Question

A circle is growing so that each side is increasing at the rate of 5 cm/min. How fast is the area of the circle changing at the instant the radius is 20 cm? Include units in your answer.

I know this is a related rates question but it seems odd is all sides refering to a linear scale? so then it would correlate with the radius, or the circumference and if circumference i am having trouble getting a dc/dt to work so i can sub in the 5cm/min. please help!

Answer 1

side of a circle = perimeter - yes ?

Answer 2

Thats what im geussing its just the all sides is odd, probably some bad coding on the test makesr part

Answer 3

The problem im having is finding a format that gets me to DC/DT so i can sub in the 5cm/min

Answer 4

the closest i have gotten is something like\[a=\pi * (\frac{ c^2 }{ 2\pi^2 })\]

Answer 5

but i think i may be wrong here

Answer 6

im not sure how to go about it once i get da/dt = pi d/dt (c^2/2pi^2)

Answer 7

Like seperating the fraction

Answer 8

i need an equation with DC/DT R and DA/DT somehow

Answer 9

i think, or im doing this all wrong would be nice to get some input

Answer 10

This is a nice "related rates" problem.

Please go back and check the wording of the original problem. You've written, "a circle is growing so that each side increases in length at the rate of 5 cm/min. But circles don't have sides, do they?

Answer 11

thats a direct copy-paste

Answer 12

it has to mean circumference

Answer 13

anything else doesnt make sense

Answer 14

If that's the case, it looks as tho' you have 2 choices:

Replace "circle" with "square"

or

solve the problem "a circle's circumference is increasing at the rate of 5 cm per min, and ... "

Answer 15

Can you draw a circle with four sides? I can't.

Answer 16

im trying to solve with circumference but am having difficulty formulating a problem

Answer 17

C = 2*Pi*r
A = Pi*r^2

Find the derivative with respect to time of each of the above. Don't forget to apply the Chain Rule.

Answer 18

A= 2 PI c/2pi^2?

Answer 19

but then would how do i derive it to get d/c/dt?

Answer 20

and if thats the case where does r go?

Answer 21

ive literally been trying to figure this out for 2 hours

Answer 22

dc/dt*

Answer 23

We're getting there. Again, A=pi*r^2.
Differentiating with respect to time, t, we get \[\frac{ dA }{ dt}=2\pi r \frac{ dr }{ dt }\]

Answer 24

Note: this is an application of the chain rule.

Answer 25

Now, consider C=circumference=2*pi*r

Answer 26

Yes, I got that on paper just didnt post it, thought it was a dead end

Answer 27

find\[\frac{ dC }{ dt }\]

Answer 28

Ahh

Answer 29

You'll need to use the Chain Rule here.

Answer 30

deriving 2 pi r gives me 1 or zero depending on if i sub in the r

Answer 31

can i just sub 2pi r for dcdt then?

Answer 32

rather
it would be dc/dt=2pi

Answer 33

is that correct?

Answer 34

Remember i do not even have a dr/dt unless i just asume 5cm/min is dr/dt, in which case this whole circumference buissness is poitnless

Answer 35

or would it be dc/dt=2pi dr/dt

Answer 36

Please let's go thru this problem again in summary fashion:

"solve the problem "a circle's circumference is increasing at the rate of 5 cm per min. How fast is the area of the circle increasing when t=3 sec?"

First, we need to relate circumference to area, since we want to know how fast the area is increasing at time t=3.

Circumference: C=2pi*r Area: A=pi*r^2

Let's solve for A in terms of C.

Are you following all right so far?

Answer 37

ok ok ok i have you know

Answer 38

What do we already know? First, the circumference, C, is increasing at the rate of 5 cm/min


Look, I have explained my strong belief that the problem as originally stated does not make sense, because a circle does not have sides.

I invented a related problem in the hope we could go thru it quickly and then apply the method to a correction of the original problem. Thus, t=3 was an invention.

I have serious issues with your "where the hell ... " question and your demand that I go back and read your problem. If you'd prefer to work with another person, fine.

Answer 39

I know you're frustrated, but I will not long tolerate your taking your frustration out on me.
Please make up your mind: do you want to continue or do you want to take a break and then let someone else try to help you?

Answer 40

When you say "let's go thru this problem again" and not "lets look at a similar example" its extremely confusing, im right there with you on the fact that it is incorrectly formatted, but that was equally as confusing as the problem itself

Answer 41

notice the phrase "this problem"

Answer 42

I would greatly apreciate the help and do apologize for my temperment but ive had a long day

Answer 43

I really really think they meant radius by "side length".
I seriously doubt they were meaning for that to be circumference.
Poorly written problem unfortunately, so it's hard to say.

