# assistance please

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• Nov 16th 2010, 07:06 PM
math321
assistance please
can any 1 help me get tru this single problem ...please
any help will be greatly accepted

A thermometer reading 11°C is brought into a room with a constant temperature of 20°C. If the thermometer
reads 18°C after 3 minutes, what will it read after being in the room for 5 minutes? Assume the cooling follows
Newton's Law of Cooling:
U = T + (U0 - T)e^kt.
• Nov 16th 2010, 09:44 PM
Unknown008
Use the curve given to you to find the value of the constants.

$U = T + (U_o - T)e^{kt}$

At time t = 3, the temperature is 18 C.

$18 = 20 + (11 - 20)e^{3k}$

Find the value of k.

Then, use the equation again with the value of k you obtained and t = 5. FInd the value of U. (Smile)

What do you get?
• Nov 17th 2010, 02:58 AM
HallsofIvy
$e^{kt}= left(e^k\right)^t$ so you really only need to find $e^k$, not k itself.
• Nov 19th 2010, 11:57 AM
math321
newton law of cooling
needs some assistance with this question please

A thermometer reading 11°C is brought into a room with a constant temperature of 20°C. If the thermometer
reads 18°C after 3 minutes, what will it read after being in the room for 5 minutes? Assume the cooling follows
Newton's Law of Cooling:
$
U = T + (U0 - T)e^{kt}
$

(Round your answer to two decimal places.)

reach as far as this
$
18 = 20 + (11 - 20)e^{3k}
$

dont no if it is correct
• Nov 19th 2010, 12:07 PM
Ackbeet
I think you'll find that, with the way you've written the equation, k < 0. Try to solve for k. What do you get?
• Nov 19th 2010, 02:25 PM
math321
• Nov 19th 2010, 02:27 PM
Ackbeet
Ok. So far, so good. Now you need to "undo" the exponential function in order to get at the exponent. How do you "undo" an exponential?
• Nov 19th 2010, 02:51 PM
math321
not sure how thats done
can u break it down futher5 for me please
• Nov 19th 2010, 05:22 PM
Ackbeet
What is the function inverse of the exponential function? If I have the equation $y=e^{x},$ how could I solve for $x?$
• Nov 20th 2010, 03:30 AM
math321
is it this

$
ln (\frac{2}{9}) = ln (e)^{3k}
$

$
3k ln (e) = ln (\frac{2}{9})
$

$
3k ln (e) = ln (\frac{2}{9})
$

$
3k = -1.5040
$

$
k = -0.501359133
$
• Nov 20th 2010, 03:34 AM
Ackbeet
Well, I would write it this way (being careful with parentheses):

$\ln\left(\dfrac{2}{9}\right)=\ln\left(e^{3k}\right ).$

But that's definitely the right idea. Now what?
• Nov 20th 2010, 03:51 AM
math321
$
-1.5040774 = 3kln (e)
$
• Nov 20th 2010, 03:53 AM
Ackbeet
Keep going!
• Nov 20th 2010, 03:56 AM
math321
$
-1.5040774 = 3k

k = -0.501359133
$
• Nov 20th 2010, 03:57 AM
Ackbeet
So, k = -0.501359133. Now what?
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