Asked by piper

With time, t, in minutes, the temperature, H, in degrees Celsius, of a bottle of water put in the refrigerator at time t=o is given by
H = 4 + 16e-.02t
How fast is the water cooling initially? After 10 minutes? Include units in your answer
Answered by Reiny
Do you have a typo?
Should it not be H = 4 + 16e^(-.02t) ?

if so, then ....

dH/dt = - .32e^(-.02t)

Initially means t = 0
dH/dt = -.32e^0 = - .32° / minute

when t=10
dH/dt = -.32e^(-.2) = -.26° / min

check my arithmetic
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