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.

1. What do you expect the temperature of the milk to be at t = 0, assuming it was left in the fridge a long time before removal?

2. Consider what the milk temperature becomes as t becomes very large.

3., calculate the derivative, observe the behavior of the function, and tell us what you think.

4. Solve the equation for t when y = 55

5. Calculate dy/dt when y = 55

You should not expect us to do these calculations for you

1. To determine the temperature of the refrigerator, we need to find the value of y when the milk is kept in the refrigerator. The equation provided is y = 72 - 30(0.98)^t, where t represents the time in minutes. The initial temperature of the milk is at t=0. From the equation, we can see that when t=0, the equation simplifies to y = 72 - 30(0.98)^0 = 72 - 30(1) = 42. Therefore, the temperature of the milk in the refrigerator is 42 degrees Fahrenheit.

2. To determine the temperature of the room, we need to find the value of y as time goes to infinity. In the equation y = 72 - 30(0.98)^t, as t approaches infinity, the term (0.98)^t becomes extremely small, approaching zero. Thus, the equation simplifies to y = 72 - 30(0) = 72. Therefore, the temperature of the room is 72 degrees Fahrenheit.

3. The milk warms up the fastest when its temperature changes at the highest rate. This corresponds to the derivative of y with respect to time, or dy/dt. We can find dy/dt by differentiating the given equation. Taking the derivative, we get dy/dt = -30(0.98)^t * ln(0.98). The negative sign indicates that the milk is cooling down. Since ln(0.98) is negative, the value of dy/dt decreases exponentially as time passes. Therefore, the milk is warming up the fastest when t = 0, and the rate decreases as time goes on.

4. To find when the temperature of the milk reaches 55 degrees Fahrenheit, we can set y equal to 55 and solve for t. Plugging in y = 55 into the equation, we get 55 = 72 - 30(0.98)^t. Rearranging the equation, we have 30(0.98)^t = 17. Dividing both sides by 30, we get (0.98)^t = 17/30. Taking the natural logarithm of both sides, we have t * ln(0.98) = ln(17/30). Dividing both sides by ln(0.98), we get t = ln(17/30) / ln(0.98). Using a calculator, we find that t โ‰ˆ 9.898 minutes. Therefore, the temperature of the milk reaches 55 degrees Fahrenheit after approximately 9.898 minutes.

5. To find the rate at which the milk is warming when its temperature is 55 degrees Fahrenheit, we need to evaluate dy/dt when y = 55. From the previous step, we found that t โ‰ˆ 9.898 minutes when y = 55. Plugging in this value into the equation for dy/dt, we get dy/dt = -30(0.98)^9.898 * ln(0.98). Calculating this using a calculator, we find that dy/dt โ‰ˆ -2.637 degrees Fahrenheit per minute. The negative sign indicates that the milk is cooling down. Therefore, at a temperature of 55 degrees Fahrenheit, the milk is cooling down at a rate of approximately 2.637 degrees Fahrenheit per minute.