Homework Help: Differential Equation Proof

1. Jan 13, 2006

I need some help.

A pot of boiling water at 100C is removed from a stove at time t = o and left to cool in the kitchen. After 5 min, the water temperature has decreased to 80C, and another 5min later it has dropped to 65C. Assume Newtons law of cooling (dT/dt = k(M - T) ) applies, determain the constant temperature (M) of the kitchen.

Okay, so the solution to the differential equation is - T= Ce^(kt) + M and we want to solve for M

T(0) = C +M
100 = C + M

Equation 1.
T(5) = Ce^(k5) + M
80 = Ce^(k5) + M

Equation 2.
T(10) = Ce^(k10) + M
65 = Ce^(k10) + M

Therefore.. the 2 equations are
80 = Ce^(k5) + M
65 = Ce^(k10) + M

and if I subtract them, I get

15 = Ce^(k5) - Ce^(k10)

And this is where im lost. I have no idea how to solve for M

Thanks

2. Jan 13, 2006

saltydog

So you have:

$$\frac{dT}{dt}=k(M-T)$$

or:

$$\frac{dT}{M-T}=kdt$$

Integrating:

$$-ln(M-T)=kt+C$$

or:

$$ln(M-T)=C-kt$$

(minus c, plus c same dif it's arbitrary)

so that:

$$M-T=e^{C-kt}$$

or:

$$T(t)=M-ce^{-kt}$$

So you have 3 conditions, and 3 unknowns, kinda' messy but try and make those substitutions and then solve for something like M in terms of c, c in terms of k, then substitute that into one of them to solve for k, then the others.

Last edited: Jan 13, 2006
3. Jan 13, 2006

HallsofIvy

You are given the temperature at 3 times because you have 3 unknown numbers (C, k, and M) and need to solve 3 equations. You wrote the third equation but didn't number it or use it: C+ M= 100.

You can use that to write the two equations 80= Ce5k+ M and 65= Ce10k+ M as 80= (100-M)e5k+ M and 65= (100-M)e10k+ M. Now, you want to eliminate k, not M, from those equations. Since e10k= (e5k)2 one way to do that is to write (80- M)2= (100-M)2e10k and 65- M= (100-M)e10k and divide one equation by the other. You get
$$\frac{(80-M)^2}{65-M}= 100- M$$.
That's a quadratic equation for M.

4. Jan 14, 2006

Salty dog, when i originally did solved the differential equation, i got the same answer as you, T= M - Ce^(-kt)

However, my text book solution gives T= Ce^(kt) + M.

Now im really confused. which one is right?

Well... I did land at the same answer for M for both answers, so i assume they are both correct

Last edited: Jan 14, 2006
5. Jan 14, 2006

Integral

Staff Emeritus
Since C, k, and M are arbitrary constants their algebraic sign is arbitrary until you have determined their value.

6. Jan 14, 2006

saltydog

Ok guys. Didn't notice that.

They are the same equation when solved for the constants:

$$T=M-Ce^{-kt};\;M=20,\;C=-80,\;k=0.0575$$

$$T=M+Ce^{kt};\;M=20,\;C=80,\;k=-0.0575$$

7. Jan 14, 2006

okay i have one more differential equation....

m(dv/dt) = mg - bv

where m is the mass, g is the acceleration of gravity, and b>0 is a constant.
if m = 100kg, g = 9.8m/s^2 and b=5 kg/sec, and v(0) = 10m/s, solve for v(t). what is the terminal velocity of the object.

So firstly, i want to solve for this differential equation.
But i cannot find a way to seperate it. (i mean, its not seperable?)

m(dv/dt) = mg - bv
(dv/dt) = g - bv/m

then i pluged in the numbers and got

(dv/dt) = 9.8 - 0.05v

and thats about as far as i could get.

could somebody please give me a little push

thanks

8. Jan 14, 2006

qbert

It is seperable, I'm not sure that's the
way i would solve it but:

(dv/dt) = g - (b/m)v => dv/(g - vb/m)= dt

then a change of variables v -> vb/m turns this
into a simple integral.

9. Jan 14, 2006

oh actually you are right...

dv/(g - vb/m)= dt and you sub in the constants to get

dv/(9.8-0.05v) = dt and solve

(-1/0.05)ln|9.8 - 0.05v| = t + c

and isolate for v to get

v(t) = 196 - 20e^(-0.05t)(c)

and given that v(0) = 10, you can solve for C to get

v(t) = 196 - 186e^(-0.05t)

Now.. the problem asks "What is the limiting (i.e, terminal) velocity of the object?"

Would that just be the limit t-> infinity ? that would get 10m/s?

10. Jan 14, 2006

qbert

the idea is right:

v_term = limit (t->infinity) 196 - 186 exp(- t/20) = 196.

given v(0) = v0 and m dv/dt = mg - bv we get:
v(t) = gm/b + (v0 - gm/b)exp(-bt/m)

and v_terminal = limit (t-> infinity) v(t).
thus v_terminal = gm/b

[ as lim t->infinity exp(-k*t) = 0 if k>0 ]

we can check the answer, since at v_terminal
there is no acceleration. that is dv/dt = 0 at v_terminal
and the equation of motion is:

m dv/dt = 0 = m g - b v_terminal
=> v_terminal = mg/b.

i always like problems that have built in checks
to see whether or not i get the right answer.