Question

@Michele_Laino

Answer 1

"A circle is growing so that each side is increasing at the rate of 5cm/min."

It makes no sense, since when do circles have sides?

Answer 2

Skip this question for now, and have a talk with teacher about it, if you can?

Answer 3

I don't know how to solve such question...

Do it means that radius increasing at the rate of 5cm/min?

I think this should be on hold. Teacher should understands.

Answer 4

I think that @geerky42 is right, namely 5 cm/min may be the rate of increasing of the radius of your circle

Answer 5

You can go ahead and assume that then.

Answer 6

ok!, please @wade123 what is the area of a circle whose radius is r?

Answer 7

ok!, then calculate the derivative:
\[\frac{ da }{ dt }\]
keeping in mind that r is a function of t, namely r=r(t)

Answer 8

derivative of pi*r^2 respect to time

Answer 9

please note that also r is a function of t, so you have to apply the chain rule

Answer 10

..

Answer 11

your derivative is:
\[\frac{ da }{ dt }=\frac{ d (\pi*r ^{2}) }{ dt }=2*\pi*r*\frac{ dr }{ dt }\]

Answer 12

is it clear? please?

Answer 13

it can't be, because as I explained before, r is a function of t, so, in order to calculate the derivative of a=pi*r^2, you have to apply the chain rule, or the rule of the derivative of composed function

Answer 14

now, is it clear? please?

Answer 15

or in other words, in order to calculate the derivative of pi*r^2, you have to apply the subsequent formula:
\[\frac{ d(\pi*r ^{2} )}{ dt }=\frac{ d(\pi*r ^{2} )}{ dr }*\frac{ dr }{ dt }\]
that's is the chain rule

Answer 16

please, apply the rule which I wrote above

Answer 17

you have to multiply your derivative by dr/dt

Answer 18

namely 2 pi r*dr/dt, that's all

Answer 19

thats it?

Answer 20

@wade123
now insert your numerical data in place of dr/dt and r into formula:
\[\frac{ d(\pi*r ^{2} )}{ dt }=2\pi*r*\frac{ dr }{ dt }\]
and you will get your ratio

Answer 21

ok! then wait a moment please!

Answer 22

@perl please check

Answer 23

yes that is correct

Answer 24

Yeah it looks good. If \(\Large A = \pi r^2\) then \(\Large \frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)

Answer 25

dA/dt = 2*Pi*r * dr/dt

Answer 26

Plug in your radius r and the speed at which the radius is changing (dr/dt) to get dA/dt. The value of dA/dt represents how fast the area is changing at that given radius.

Answer 27

so you plug in r = 20

Answer 28

if the radius is increasing at a rate of 5 cm/s, then dr/dt = 5

Answer 29

sorry 5 cm/min, but dr/dt = 5 still

Answer 30

so now where do i plug these in??

Answer 31

into \(\Large \frac{dA}{dt} = 2\pi r \frac{dr}{dt}\)

Answer 32

your goal is to find dA/dt because it asks " How fast is the area of the circle changing"

Answer 33

correct

Answer 34

yep

Answer 35

close, square cm/min

Answer 36

the area is in square cm, not just cm

Answer 37

cm/min^2 is an acceleration, so that is not what you want

Answer 38

you want cm^2 per min

Answer 39

area is in cm^2 and that changes per min