Answer 44

Sorry, but I am not responsible if the original problem I posted makes no sense. If you have a flesh-and-blood teacher, I'd suggest you take that up with him / her.

Also sorry, but with the tone of this conversation being what it is, I'd prefer not to continue working with you, for the time being, at least. zepdrix has chosen to become involved. He is very capable and should be able to help you. Thank you, @zepdrix.

Answer 45

:3

Answer 46

Okay so we are dealing with a square then

Answer 47

that makes a whole lot more sense

Answer 48

Probably a squarcle
jk XD

Answer 49

yeah lol, honestly for all the work i put in i have a 1/3 chance of choosing the right starting point and there are 6 total questions and i have a chapter exam next lol

Answer 50

im ganna leave it with a note that the question makes little logical sense and move on, i have alot more ahead of me, so its not worth me wasting anymore of mine or especailly your time for a fruitless endevour

Answer 51

If it was meant to be the rate of change of circumference, \(\large\rm C'=5\)
Then we first need to get the C value, \(\large\rm C=2\pi r=40\pi\)

And replace the r with something involving C,
\(\large\rm A=\pi r^2=\pi\left(\frac{C}{2\pi}\right)^2\)

Then differentiate and plug in the pieces, yes? :)

Answer 52

fine fine fine :) Whatever you gotta do hehe

Answer 53

i got that exactly, but then realised i was missing r

Answer 54

If it was meant to be radial change, \(\large\rm r'=5\)
then of course the problem works out a little bit easier,\[\large\rm A=\pi r^2\]\[\large\rm A'=2\pi r\cdot r'\]

Answer 55

also how would you preform the d/dt for a fraction like that im a bit rusty

Answer 56

yeah that one is easy butt then i get a huge number

Answer 57

626 per min

Answer 58

628*

Answer 59

I don't know why you think having no r is problem.
We replaced it with C.

But anyway, notice the denominator is constant :)
Nothing to worry about there,
just pull it aside before differentiating,\[\large\rm A=\frac{\pi}{4\pi^2}C^2\]And apply simple power rule to the C.

Answer 60

@memman, it's hard to address your "how do I perform" question without actually seeing your work. It may take a while to share your work, but save time in the long run.

Answer 61

\[\large\rm A=\pi\left(\frac{C}{2\pi}\right)^2\quad=\quad\pi\left(\frac{C^2}{2^2\pi^2}\right)\quad=\quad \frac{\pi}{2^2\pi^2}C^2\quad=\quad number\cdot C^2\]Just power rule, yes? :)

Answer 62

You can do it merman!
Just dive deep into the ocean!
Find the pearl that you need!

Answer 63

Long day?
Humor gone?
Ok my bad.

Answer 64

So then after deriving i get ac/dt=2c*dc/dt

Answer 65

no i apreciate it anything to lighten the mood

Answer 66

I'm not sure what ac/dt is.
Is that supposed to be our area derivative, dA/dt?

Answer 67

I like the sneaky shortcut derivative notation myself,
those primes are so clean.

Whatever works for you though.
Probably good to get comfortable with Leibniz notation for these types of problems.

Answer 68

whoops da/dt

Answer 69

yeah MyB

Answer 70

\[\large\rm A=\frac{\pi}{2^2\pi^2}C^2\]So then,\[\large\rm A'=\frac{\pi}{2^2\pi^2}\cdot2C\cdot C'\]And earlier we determined that when the radius is 20,
C=20 and C'=5
plug and chug

Answer 71

woops, C=40pi*

Answer 72

So ya, your derivative looks good,
just don't forget about the junk in front.

Answer 73

But again, I really really think they meant for r'=5
XD

Answer 74

When I think of something growing in every direction,
the "sides" growing,
I'm thinking of it growing radially, outward.

It certainly could be interpreted as circumference though :)

Answer 75

ok well atleast i know now, i already submitted it like 5 min ago but i needed closure

Answer 76

I guess you would want to compare it to similar problems that you've been working on. It should match the level of difficulty of those problems.

Answer 77

Exactly its still useful material you guys are great, i also havent eaten since late lunch/dinner yesterday

Answer 78

i got a little involved in this....

Answer 79

been going since like 8 am

Answer 80

oo that's no good.
swim around, go find an eel or something D:
gotta stay fed

Answer 81

yikes

Answer 82

you deserve a million medals my freind

Answer 83

both of you

Answer 84

np :D
Study, do well, be blessed.

Answer 85

LOL that picture XD