# Calculas HS

A glass of cold milk from the refrigerator is left on the counter on a warm summer day. its temperature y (in degrees Fahrenheit) after sittin gon the counter t minutes is

y= 72-30(0.98)^t

Answer the question by interpreting y and dy/dt.

1. what is the temperature of the refrigerator? how can you tell?
2. what is the temperature of the room? how can you tell?
3. when is the milk warming up the fastest? how can you tell?
4. determine algebraically when the temperature of the milk reaches 55F(degree)
5. at what reate is the ilk warming when its temperature is 55F(degree)? answer with an appropriate unit of measure.

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1. 1. When t = 0, we are at refrigerator temp
anything^0 = 1
72 - 30 = 42

2. when t gets big, the second term gets vanishingly small so Troom = 72

3. dy/dt = -30 *.98^t * ln(.98)
= .606 * .98^t
That is maximum (.606) when t = 0 because .98 is less than 1

4. 55 = 72 - 30*.98^t
- 17 = -30 *.98^t
.98^t = .5667
t ln .98 = ln .5667
t = -.568/-.0202 = 28.1 minutes

5. dy/dt = .606 * .98^t
.606 * .98^28.1
= .343 degrees/minute

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posted by Damon

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