Differential Equations: Translate Word Problem

[csmajor86]

The problem states:

"Write a differential equation for each of the following scenarios.
a. The rate of change of the area of a disk over time is proportional to the product of its radius and the rate of change of its radius.

b. The rate at which heat is lost from an object is proportional to the difference between the temperature of the object and the constant ambient temperature."

I don't think these problems should be that hard but I am kind of lost as where to even start.

Part a.
I obviously know that the area (I'll call it a) is equal to (pi)r[sup:2vd75kle]2[/sup:2vd75kle].
a = (pi)r[sup:2vd75kle]2[/sup:2vd75kle]
The "rate of change of the area of a disk over time" I know is da/dt
but I don't know what da/dt should equal. I'm thinking it's Logistic Growth problem because the rate at which a(t) grows is dependent on the current size of the radius. What confuses me is that although it appears to be a Logistic Growth problem I can't figure out what the "K" term would be because unlike population there is no "saturation level" Any help would be much appreciated.

Part b. I'm guessing that this one is also some how logistic growth. Because it is the only kind of problem we have learned about where the rate of growth changes over time. If f(t) = y is the function where y = the temperature of an object at time t, then dy/dt is the rate at which heat is lost from that object. I'm thinking the ambient temperature is "K" term. I'm thinking maybe the "k" term is just f(t) - K.

Again, any help at all would be very helpful.

Thanks,
Travis

 

[soroban]

Hello, Travis!

Write a differential equation for each of the following scenarios..

(a) The rate of change of the area of a disk over time is proportional to
the product of its radius and the rate of change of its radius.

\(\displaystyle \text{The rate of change of the area is: }\:\frac{dA}{dt}\)

\(\displaystyle \text{The radius is }r\text{ and the rate of change of the radius is }\frac{dr}{dt}\)

\(\displaystyle \text{Therefore: }\;\frac{dA}{dt} \;=\;k\!\cdot\!r\!\cdot\!\frac{dr}{dt}\)

(b) The rate at which heat is lost from an object is proportional to the difference
between the temperature of the object and the constant ambient temperature.

\(\displaystyle \text{The rate of heat loss is: }\:\frac{dH}{dt}\)

\(\displaystyle \text{The termperature of the object is: }T\)
\(\displaystyle \text{The ambient temperature is: }T_o\)

\(\displaystyle \text{Therefore: }\;\frac{dH}{dt} \;=\;k\left(T - T_o\right)\)