Differentiation

Use differentials to find the change in mass m, as a percentage, when the radius r of a solid sphere increases by 2%. Note that the mass of a sphere if uniform density p is given by:

m = (4/3)p(pi)r³

I can't remember how to do these...I think you do dm/dr and then change r to 1.02 r...

Can someone please help me out?

Thanks!

Reply 1

Jonny W

The increase in mass is (approximately) 0.02r dm/dr.

As a percentatage of m, that's (100/m)*(0.02r dm/dr).

You should get a simple answer, not involving r or m.

Reply 2

InTeRRoGaToR

id like to help....but ur on the rong end of the kashmir dispute
and plus i dont know anyway
lol
nothin personal
me n my indian friends love the banter
tis well interesting

Reply 3

devesh254

anyone else have any ideas with this one? any help is appreciated! :smile:

Reply 4

Jonny W

This post fills in some details that I left out last time.

The increase in mass is (approximately)

0.02r dm/dr
= 0.02r * 4p pi r^2
= 0.08p pi r^3

As a percentatage of m, that's

(100/m)*(0.02r dm/dr)
= 8p pi r^3/m
= 8p pi r^3/[(4/3)p pi r^3]
= 6

Answer: 6%

[Alternative method. By the binomial theorem, (1.02)^3

\approx

1.06]

Reply 5

devesh254

Jonny W

[Alternative method. By the binomial theorem, (1.02)^3

\approx

1.06]

yeah thats what i was thinking...but it seemed too easy...

cheers mate!

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