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course MTH 173
6/16 1:00
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?
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.81 seconds?
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.801 seconds?
What do you estimate is the rate at which temperature is changing at clock time t = 6.8 seconds?
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?
F(t final)-f(t initial)/ interval
T=6.8 &6.9
(f( 62.68300844)- f( 62.70178211))/.1= -.1877367 C
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It's not clear why you're evaluating f at 62.683, 62.702, etc.. Nor is it clear what your function f is, since no f function is defined in the problem.
It's fine to refer to a function f, as long as you've first defined it.
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t=6.8 &6.81
(f( 62.69990433)- f( 62.70178211))/.1= -.187778 C
t=6.8 &6.801
(f( 62.70159433)- f( 62.70178211))/.1= -.187782158 C
####I do not understand how we find the derivative function of t=40*2^-.007t+24 to find the rate at which the temp is changing at 6.8 secs????
(f(.0383818409) -f(.0396693267))/ 6.8= -1.893 C
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T = 40* 2- .007 t + 24 applied to the first problem.
This question applies to the new situation, in which the rate is
Rate = .041 * 2- .007 t,
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