Question

Related Rates problem...

A puddle is evaporating in such a way that its diameter is decreasing at a rate of 0.1 cm/min. At what rate is the area of the puddle decreasing when the diameter is 8cm.

Answer 1

ooohhh funn =]

Answer 2

noo.. I'm horrible at these and word problems in general.

Answer 3

The area of a circle is
\[A = \Pi r^2\]

Answer 4

relevant equation

\[A=\frac{\pi}{4}d^2\]

Answer 5

you're trying to fnd

\[\frac{dA}{dt}=?\]
you know the rate of change of thediameter

\[\frac{dD}{dt}\] lol double d=]

Answer 6

if you take the derivative of the original equation.. .what do you get =]

Answer 7

Your answer should be:
\[-4\Pi\]

try it first and let me know how it goes.

Answer 8

\pi is the code for pi =]...

Answer 9

ahh, hold on ;/

Answer 10

\[ \pi \ ] without the bracket being out in the middle of nowhere

Answer 11

It might help if you write down all your givens first, monroe. That will help guide you in the right direction.

Answer 12

A'=(pi/4)*2d ?

Answer 13

You already know your dr/dt

Answer 14

almost when you took the derivative of \[d^2\] in terms of time... you forgot something... there was no t.... soooooo

Answer 15

in terms of time>?

Answer 16

ok...

Answer 17

yes \[\frac{d}{dt}\]if you had something like this

\[\frac{d}{dt}[t^2]=2t\].

however you have

\[\frac{d}{dt}(d^2)\] we can't just say it's 2d ... think about when you implicitly differentiated before

Answer 18

so because it's d/dt the d changes to t?

Answer 19

ughhh.. okay soo.. 2d(d/dt)

Answer 20

\[\frac{dD}{dt}\]...

Answer 21

ughhh.. ;[

Answer 22

A'=\[\frac{ \pi }{ 4 }*2d \frac{ dD }{ dt }\]

Answer 23

I didn't see that you wrote "you know the rate of change of thediameter dD/dt" at the top.

Answer 24

think back when you did chain rule

what we are noting is that the d in our equation is some sort of function of time otherwise

\[d=f(t)\]

so going back to chain rule and the way i showed you earlier
if \[d=f(t)\] and \[y=cd^2 \]

using chain rule, \[\frac{dy}{dt}=\frac{dd}{dt}*\frac{dy}{dd}\]

Answer 25

so whenever you differentiate in respects to a term that is not within the function itself you have to think of all other terms as a function of time

Answer 26

and yep what you said was correct.. and now you have everything you need to solve ... just plug in and you'll get the rate at which your area is changing =] fun huh?

Answer 27

what did you get

Answer 28

I have no idea what he's talking abt. Lol

Answer 29

A'=\[\frac{ \pi }{ 4 }2*8*\]
wait is 0.1 dD/dt?

Answer 30

\[\frac{ dA }{ dt } = 2\pi4(0.1)\]

Answer 31

-.1

Answer 32

yeahh yea.. because it's decreasing right>?

Answer 33

yes dD/dt = the rate of change of your Diameter in respect to time or

\[\frac{\Delta D}{\Delta t}\]

Answer 34

-0.4pi cm^2/min :)

Answer 35

and btw i do not know how you get 4 for 2d thats why you don't do it with radius you have to multiply by 2 to get it back to diameter =]

Answer 36

... me?

Answer 37

oh mkay.

Answer 38

lol

Answer 39

actually i take that back you can't even do that... the rate of radius is much different than the rate of diameter