This kid sets out milk and cookies for Santa. He finds the temperature of the milk y in degrees Fahrenheit after sitting on the counter t minutes is y=72-30(0.98)^t
1) what is the temperature of the refrigerator and how can you tell?
2) what is the temperature of the room and how can you tell?
3) when is the milk warming up fastest and how can you tell?
4) when does the temperature of the milk reach 55 degrees F?
5) at what rate is the milk warming when its temperature is 55 degrees F?
6) at 3:00 am will the milk be at room temperature? assume the kid goes to bed at 11:00 pm. justify your answer by determining when the milk will be within 0.1 degree F of room temperature
1) what is the temperature of the refrigerator and how can you tell?
2) what is the temperature of the room and how can you tell?
3) when is the milk warming up fastest and how can you tell?
4) when does the temperature of the milk reach 55 degrees F?
5) at what rate is the milk warming when its temperature is 55 degrees F?
6) at 3:00 am will the milk be at room temperature? assume the kid goes to bed at 11:00 pm. justify your answer by determining when the milk will be within 0.1 degree F of room temperature
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1) The temperature of the refrigerator is the initial temperature of the milk at time t = 0.
Thus, it is y = 72 - 30(0.98)^0 = 72 - 30 = 42 degrees Fahrenheit.
2) The temperature of the room is the limit of the milk's temperature as t goes to infinity.
Since limit 0.98^t as t tends to infinity is 0, we have y = 72 degrees Fahrenheit.
3) The milk warms up fastest initially at time t = 0, since the difference in temperature between 72 and 42 is the greatest at time t = 0.
4) Let y = 55. Solving for t yields:
55 = 72 - 30(0.98)^t
30(0.98)^t = 17
0.98^t = 17/30
t = log(17/30)/log(0.98) ≈ 28.11 minutes.
5) Since y'(t) = -30ln(0.98)(0.98)^t, we have:
y'(55) = -30ln(0.98)(0.98)^55 ≈ 0.20 degrees per minute
6) Since 3:00am is 4 hours after 11:00pm, we have t = 240 minutes.
So the temperature will be y = 72 - 30(0.98)^240 ≈ 71.76 degrees Fahrenheit.
This is almost but not quite the room temperature of y = 72.
The milk will be within 0.1 degrees when y = 71.9 degrees. Solving for t gives:
71.9 = 72 - 30(0.98)^t
30(0.98)^t = 0.1
0.98^t = 0.1/30
t = log(0.1/30)/log(0.98) ≈ 282.33 minutes
So the milk will be within 0.1 degree of room temperature at around 3:42am.
Thus, it is y = 72 - 30(0.98)^0 = 72 - 30 = 42 degrees Fahrenheit.
2) The temperature of the room is the limit of the milk's temperature as t goes to infinity.
Since limit 0.98^t as t tends to infinity is 0, we have y = 72 degrees Fahrenheit.
3) The milk warms up fastest initially at time t = 0, since the difference in temperature between 72 and 42 is the greatest at time t = 0.
4) Let y = 55. Solving for t yields:
55 = 72 - 30(0.98)^t
30(0.98)^t = 17
0.98^t = 17/30
t = log(17/30)/log(0.98) ≈ 28.11 minutes.
5) Since y'(t) = -30ln(0.98)(0.98)^t, we have:
y'(55) = -30ln(0.98)(0.98)^55 ≈ 0.20 degrees per minute
6) Since 3:00am is 4 hours after 11:00pm, we have t = 240 minutes.
So the temperature will be y = 72 - 30(0.98)^240 ≈ 71.76 degrees Fahrenheit.
This is almost but not quite the room temperature of y = 72.
The milk will be within 0.1 degrees when y = 71.9 degrees. Solving for t gives:
71.9 = 72 - 30(0.98)^t
30(0.98)^t = 0.1
0.98^t = 0.1/30
t = log(0.1/30)/log(0.98) ≈ 282.33 minutes
So the milk will be within 0.1 degree of room temperature at around 3:42am.