#$&*
course Mth 173
6/13/13~12:50AM
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?
40*2^.007(6.8)+24→ 40*2^-.0476+24→ 62.70178degrees C
40*2^.007(6.9)+24→ 40*2^-.0483+24→ 62.68300degrees C
62.68300-62.70178=-.01878degrees
6.9-6.8=.1second → .01878 degrees/.1 second
= -.1878deg/sec
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.81 seconds?
40*2^-.007(6.81)+24= 62.69990degrees
62.69990-62.70178=.001875667degrees
6.81-6.8=.01sec → -.001875667/.01=.1875667degrees per second
At what average rate is the temperature of the potato changing between clock times t = 6.8 and t = 6.801 seconds?
T=40*2^-.007(6.801)+24→ 62.70159
6270159-62.70178=-.000185672 degrees
6.801-6.8=.001seconds
→ -.000185672/.001=-.1856722degrees per second
What do you estimate is the rate at which temperature is changing at clock time t = 6.8 seconds?
Around -.18degrees per second
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?
13.6-6.8=6.8
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You haven't identified this, but it's the change in the rate.
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.041*2^-.007(6.8)
-.039669
.041*2^-.007(13.6)
-.0078064
average rate = -.039669+-.0078064/2=-.0237377
average rate*change in time interval → -.0237377*6.8 =
-.16141636 degrees temperature change
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Fortunately you didn't use the change in the rate to calculate your result. I was worried for a minute that you would.
Very good solution.
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&#Your work looks very good. Let me know if you have any questions. &#