Need help with 3 Calculus Extra Credit Problems

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November 21st 2008, 03:48 PM
Sm10389

Need help with 3 Calculus Extra Credit Problems

I'm in intro to calculus and I need help setting this equation up:

Question 1-(Additional Topics in Integration)-

Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80°F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40°F when it left the refrigerator and was 50°F after 20 minutes in the room.

Thanks!
November 21st 2008, 04:47 PM
skeeter

Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.

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

A = room temperature (a constant)
T = temperature of the cold drink at any time t in minutes
k = proportionality constant

separate variables and integrate.
November 21st 2008, 05:46 PM
Sm10389

Ok, with your help and help from another problem, I have this. Is it the next step?

dT/dt= -k(T-TM)

Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.

So:

dT/dt= -(T-80)

This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus :(
November 21st 2008, 06:36 PM
skeeter

$\frac{dT}{dt} = k(T-80)$

separate variables ...

$\frac{dT}{T-80} = k \, dt$

integrate ...

$\ln|T-80| = kt + C_1$

$|T-80| = e^{kt + C_1}$

$|T-80| = e^{C_1} \cdot e^{kt}$

$T - 80 = C_2 \cdot e^{kt}$

$T = 80 + C_2 \cdot e^{kt}$

The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time.
November 21st 2008, 06:53 PM
Sm10389

I greatly appreciate your help. I have:

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
How do I find k?

Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= Ce^20k
Again finding k has stumped me.

What does k represent and how do i find it?
November 21st 2008, 06:57 PM
skeeter

Quote:

Originally Posted by Sm10389

I greatly appreciate your help. I have:

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
good.

How do I find k?
Second value (50 degrees at time 20)
how about substituting in -40 for C ? then find k with the second value.
T= 80 + Ce^kt

do it.
November 21st 2008, 07:04 PM
Sm10389

you my friend are a genius.

so....

Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40

Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= -40e^20k
3/4= e^20k
ln3/4=20k
k= ln(3/4)/80

so would this be my final answer?

T= 80 + -40e^ln((3/4)/20)t
November 22nd 2008, 10:43 AM
Sm10389

skeeter

how does that look
November 22nd 2008, 11:37 AM
skeeter

Quote:

Originally Posted by Sm10389

how does that look

check it out yourself ... graph the result in your graphing utility and see if the given info matches up.
November 22nd 2008, 08:05 PM
Sm10389

i just did, and it is not correct. where did i goof up?
November 23rd 2008, 03:14 AM
skeeter

Quote:

so would this be my final answer?

T= 80 + -40e^ln((3/4)/20)t

looks like you have ...

$k = \ln\left(\frac{\frac{3}{4}}{20}\right)$

... which it ain't.

should be ...

$k = \frac{\ln\left(\frac{3}{4}\right)}{20}$
November 23rd 2008, 09:31 AM
Sm10389

Thank you, I had that too, I just did not use the parentheses correctly in my calculator.

The next step of the problem is to calculate it if the drink were warmer then room temperature (>80).

I know how to set it up like the last one, but we would only be given one point (0, 85) for example.

How would I calculate k here?
November 23rd 2008, 11:36 AM
skeeter

$k$ will remain the same because the rate of heat transfer will remain the same

... however, you'll have to recalculate $C_2$ .
November 23rd 2008, 01:12 PM
Sm10389

QUESTION #2

More from the Introduction to Differential Equations-

Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.

A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.

I came up with this:
V(t)= (C/r)*e^rt - (D/r)

I am sure it is correct. This is the next part:
Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of $8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?

I know how to do everything but:
Find C
Figure out how compounding continuously would affect the equation.

Please help me.