Certainly! Let's proceed with finding the solutions for parts (b), (c), and (d).

### Part (b): The rate of change of the mineral level at \( t = 3 \) days

We already have the derivative of \( C(t) \):

\[ \frac{dC}{dt} = -\frac{3}{2} (6t + 3)^{-\frac{3}{2}} \]

Substitute \( t = 3 \):

\[ \frac{dC}{dt} \Bigg|_{t=3} = -\frac{3}{2} (6(3) + 3)^{-\frac{3}{2}} = -\frac{3}{2} (18 + 3)^{-\frac{3}{2}} = -\frac{3}{2} (21)^{-\frac{3}{2}} \]

Calculate \( (21)^{-\frac{3}{2}} \):

\[ (21)^{-\frac{3}{2}} = \frac{1}{21^{\frac{3}{2}}} = \frac{1}{\sqrt{21^3}} = \frac{1}{\sqrt{9261}} \approx \frac{1}{96.23} \]

So:

\[ \frac{dC}{dt} \Bigg|_{t=3} \approx -\frac{3}{2} \cdot \frac{1}{96.23} \approx -\frac{3}{192.46} \approx -0.016 \]

Rounded to two decimal places:

\[ \frac{dC}{dt} \Bigg|_{t=3} \approx -0.02 \text{ milligrams per cubic centimeter per day} \]

### Part (c): The rate of change of the mineral level at \( t = 6.5 \) days

Substitute \( t = 6.5 \):

\[ \frac{dC}{dt} \Bigg|_{t=6.5} = -\frac{3}{2} (6(6.5) + 3)^{-\frac{3}{2}} = -\frac{3}{2} (39 + 3)^{-\frac{3}{2}} = -\frac{3}{2} (42)^{-\frac{3}{2}} \]

Calculate \( (42)^{-\frac{3}{2}} \):

\[ (42)^{-\frac{3}{2}} = \frac{1}{42^{\frac{3}{2}}} = \frac{1}{\sqrt{42^3}} = \frac{1}{\sqrt{74088}} \approx \frac{1}{271.93} \]

So:

\[ \frac{dC}{dt} \Bigg|_{t=6.5} \approx -\frac{3}{2} \cdot \frac{1}{271.93} \approx -\frac{3}{543.86} \approx -0.0055 \]

Rounded to two decimal places:

\[ \frac{dC}{dt} \Bigg|_{t=6.5} \approx -0.01 \text{ milligrams per cubic centimeter per day} \]

### Part (d): Are the slopes of the tangent lines to \( C(t) \) always positive or always negative? What does this indicate about the rate of change of the mineral level?

The derivative we calculated is:

\[ \frac{dC}{dt} = -\frac{3}{2} (6t + 3)^{-\frac{3}{2}} \]

Since \( (6t + 3) \) is always positive for all \( t \geq 0 \), \( (6t + 3)^{-\frac{3}{2}} \) is also always positive. The factor \( -\frac{3}{2} \) ensures that the derivative \( \frac{dC}{dt} \) is always negative.

Thus, the slopes of the tangent lines to \( C(t) \) are always negative.

### Interpretation:

This indicates that the mineral level in the bloodstream is always decreasing over time. The negative rate of change suggests that as time progresses, the amount of the mineral in the bloodstream decreases consistently.

Do you have any questions or would you like more details?

Here are five questions you can ask next:
1. How do you interpret the meaning of the derivative in the context of this problem?
2. Can you explain how the chain rule is applied in this differentiation?
3. What happens to the rate of change as \( t \) increases?
4. How does the initial amount of the mineral affect the rate of change?
5. What are some real-world implications of this decreasing rate of change?

**Tip:** When dealing with negative rates of change, it often indicates a decrease or reduction in the quantity being measured.

Explore solutions for calculating the rate of change of mineral levels over time. Includes detailed derivations and calculations for specific time points. Suitable for advanced high school students studying calculus.

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