]
\[\frac{du}{dt} = a\frac{dx}{dt}\]
\[\frac{dy}{dt} = -2\frac{du}{dt}(u)^{-3}\]
you'd end up with
\[\frac{dy}{dt} = -2a\frac{dx}{dt}(3+ax)^{-3}\]

Answer 17

So it would like this:
\[\frac{ dw }{ dt } = -200(1+\frac{ 1 }{ 4000}r)^{-3}\]

Answer 18

Almost. You still need the dr/dt out front. In the example
y = w
a = 1/4000
x = r
and we have 100 as a coefficient outright, so
\[\frac{dw}{dt} = 100*\left(\frac{-2}{4000}\right)\frac{dr}{dt}\left(1+\frac{r}{4000}\right)^{-3}\]

Answer 19

Oh okay. Sorry, the equation looks weird to me, so that's why I didn't know how to take the derivative out of it. >.<

Answer 20

It's strange, for sure, and the chain rule applied to functions of a variable are kinda strange in any case ^_^ It will be drilled into your head in no time ^_^

Answer 21

okay, what do we do next?

Answer 22

That was the hard part! Now you just have to plug in your knowns - the instantaneous radius and the dr/dt (the rate at which you're moving away from the planet)

Answer 23

So it would be -125/288 kg/sec??

Answer 24

that's what I got ^_^

Answer 25

YAY! I try redoing the derivative of the equation, and I think I got now. :D
So first we take the devrivatve of the equation, and second, sub in the known values to get the rate.

Answer 26

Thank you so much! ^_^

Answer 27

Exactly yup yup ^_^ Since it already gives you how the weight varies with the radius, taking the derivative wrt time gives you the rate of change of the weight in relation to the rate of change of the radius.

Very very welcome ^_^