Rates of Change Differentiation Problem.

[onyx topaz]

I am stuck at part (iii). I found dC/dr and dA/dr. I used the formula want = have x need. But it doesnt get me the right answer.

[tulip mirage]

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[onyx topaz]

$dC/dr = 2pi$

[warped grotto]

Hanzie

[onyx topaz]

$dA/dr = 2pir$

[warped grotto]

Hanzie

[strong lake]

By the Chain rule, $\dv{C}{r}=\dv{C}{A}\dv{A}{r}$

[warped grotto]

Omegabet_

[strong lake]

to which it follows

[onyx topaz]

i thought it was $dC/dA = dC/dr x dr/dA$

[warped grotto]

Hanzie

[strong lake]

that's equivalent

also please learn LaTeX before using it

[onyx topaz]

i did that but i just cancels out and gives me a whole number

[onyx topaz]

mb

[strong lake]

it shouldnt

$2\pi=\dv{C}{A}\times(2\pi r)\to\dv{C}{A}=\frac{1}{r}$

[warped grotto]

Omegabet_

[onyx topaz]

im a little bit confused with that

[strong lake]

what about it?

[onyx topaz]

which chain rule is this

[strong lake]

... the chain rule?

there's only 1

[onyx topaz]

is it this one

[onyx topaz]

or this

[strong lake]

they're both the chain rule

but I used the one I wrote cause it has the stuff we know about

namely has dC/dr and dA/dr without having to invert anything

[onyx topaz]

ohhhh right i see it now

[strong lake]

again, your version of applying the chain rule is equivalent

[onyx topaz]

my one doesnt work for some reason

and thats the way we learnt it

WANT = HAVE x NEED

[strong lake]

$\dv{C}{r}\times\dv{r}{A}=\frac{\dv{C}{r}}{\dv{A}{r}}$

[warped grotto]

Omegabet_

[strong lake]

since $\dv{r}{A}=\frac{1}{\dv{A}{r}}$

[warped grotto]

Omegabet_

[strong lake]

to which you then get (2pi)/(2pi r) = 1/r

[onyx topaz]

ohhhh i see now