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course mth 173
7/15 11
The Celsius temperature of a hot potato placed in a room is given by the function T = 40* 2- .007 t + 24 , where t is clock time in seconds and T is temperature in Celsius. At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.9 seconds seconds?
T(6.8) = 40 * 2^(-.007(6.8)) + 24 = 62.70178
T(6.9) = 40 * 2^(-.007(6.9)) + 24 = 62.68301
(62.683-62.70178)/.1 = -.1878
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.81 seconds?
T(6.8) = 40 * 2^(-.007(6.8)) + 24 = 62.70178
T(6.81) = 40 *2^(-.007(6.81))+24 = 62.6999
(62.6999-62.70178)/.01 = -.1875
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.801 seconds?
T(6.8) = 40 * 2^(-.007(6.8)) + 24 = 62.70178
T(6.801) = 40 * 2^(-.007(6.801)) + 24 = 62.70159
(62.10159-62.70178)/ .001 = -.18567
What do you estimate is the rate at which temperature is changing at clock time t = 6.8 seconds?
-.18
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The result is certainly getting closer to -.18, but I don't think, from the trend of your numbers, that it's going to make it that far.
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The rate at which the Celsius temperature of a hot potato placed in a room is given by Rate = .041 * 2- .007 t, where R is rate of change in Celsius degrees per second and t is clock time in seconds. How much temperature change do you estimate would occur between t = 6.8 and t = 13.6 seconds?
Temperature is approximately equal to the average rate of change * dt.
R(6.8) = .041 * 2 ^ (-.007*6.8) = .0397
R(13.6) = .041 * 2 ^ (-.007*13.6) = .0384
average rate = (.0397+.0384)/2 = .0391
Temperature change = .0391 * (13.6-6.8) = .26 celsius degrees
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Very good.
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Good work on both problems, but do see my note.
I actually don't think the rate is going to make it as far as -.18560.
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