Question

A can of soda is removed from a refrigerator at time (in minutes) and its temperature, , in degrees Fahrenheit, is computed at regular intervals. Based on the data, a model is formulated for the object’s temperature, given by f(T)=74.4-38.8e^-0.05t

a)Determine the exact time when the soda’s temperature is 50 F .
b)Is there ever a time when the soda’s temperature is 36F ? Why or why not?
c)For the model, its domain is the set of all positive real numbers, . What is its range?
d)Find a formula for the inverse of the function y=f(t). What is the meaning of this function?

Answer 1

Answer:

a) 9.28 minutes

b) Yes it is time when the soda is just removed from the refrigerator

c) The range is 35.6°F ≤ T ≤ 74.4°F

d) The formula is given as follows;

$t = \dfrac{ln \left (\dfrac{74.4 - T}{38.8} \right )}{-0.05}$

The meaning of the function is that it shows the relationship between the temperature of the soda and the duration since the soda was removed from the fridge

Step-by-step explanation:

a) Given that the temperature T as a function of time is given by the following relation;

$f(T) = 74.4 - 38.8 \cdot e^{-0.05 \cdot t}$

When the temperature = 50°F, we have

$50 = 74.4 - 38.8 \cdot e^{-0.05 \cdot t}$

$50 - 74.4 = - 38.8 \cdot e^{-0.05 \cdot t} = -24.4$

$24.4 = 38.8 \cdot e^{-0.05 \cdot t}$

$e^{-0.05 \cdot t} = \dfrac{24.4 }{38.8 } = \dfrac{61}{97} = 0.629$

-0.05·t = ㏑(0.629)

t = ㏑(0.629) ÷(-0.05) = 9.28 minutes

b) At 36°F, we have;

$e^{-0.05 \cdot t} = \dfrac{74.4 - 36}{38.8 } = \dfrac{96}{97} = 0.99$

t = ㏑(0.99) ÷(-0.05) = 0.207 minutes = 0.207 × 60 = 12.44 seconds

Therefore, the time when the soda's temperature is 36°F is when it has just been removed from the refrigerator

The answer is yes it is time when the soda is just removed from the refrigerator

c) Given the domain is the set of positive real numbers, we have that the minimum value is given by t = 0 which gives;

$f(T)_{minimum} = 74.4 - 38.8 \cdot e^{-0.05 \times 0} = 74.4 - 38.8 \times 1 = 35.6^ {\circ} F$  

The maximum value is given by the expression $38.8 \cdot e^{-0.05 \times t}$ = 0, therefore we have;

$f(T)_{maximum} = 74.4 - 38.8 \cdot e^{-0.05 \times t} = 74.4 - 0 = 74.4^ {\circ} F$

The range is 35.6°F ≤ T ≤ 74.4°F

d) If the time taken for the soda to reach maximum temperature of 74.4°F is 100 minutes, we have;

$T = 74.4 - 38.8 \cdot e^{-0.05 \cdot t}$

$t = \dfrac{ln \left (\dfrac{74.4 - T}{38.8} \right )}{-0.05}$

The above formula gives the temperature of the soda at a particular time.