Question

Can you check my answer?
A circular oil slick of uniform thickness is caused by a spill of 1 cubic meter of oil. The thickness of the oil is decreasing at the rate of 0.1 cm/hr as the slick spreads. (Note: 1 cm = 0.01 m.) At what rate is the radius of the slick increasing when the radius is 8 meters? (You can think of this oil slick as a very flat cylinder; its volume is given by V = r2h, where r is the radius and h is the height of this cylinder.)

My Answer; 0.804 meters per hour

Answer 1

I don't think the volume of a cylinder is r2h

Answer 2

@ArkGoLucky, they say for this question, assume the volume of the oil slick cylander is \(2rh\)

Answer 3

opps I meant centimeters per hour..

Answer 4

? so is it \(V=r\,\,\,c/h\)

Answer 5

I like your name-thingy by the wayy

Answer 6

you know what, I think I just completely messed up on this problem..

Answer 7

Well, the volume is constant. You can start there.

Answer 8

Name thingy?

Answer 9

ahsome, she means your username

Answer 10

Oh, K ;)

Answer 11

Yeah sometimes I forgot certain words for things in english

Answer 12

*forget...

Answer 13

Um if the height of the "cylinder" is decreasing at a constant rate and the volume is really 2rh, then the radius should also decrease at the same constant rate

Answer 14

Well maybe not the same, but at a constant rate

Answer 15

I would be the reciprocal of the rate of decrease of the height so 10cm/hr

Answer 16

Wait nevermind. That's wrong

Answer 17

Okay so know I got 1.234

Answer 18

Ohhh

Answer 19

Volume of cylinder of radius \(r\) and height \(h\) is \(\pi r^2 h\) no matther who says what

Answer 20

So radius does change at changing rate. The problem is confusing.

Answer 21

I know right -_-

Answer 22

Are you sure that V=2rh? the units don't even work

Answer 23

cm^2 is not a unit of volume

Answer 24

maybe it's a mess up?

Answer 25

think so

Answer 26

I guess we can solve it assuming V=pi*r^2*h

Answer 27

Yeah

Answer 28

dh/dt = 0.01

decreasing sign
-0.1/100 = -2dr/8^3dt into pi

Answer 29

Ugh this problem is stressful. I will attack it again later. Good luck.

Answer 30

Think of one little cubic inch of the oil as a little circular disk with a certain radius and height.

Answer 31

So this is a rate of change problem in that each little cubic meter of this oil slick is changing at a constant rate (think of this cylinder decreasing in size). That means the volume of the slick is changing with respect to time, as well as the height and the radius of each little cylinder this slick is made up of.

Your problem messed up messed up in the format of the volume of a cylinder. It should be \[V_{cyl}=\pi r^2 h\]

Answer 32

wow @Jhannybean thanks for the medal

Answer 33

first medal i got from a mathlete

Answer 34

appreciated thanks

Answer 35

It says the `thickness` is changing of the cylinder. Here, the thickness is represented by the height of the cylinder. so in order to represent the change we label it \(\dfrac{dh}{dt}\) The value of this decrease is given as \(\dfrac{dh}{dt} =-0.1 \dfrac{\text{cm}}{\text{hr}}\)\[\sf r=8 m\]So in order to solve this, we have to take the derivative of the function with respect to time.\[V=\pi r^2h\]\[h=\frac{V}{\pi r^2}\]the \(\pi\) in this case is insignificant, therefore can be removed. Also, the volume is changing at a constant rate, so \(\dfrac{dV}{dt}=1\)\[\frac{d}{dt}\left(h=\frac{V}{r^2}\right)\]

Answer 36

\[(h)'=(V)' \cdot \left(\frac{1}{r^2}\right)'\]\[\frac{dh}{dt}=1 \cdot -\frac{2}{r^3}\cdot \frac{dr}{dt}\]Now just solve for \(\dfrac{dr}{dt}\)

Answer 37

Thanks everyone for the help :) I really appreciate it. And sorry I fell a sleep on you guys last night :0