1. ## Differentials

Newton's law of gravitation states that the force of F of attraction between tow particles having masses m and M is given by F = (gmM)/(s^2) for a constant g and the distance s between the particles. If s = 20cm, use differentials to approximate the change in s that will increase F by 10%.

2. ## Ok

Originally Posted by smschaefer
Newton's law of gravitation states that the force of F of attraction between tow particles having masses m and M is given by F = (gmM)/(s^2) for a constant g and the distance s between the particles. If s = 20cm, use differentials to approximate the change in s that will increase F by 10%.
You have $\displaystyle \frac{dF}{ds}=\frac{-2(gMm)}{s^3}$..so $\displaystyle dF=\frac{ds\cdot{-2(gMm)}}{s^3}$ so imput yoru values and calculate $\displaystyle ds$

3. What?!?!?

4. ## ..................

Originally Posted by smschaefer
What?!?!?
What do you mean what...that is what you asked for isnt it? Using differentials?

5. i just dont get it... is there another way you could explain it or expand more on what you did.

6. ## ok

Originally Posted by smschaefer
i just dont get it... is there another way you could explain it or expand more on what you did.
Most of what I am about to say is not technically correct but it will help you understand....since a derivative is measuring slope you have the derivative is $\displaystyle \frac{dy}{dx}\Rightarrow{\frac{\Delta{y}}{\Delta{x }}}$...therefore $\displaystyle dy\approx{\Delta{y}},dx\approx{\Delta{x}}$...so you take the derivative so you can get your changes in x and y...now that you have that seperate...so you are trying to find the change in s($\displaystyle ds$) that makes F change 10%...make sense?

7. that sounds a little bit more familiar, but how did you get -2 and s^3 in the problem?

8. ## Because

Originally Posted by smschaefer
that sounds a little bit more familiar, but how did you get -2 and s^3 in the problem?
you have $\displaystyle F=\frac{gMm}{s^2}$...now to get $\displaystyle \frac{dF}{ds}$ which is what we want we have to differentiate both sides which is how I got that

9. ok i think i got it...thanks

10. ## Yep

Originally Posted by smschaefer
ok i think i got it...thanks
No problem...thanks

11. ## more help

if i am suppose to plug in s=20 to the equation how do i factor in the 10%.... dF= (ds(-2(gmM))) / 20