A weight loss model for someone on a restrictive 30 week diet can be modeled by the function  w(t) = (160t + 1200) / (t+6)

Where w(t) is the weight of the person after t weeks  of beginning the diet program

determine w'(t) 

determine w'(12) in context of the problem 

Let's first compute the derivative of the given function \( w(t) \) using the quotient rule.

The weight function is given by:

\[
w(t) = \frac{160t + 1200}{t + 6}
\]

### Step 1: Use the Quotient Rule
The quotient rule states:

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

Where:
- \( f(t) = 160t + 1200 \)
- \( g(t) = t + 6 \)

Now, we need to calculate \( f'(t) \) and \( g'(t) \):

\[
f'(t) = 160 \quad \text{(since the derivative of a linear function is its slope)}
\]
\[
g'(t) = 1 \quad \text{(the derivative of \( t + 6 \) is 1)}
\]

### Step 2: Apply the Quotient Rule
Now, applying the quotient rule:

\[
w'(t) = \frac{(160)(t+6) - (160t+1200)(1)}{(t+6)^2}
\]

Simplifying the numerator:

\[
w'(t) = \frac{160(t+6) - (160t + 1200)}{(t+6)^2}
\]
\[
w'(t) = \frac{160t + 960 - 160t - 1200}{(t+6)^2}
\]
\[
w'(t) = \frac{-240}{(t+6)^2}
\]

Thus, the derivative of the weight function is:

\[
w'(t) = \frac{-240}{(t+6)^2}
\]

### Step 3: Calculate \( w'(12) \)
Now, let's determine the value of \( w'(12) \). Substitute \( t = 12 \) into \( w'(t) \):

\[
w'(12) = \frac{-240}{(12+6)^2} = \frac{-240}{18^2} = \frac{-240}{324}
\]
\[
w'(12) = -\frac{20}{27}
\]

### Interpretation in the context of the problem:
The derivative \( w'(12) = -\frac{20}{27} \approx -0.74 \) represents the rate of change of the person's weight after 12 weeks on the diet. Specifically, it means that at week 12, the person's weight is decreasing at a rate of approximately 0.74 pounds per week.

---

Would you like further details on this calculation, or any clarification on the context?

### Related Questions:
1. What is the significance of \( w'(t) \) when \( t = 0 \)?
2. How would the rate of weight loss change if the diet lasts for 40 weeks instead of 30?
3. What is the long-term behavior of \( w'(t) \) as \( t \to \infty \)?
4. How much weight does the person lose in the first 12 weeks?
5. Can the model predict a week where the rate of weight loss is less than 0.5 pounds per week?

#### Tip:
Always check the unit or context when interpreting derivatives; it often provides meaningful insights into real-world changes.

A weight loss model for someone on a restrictive 30 week diet can be modeled by the function w(t) = (160t + 1200) / (t+6). Determine w'(t) and w'(12) in the context of the problem.

Explore how to compute the derivative of a weight loss model using the quotient rule. Learn how to apply calculus to real-world scenarios like weight loss over time, with a step-by-step calculation for w'(12) and its interpretation. Suitable for Grades 11-12.

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