math - calculus

Question

You pour yourself a cup of coffee. The function that models the cooling of 73◦C degree cup of coffee in a room of temperature 20◦C is given by T(t) = 20 + 53e^kt .

where the temperature T is (C), time t is in minutes (min), and k < 0 is an unknown constant.

Suppose that 5 minutes later, the cup of coffee is at a temperature of 65◦C.

(a) Find the unknown constant k.

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(b) How long will it take to for the cup of coffee to cool to 45◦C? Be accurate to the minute.

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(c) How fast is the coffee cooling 10 minutes after pouring? Be accurate to two decimal places

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(d) How many minutes after pouring will the coffee cooling at a rate of 1◦C/min? Be accurate to the minute

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Answer 1

a) Use the given time / temperature coordinate pair, (5 , 65), to solve for k:

T(5) = 20 + 53e^5k = 65

53e^5k = 45

e^5k = 45/53

5k = ln(45/53)

k = ln(45/53) / 5

k will be a negative decimal, which you will want to keep to 3 or 4 decimal places

Now that we have the equation for T(t), the remaining parts of the question are straightforward:

b) Set T(t) = 45 and solve for t. You will need to take the ln of both sides at some point.

c) T^'(t) = 53k·e^kt then plug in t = 10 to evaluate the derivative.

d) Set T^'(t) (above) = - 1 and solve for t. You will need ln again here.