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C3

"The force F newtons between two magnetic poles is given by the formula

F=1500r2 F = \displaystyle\frac{1}{500r^{2}} where r m is their distance apart.

Find the rate of change of the force when the poles are 0.2 m apart and the distance between them is increasing at a rate of 0.03 ms^-1"

dFdt=drdtdFdr \displaystyle\frac{dF}{dt} = \frac{dr}{dt}*\frac{dF}{dr}

dr/dt = 0.03 ms^-1 and df/dr = -1000r^-3.....where r = 0.2m so don't i just do for df/dr, -1000*(0.2^-3) and then multiply it by 0.03???

Thanks in advance.
boromir9111
"The force F newtons between two magnetic poles is given by the formula

F=1500r2 F = \displaystyle\frac{1}{500r^{2}} where r m is their distance apart.

Find the rate of change of the force when the poles are 0.2 m apart and the distance between them is increasing at a rate of 0.03 ms^-1"

dFdt=drdtdFdr \displaystyle\frac{dF}{dt} = \frac{dr}{dt}*\frac{dF}{dr}

dr/dt = 0.03 ms^-1 and df/dr = -1000r^-3.....where r = 0.2m so don't i just do for df/dr, -1000*(0.2^-3) and then multiply it by 0.03???

Thanks in advance.


Yes, but you've made a mistake in your differentiation for df/dr.

PS: "\times" will give you the multiply sign in LaTex
Reply 2
ghostwalker
Yes, but you've made a mistake in your differentiation for df/dr.

PS: "\times" will give you the multiply sign in LaTex


Thanks for your message but i don't see where i have gone wrong.....

1/500r^2 = 500r^-2 so df dr = -2*500r^-3 = -1000r^-3????
boromir9111
Thanks for your message but i don't see where i have gone wrong.....

1/500r^2 = 500r^-2 so df dr = -2*500r^-3 = -1000r^-3????


I could just tell you, but you make a lot of these little mistakes; so just telling you isn't helping.

So, go back over what I've quoted one character at a time if necessary.
Reply 4
ghostwalker
I could just tell you, but you make a lot of these little mistakes; so just telling you isn't helping.

So, go back over what I've quoted one character at a time if necessary.


oh yeah, 500 isn't a function.... just a constant, so that won't differentiate so we just have -2r^-3 = dF/dr???

edit - that still doesn't give me the answer so i am still going wrong somewheree
boromir9111
oh yeah, 500 isn't a function.... just a constant, so that won't differentiate so we just have -2r^-3 = dF/dr???


Not quite.

However it is the constant term that is incorrect.
Reply 6
ghostwalker
Not quite.

However it is the constant term that is incorrect.


Oh don't tell me i forgot to add the - to the 500 when flipping the fraction over??? making it 1000 instead of -1000
boromir9111
Oh don't tell me i forgot to add the - to the 500 when flipping the fraction over??? making it 1000 instead of -1000


No.

Doesn't look like you're going to get it, so it's in the spoiler.

Spoiler

Reply 8
boromir9111
Oh don't tell me i forgot to add the - to the 500 when flipping the fraction over??? making it 1000 instead of -1000


d/dr (r^-2) = -2r^-3

So it should be -2/500r^2 = -1/250r^3

I think.

I told you this because you don't seem to be getting anywhere with your constant :P:
boromir9111
"The force F newtons between two magnetic poles is given by the formula

F=1500r2 F = \displaystyle\frac{1}{500r^{2}} where r m is their distance apart.

Find the rate of change of the force when the poles are 0.2 m apart and the distance between them is increasing at a rate of 0.03 ms^-1"

dFdt=drdtdFdr \displaystyle\frac{dF}{dt} = \frac{dr}{dt}*\frac{dF}{dr}

dr/dt = 0.03 ms^-1 and df/dr = -1000r^-3.....where r = 0.2m so don't i just do for df/dr, -1000*(0.2^-3) and then multiply it by 0.03???

Thanks in advance.

Basically when calculating dFdr\dfrac{dF}{dr} rewrite F as F=r2500 F=\dfrac{r^{-2}}{500} and try differentiation again. In other words leave the constant alone whilst you're differentiating and simplify the resulting derivative.
AnonyMatt
d/dr (r^-2) = -2r^-3

So it should be -2/500r^2 = -1/250r^3

I think.

I told you this because you don't seem to be getting anywhere with your constant :P:

Almost right, the bold term has to be cubed. Typo most probably.
Reply 11
thanks guys for your help...... one little mistake and changed my entire result..... i have the correct answer now anways, much appreciated your help guys :smile:
Original post by Farhan.Hanif93
Basically when calculating dFdr\dfrac{dF}{dr} rewrite F as F=r2500 F=\dfrac{r^{-2}}{500} and try differentiation again. In other words leave the constant alone whilst you're differentiating and simplify the resulting derivative.

sounds like a silly question but why do we leave the constant whilst differentiating?
Reply 13
Original post by Lebkuchen
sounds like a silly question but why do we leave the constant whilst differentiating?


This is 13 years old (!) but the derivative basically measures the slope of the (tangent to the) function, so a constant multiplier in front of the function just multiplies the slope / gradient at each point.

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