A can of soft drink is initially at a temperature of 72.3◦F. Let t be the number of minutes since the can is placed into a fridge, set at 37.7◦F. The temperature of the drink, 90 minutes after being placed in the fridge, is 59.5◦F.
Suppose the formula for the temperature T , in ◦F, is T = ae−k t +c where a, k, and c are constants.
(a) Determine the units, if possible, of the constants a and k.
(b) Use algebraic reasoning, and your understanding of the physical situation to determine the exact values of a, k, and c. Explain your reasons.
(c) The formula we have ceases to be useful when the temperature is less than 1◦F away from the temperature of the fridge. What time does this occur?
(d) Calculate the rate of change of the drink over the interval 25 É t É 30. What are the units? What is the meaning of this number?
(e) If the temperature of the can of drink started at 65◦F but the other data in the question stayed the same, would the value of k be larger or smaller?
Thank you so much, why is a + 37.7 = 72.3 instead of the initial temp?
I also don't know how to attempt c and d
(a) well you have ae^-kt is °F. And you know that t is in °F, so a is just a number if e^-k(°F) is also °F. so k is -1/°F
we know that as t→∞, f(t) → 37.7 so c=37.7
(b) using the given data, we have the points (0,72.3) and (90,59.5)
a + 37.7 = 72.3
so a = 34.6
Now we have
34.6 e^(-90k) + 37.7 = 59.5
k = 0.00513
I expect you can answer the rest of the questions, ok?
(a) The units of the constant a would be in ◦F, as it represents the initial temperature of the drink. The units of the constant k would be in inverse minutes (min^(-1)), as it represents the rate at which the temperature decreases over time.
(b) To determine the values of a, k, and c, we can use the given information. We know that after 90 minutes, the temperature is 59.5◦F. Plugging this into the formula for T, we have:
T = ae^(-kt) + c
59.5 = ae^(-90k) + c (equation 1)
We also know that the initial temperature of the drink is 72.3◦F, and after 90 minutes in the fridge, the temperature is 59.5◦F. Plugging these values into the formula:
T = ae^(-kt) + c
72.3 = ae^(-0k) + c (equation 2)
59.5 = ae^(-90k) + c (equation 3)
From equations 2 and 3, we can see that c = 72.3 and ae^(-0k) = 72.3.
From equation 1, we can substitute ae^(-0k) with 72.3:
59.5 = 72.3e^(-90k) + c
59.5 = 72.3e^(-90k) + 72.3
Simplifying:
e^(-90k) = (59.5 - 72.3) / 72.3
e^(-90k) = -0.174825
Taking the natural logarithm of both sides:
-90k = ln(-0.174825)
k = ln(-0.174825) / -90
The value of k that satisfies this equation will depend on the natural logarithm of a negative number, which is undefined. Therefore, the exact value of k cannot be determined with the given information.
(c) The formula ceases to be useful when the temperature is less than 1◦F away from the temperature of the fridge, which is 37.7◦F. Therefore, we can set up the inequality:
T - 37.7 ≤ 1
ae^(-kt) + c - 37.7 ≤ 1
Simplifying:
ae^(-kt) + c ≤ 38.7
Substituting c = 72.3:
ae^(-kt) + 72.3 ≤ 38.7
ae^(-kt) ≤ -33.6
Since the exponential function is always positive, there is no value of t that would satisfy this inequality. Therefore, the formula ceases to be useful for any value of t.
(d) To calculate the rate of change of the drink over the interval 25 ≤ t ≤ 30, we need to find the derivative of T with respect to t:
dT/dt = -ake^(-kt)
The units of this derivative would be in ◦F/min. The meaning of this number is the rate at which the temperature is changing with respect to time, in terms of degrees Fahrenheit per minute, over the given interval.
(e) If the temperature of the can of drink started at 65◦F but the other data in the question stayed the same, the value of k would be smaller. This is because a larger initial temperature difference would result in a faster cooling rate, requiring a smaller value of k to reach the target temperature of 59.5◦F in 90 minutes.
(a) The units of the constants a and k can be determined from the equation T = ae^(-kt) + c. Since T represents temperature in degrees Fahrenheit (°F), a, k, and c should have units that can be combined with the units of t, which represents time in minutes.
(b) To determine the exact values of a, k, and c, we can use the given information at time t = 90 minutes, where T = 59.5°F.
Plugging these values into the equation, we get:
59.5 = a * e^(-k * 90) + c
We also know that the initial temperature at t = 0 minutes is 72.3°F:
72.3 = a * e^(-k * 0) + c
From this equation, we can conclude that a * e^0 = a, so a = 72.3 - c.
Substituting this value of a into the first equation:
59.5 = (72.3 - c) * e^(-k * 90) + c
Now we have an equation with two unknowns, c and k. We need another equation to solve for both.
We know that the temperature of the drink, 90 minutes after being placed in the fridge, is 59.5°F:
59.5 = (72.3 - c) * e^(-k * 90) + c
Now we can solve this equation using algebraic methods or numerical approximation techniques to find the exact values of c and k.
(c) The formula ceases to be useful when the temperature is less than 1°F away from the temperature of the fridge, which is 37.7°F in this case. So, we need to find the time at which T = 37.7 ± 1:
37.7 - 1 < T < 37.7 + 1
36.7 < T < 38.7
Substitute T = ae^(-kt) + c into the inequality:
36.7 < ae^(-kt) + c < 38.7
Since c represents the temperature of the fridge (37.7°F) and a and k are constants, we can focus on the exponential term ae^(-kt):
ae^(-kt) < 1
e^(-kt) < 1/a
For simplicity, assume a > 0, which means the drink is cooling down in the fridge.
Take the natural logarithm of both sides of the inequality:
-ln(a) > -kt
Since k and t are both positive values, we can divide both sides by -1 without changing the inequality:
ln(a) < kt
Now we have an expression for t:
t > ln(a) / k
Using the values of a and k obtained in part (b), we can plug them into this equation to find the specific value of t when the temperature is less than 1°F away from the fridge temperature.
(d) To calculate the rate of change of the drink's temperature over the interval 25 ≤ t ≤ 30, we need to find the derivative of the temperature equation with respect to time (t). Then we can plug in the appropriate values to calculate the rate of change.
Differentiate the temperature equation T = ae^(-kt) + c with respect to t:
dT/dt = -ake^(-kt)
Substituting the values a, k, and t from the given information:
dT/dt = -ae^(-kt) * k
Since a, k, and t have units, the rate of change dT/dt will have units of temperature (°F) divided by the unit of time (minutes). Therefore, the units of the rate of change will be °F/minute.
The meaning of this number is that it represents the amount by which the temperature of the drink changes per unit of time, specifically, the rate at which it cools down.
(e) If the temperature of the can of drink started at 65°F but the other data in the question stayed the same, the value of k would be larger.
This is because in the given temperature equation T = ae^(-kt) + c, a larger value of k makes the exponential term decay faster, leading to a faster cooling rate. So, starting at a higher initial temperature would require a larger value of k to produce the same temperature change over time.