As a bowl of soup cools, the temperature of the soup is given by the twice-differentiable function H for 0[tex]\leq[/tex]t[tex]\leq[/tex]12, where H(t) is measured in degrees Celsius (C), and time t is measured in minutes. Values of H(t) at selected values of time t are shown in the table above.

a) Use the data in the table to approximate H'(5). Show the computations that led to your answer. Using correct units, explain the meaning of H'(5) in the context of the problem.

b) Is there a time t for 0[tex]\leq[/tex]t[tex]\leq[/tex]12 at which the temperature of the soup is 60C? Justify your answer.

c) The temperature of the soup is also modeled by the twice-differentiable function C for 0[tex]\leq[/tex]t[tex]\leq[/tex]12, where C(t) is measured in degrees Celsius (C), and time tis measured in minutes. It is known C(3)= 80 and the derivative of C is given by C'(t)= -3.6e-0.05t. Write an equation for the line tangent to the graph at t=3 and use it to approximate C(5).

d) Based on the model given in part c, at what rate, in degrees Celsius per minute minute, is the rate of change of the temperature of the soup changing at t=3 minutes?

The rate of change of temperature with time at a point in time is given by

the derivative of the function for the temperature of the soup.

The correct responses are;

Reasons:

a) From the data in the table, we have;

The approximate value of H'(5) is given by the average value of the rate of

change of the temperature with time between points, t = 3, and t = 8

Therefore;

[tex]\displaystyle H'(5) = \mathbf{\frac{H(8) - H(3)}{8 - 3}}[/tex]

Which gives;

[tex]\displaystyle H'(5) = \frac{80 - 67}{8 - 3} = \mathbf{2.6}[/tex]

Therefore, H'(5) = -2.6°C per minute

b) Given that the function is twice differentiable over the interval, 0 ≤ t ≤ 12, the function for the change in temperature is continuous in the interval 0 ≤ t ≤ 12

At t = 0, H(0) = 90 °C

At t = 12, H(12) = 58 °C

Therefore, there exist a temperature, of 60 °C between 90° C and 58 °C

c) The given derivative of C is, [tex]C'(t) = \mathbf{-3.6 \cdot e^{-0.05 \cdot t}}[/tex]

At t = 3, we have;

[tex]The \ slope \ at \ t = 3 \ is \ C'(3) = -3.6 \cdot e^{-0.05 \times 3} \approx -3.1[/tex]

Therefore, we have;

y - 80 ≈ -3.1 × (x - 3)

The equation for the tangent is; y = -3.6 × (x - 3) + 80

y = -3.6·x + 10.8 + 80 = -3.6·x + 90.8

The  value of C(5) is approximately, C(5) ≈ -3.6 × 5 + 90.8 = 72.8

d) Based on the the model above, the rate at which the temperature of the

soup is changing at t = 3 minutes is -3.6 degrees per minute.

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