Question

The temperature T (in degrees Fahrenheit) of food placed in a refrigerator is modeled by

T = 10 (3t^2 + 17t + 76/ t^2 + 3t + 10)
Find the rate of change of T with respect to t
(in °F/hr) at the following time points:
t = 1/2 h
t = 1 h
t = 2 h

Answer 1

The rate of change of T with respect to t at t = 1/2 h is -754.8 °F/hr, at t = 1 h is -1597.6 °F/hr, and at t = 2 h is -678.8 °F/hr.

Step-by-step explanation:

To find the rate of change of temperature, we need to differentiate the given function with respect to t. Let's first simplify the function:

T = 10(3t^2 + 17t + 76 / t^2 + 3t + 10)

T = 10(3t^2 + 17t + 76) / (t^2 + 3t + 10)

Now, differentiate both the numerator and denominator:

T' = (30t^2 + 170t + 760) / (t^2 + 3t + 10) - (10(6t + 17)(t^2 + 3t + 10)) / (t^2 + 3t + 10)^2

Simplify further:

T' = (30t^2 + 170t + 760 - 60t^3 - 170t^2 - 600t - 1700) / (t^2 + 3t + 10)^2

T' = (- 60t^3 - 140t^2 - 600t - 940) / (t^2 + 3t + 10)^2

Now, substitute the values of t to find the rate of change of T:

At t = 1/2 h: T' = (-60(1/2)^3 - 140(1/2)^2 - 600(1/2) - 940) / ((1/2)^2 + 3(1/2) + 10)^2 = -754.8 °F/hr

At t = 1 h: T' = (-60(1)^3 - 140(1)^2 - 600(1) - 940) / ((1)^2 + 3(1) + 10)^2 = -1597.6 °F/hr

At t = 2 h: T' = (-60(2)^3 - 140(2)^2 - 600(2) - 940) / ((2)^2 + 3(2) + 10)^2 = -678.8 °F/hr