The temperature T (in degrees Fahrenheit) of food placed in a refrigerator is modeled by
T = 10

4t2 + 16t + 60
t2 + 4t + 10

where t is the time (in hours).
(a) What is the initial temperature of the food?
60
 
Correct: Your answer is correct.
 deg

(b) Find the rate of change of T with respect to t when t = 1. (Round your answer to two decimal places.)
T '(1) ≈ 
-5.33
 
Correct: Your answer is correct.
 deg/hr

(c) Find the rate of change of T with respect to t when t = 3. (Round your answer to two decimal places.)
T '(3) ≈ 
-2.71
 
Incorrect: Your answer is incorrect.
 deg/hr

(d) Find the rate of change of T with respect to t when t = 5. (Round your answer to two decimal places.)
T '(5) ≈ 
-5.04
 
Incorrect: Your answer is incorrect.
deg/hr

(e) Find the rate of change of T with respect to t when t = 10. (Round your answer to two decimal places.)
T '(10) ≈ 
1.07
 
Incorrect: Your answer is incorrect.
deg/hr

We are given the temperature function of the food placed in a refrigerator:

\[
T(t) = \frac{4t^2 + 16t + 60}{t^2 + 4t + 10}
\]

We are tasked with finding the rate of change of the temperature \( T \) with respect to time \( t \), i.e., the derivative \( T'(t) \), at specific times \( t = 1, 3, 5, 10 \).

### Derivative of \( T(t) \)
The derivative \( T'(t) \) can be found using the quotient rule for differentiation, which is applicable because \( T(t) \) is a ratio of two functions. The quotient rule states:

\[
T'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}
\]

where \( f(t) = 4t^2 + 16t + 60 \) and \( g(t) = t^2 + 4t + 10 \).

#### Steps:
1. **Differentiate the numerator** \( f(t) \):
   \[
   f'(t) = \frac{d}{dt}(4t^2 + 16t + 60) = 8t + 16
   \]
   
2. **Differentiate the denominator** \( g(t) \):
   \[
   g'(t) = \frac{d}{dt}(t^2 + 4t + 10) = 2t + 4
   \]
   
3. **Apply the quotient rule**:
   \[
   T'(t) = \frac{(8t + 16)(t^2 + 4t + 10) - (4t^2 + 16t + 60)(2t + 4)}{(t^2 + 4t + 10)^2}
   \]
   
4. **Simplify the expression**:
   To find the values at specific points, we can either directly compute the derivatives or use the simplified form of the expression for each given \( t \).

Let's now compute the derivative at \( t = 3, 5, 10 \) using this approach.

### (c) \( T'(3) \)
I will now calculate the value of \( T'(3) \).Here are the correct values of the rate of change of the temperature \( T \) with respect to time \( t \):

- \( T'(3) \approx -0.21 \) degrees per hour
- \( T'(5) \approx -0.09 \) degrees per hour
- \( T'(10) \approx -0.02 \) degrees per hour

These results differ from your previously calculated values, and the temperature changes more slowly at later times.

Would you like to go over the steps in detail or discuss any of the values further?

Here are 5 related questions to extend your understanding:
1. How does the temperature change over longer periods of time as \( t \to \infty \)?
2. What is the physical significance of a positive or negative \( T'(t) \)?
3. How would the graph of \( T(t) \) look over time, based on these derivatives?
4. How would you estimate the time when the temperature stops changing significantly?
5. Can we infer when the food reaches the refrigerator's internal temperature?

**Tip:** Always check the units of the derivative. In this case, \( T'(t) \) gives degrees per hour, which is important when interpreting results.

The temperature T (in degrees Fahrenheit) of food placed in a refrigerator is modeled by T = 4t^2 + 16t + 60 / t^2 + 4t + 10, where t is the time (in hours). (a) What is the initial temperature of the food? (b) Find the rate of change of T with respect to t when t = 1. (c) Find the rate of change of T with respect to t when t = 3. (d) Find the rate of change of T with respect to t when t = 5. (e) Find the rate of change of T with respect to t when t = 10.

Explore how to calculate the rate of change of temperature over time using the derivative of a function with the quotient rule. This example uses the function T(t) = (4t^2 + 16t + 60) / (t^2 + 4t + 10), demonstrating the application of calculus concepts to real-world problems. Suitable for high school students in Grades 11-12.

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