# Find the limit as t approaches infinity of T(t).

• Apr 4th 2010, 10:23 AM
Find the limit as t approaches infinity of T(t).
I have this problem and it seems pretty easy. I think I'm just overthinking it.

At time t = 0 minutes, the temperature of a cup of coffee is 180 degrees Fahrenheit. Left in a room whose temperature is 70 degrees Fahrenheit, the coffee cools so that its temperature function T(t), also measure in degrees Fahrenheit, satisfies the differential equation: dT/dt = -1/2(T) + 35.

Find $\lim_{t \to \infty} T(t)$. Explain what this means in the context of the problem.

What is it asking me to do when it says " $\lim_{t \to \infty} T(t)$?" Is it asking me to set up an equation and solve (how would I do that?), or just explain what's happening? Also, can someone briefly explain what is happening? I'm stumped. Thanks for any help.

the limit as t approaches infinity of T(t)
• Apr 4th 2010, 10:50 AM
shawsend
Quote:

I
What is it asking me to do when it says "Find the limit as t approaches infinity of T(t)?" Is it asking me to set up an equation and solve (how would I do that?), or just explain what's happening? Also, can someone briefly explain what is happening? I'm stumped. Thanks for any help.

the limit as t approaches infinity of T(t)

It's asking you to solve the differential equation, obtain the function $T(t)$, and then take it's limit as $t\to\infty$ right?

You can solve $\frac{dT}{dt}=-1/2 T+35$ by separating the variables: $\frac{dT}{35-1/2 T}=dt$

See if you can get to $\ln(35-1/2T)=-t/2+c$

then plug in the initial values $T(0)=180$ to determine what $c$ is, then take exp of both sides, (or take exp of both side then figure out what c is) to get it to the form $T(t)$, then take the limit $\lim_{t\to\infty} T(t)$.
• Apr 4th 2010, 11:00 AM
So if I plug in 180 for T, I'd get $\ln(35-90)=-180/2+c$?

Then solve for c, which becomes 94.828, I think. Unless I skipped a step.

Then do I take the e of both sides? Is that what you mean by exp?
• Apr 4th 2010, 12:46 PM
shawsend
Dawg, you're getting the function name $T$ confused with the variable name $t$. Keep them straight. So if I have $\ln(35-1/2 T)=-t/2+c$ and $c$ is an arbitrary constant from the integration, then if I add, subtract, multiply anything to it, it's still a constant. Then doing a little algegra, I get:

$35-1/2T=ce^{-t/2}$

(still just an arbitrary constant c even though I took exponentials.

little more:

$T(t)=70-ce^{-t/2}$

Now substitute the initial conditions $T(0)=180$ to figure out that c=-110 to get:

$T(t)=70+110 e^{-t/2}$

Now, you take that limit of that as $t\to\infty$.
• Apr 4th 2010, 03:41 PM
Ok, that makes sense. I'm new to calc, so bear with me.

I can see you took the e of both sides, so the ln cancels with the e on left side, and on the right side the -t/2 becomes the exponent of e. But why is c multiplied to that?

After that I can see you moved 35 over and divided by -1/2 to get 70. I get that and I understand the rest.
• Apr 4th 2010, 04:00 PM
shawsend
$\ln(35-1/2 T)=-t/2+c$

so taking exponents of both sides:

$35-1/2 T=e^{-t/2+c}=e^c e^{-t/2}$

but $e^c$ is still just some arbitrary constant if c is so I can replace it with just the arbitrary constant c and get:

$35-1/2 T=ce^{-t/2}$
• Apr 4th 2010, 11:26 PM
Wasn't it obvious from the start that $\lim_{t\to\infty} T(t)= 70$ degrees?
Mathematically, that " $-\frac{1}{2}$ multiplying the T in the differential equation tells you that the solution will involve an exponential with negative exponent that will go to 0 as t goes to infinity.