Forming differential equationsfrom words to formula

[rock.freak667]

Homework Statement

The rate of increase of thickness of ice on a pond is inversely proportional ro the thickness of ice already present. It is known that,when the thickness is x cm and the temperature is of the air above the ice is -a degrees celsius, the rate is $\frac{ax}{14400} cms^{-1}$ . Form an appropriate diff. eq'n and show that, if the air temp is -10 degrees Cel. , the time taken for the thickness to increase from 5 com to 6 cm is a little more than 2 hours

Homework Equations

The Attempt at a Solution

"The rate of increase of thickness of ice on a pond is inversely proportional or the thickness of ice already present."

$$\frac{dx}{dt} \alpha x => \frac{dx}{dt}=kx$$

therefore $$lnx=Ae^{kt}$$ and that looks wrong as I have yet to use the given rate...I am confused about the thickness and the given rate

 

[EnumaElish]

Homework Statement

The rate of increase of thickness of ice on a pond is inversely proportional ro the thickness of ice already present. It is known that,when the thickness is x cm and the temperature is of the air above the ice is -a degrees celsius, the rate is $\frac{ax}{14400} cms^{-1}$ . Form an appropriate diff. eq'n and show that, if the air temp is -10 degrees Cel. , the time taken for the thickness to increase from 5 com to 6 cm is a little more than 2 hours

Homework Equations

The Attempt at a Solution

"The rate of increase of thickness of ice on a pond is inversely proportional or the thickness of ice already present."

$$\frac{dx}{dt} \alpha x => \frac{dx}{dt}=kx$$

therefore $$lnx=Ae^{kt}$$ and that looks wrong as I have yet to use the given rate...I am confused about the thickness and the given rate

How do you get $$\frac{dx}{dt} \alpha x => \frac{dx}{dt}=kx$$? Is that really inversely proportional?

 

[Kurdt]

The rate at a given point is merely there so you can determine your constant.

 

[rock.freak667]

oh my bad..i did the question wrong.supposed to be $$x^2=Kt+C$$..but I only have one piece of info and 2 constants 'a' and 'k'

 

[Kurdt]

You've missed a factor of a half. What you do know is that:

$$\frac{dx}{dt}=kx=\frac{ax}{14400} cms^{-1}$$ at some thickness x and temperature -a degrees C.

 

[rock.freak667]

You've missed a factor of a half. What you do know is that:

$$\frac{dx}{dt}=kx=\frac{ax}{14400} cms^{-1}$$ at some thickness x and temperature -a degrees C.

I multiplied through out by 2..2k=K and 2c=C ...so then k=a/14400 and then sub that in the eq'n

 

[HallsofIvy]

You've missed a factor of a half. What you do know is that:

$$\frac{dx}{dt}=kx=\frac{ax}{14400} cms^{-1}$$ at some thickness x and temperature -a degrees C.

And you have missed the point that EnumaElish raised: the rate is inversely proportional to the thickness, not directly proportional. Assuming x is the thickness of the ice at time t (I personally would insist upon that being explicitely stated), then
$$\frac{dx}{dt}= \frac{k}{x}$$
That gives a whole different situation!

 

[Kurdt]

And you have missed the point that EnumaElish raised: the rate is inversely proportional to the thickness, not directly proportional. Assuming x is the thickness of the ice at time t (I personally would insist upon that being explicitely stated), then
$$\frac{dx}{dt}= \frac{k}{x}$$
That gives a whole different situation!

Argh! I apologise. I did this myself at 3 am and was just about to post that he hadn't written it inversely proportional when enuma elish beat me to it, then I go and make the same mistake. Can't believe I did that. Anyway sorry to the OP for any confusion.