To solve the problem, we'll first address each part individually.

Part (a): Showing that $h = r(\csc \theta - 1)$

Let's start by analyzing the relationship between the satellite, the Earth's surface, and the angle $\theta$.

1. Define the Elements of the Problem:

   • $h$: the distance from the satellite to the Earth's surface.
   • $r$: the radius of the Earth.
   • $\theta$: the angle between the line from the center of the Earth to the satellite and the line from the center of the Earth to a point directly below the satellite on the Earth's surface (this is the complement of the angle of incidence of the satellite's sensor).

2. Forming the Right Triangle:
   Imagine a right triangle where:

   • One vertex is at the center of the Earth.
   • One vertex is at the satellite.
   • The third vertex is on the surface of the Earth, directly below the satellite.

   The hypotenuse of this triangle is $r + h$, and one leg (from the center of the Earth to the point on the Earth's surface) is $r$. The angle opposite the leg of length $r$ is $\theta$.

3. Relating $h$, $r$, and $\theta$:
   By definition, the cosecant function is the reciprocal of the sine function: $\csc \theta = \frac{1}{\sin \theta}$

   In this right triangle, $\sin \theta = \frac{r}{r+h}$. Hence, $\csc \theta = \frac{r + h}{r}$

   Rearranging this equation to solve for $h$: $\csc \theta = \frac{r + h}{r} \implies r \csc \theta = r + h \implies h = r \csc \theta - r$

   Therefore, we have shown that: $h = r (\csc \theta - 1)$

Part (b): Finding the rate at which $h$ is changing with respect to $\theta$

Now, we need to differentiate $h$ with respect to $\theta$ to find $\frac{dh}{d\theta}$.

Given: $h = r (\csc \theta - 1)$

1. Differentiate both sides with respect to $\theta$: $\frac{dh}{d\theta} = r \frac{d}{d\theta}(\csc \theta)$

   The derivative of $\csc \theta$ with respect to $\theta$ is $-\csc \theta \cot \theta$. Therefore: $\frac{dh}{d\theta} = -r \csc \theta \cot \theta$

2. Substitute $\theta = 30^\circ$ and $r = 4000$ miles:

   • $\csc 30^\circ = 2$
   • $\cot 30^\circ = \sqrt{3}$

   Now plug these values into the expression for $\frac{dh}{d\theta}$: $\frac{dh}{d\theta} = -4000 \times 2 \times \sqrt{3}$ $\frac{dh}{d\theta} = -8000\sqrt{3} \approx -13856.41 \text{ miles per radian}$

The negative sign indicates that $h$ decreases as $\theta$ increases.

Conclusion

• For part (a), we showed that $h = r(\csc \theta - 1)$.
• For part (b), we calculated the rate of change of $h$ with respect to $\theta$ as $-8000\sqrt{3} \approx -13856.41$ miles per radian when $\theta = 30^\circ$.

Would you like further details or have any questions?

Follow-up Questions

1. How does the height $h$ change as the angle $\theta$ approaches 0?
2. What is the geometric interpretation of the cosecant function in this context?
3. How would the formula change if the satellite is orbiting at a higher altitude?
4. Can the rate of change $\frac{dh}{d\theta}$ be positive, and if so, under what conditions?
5. How does the radius of the Earth, $r$, affect the height $h$ and its rate of change?

Tip

When analyzing problems involving right triangles and circles, remember that trigonometric functions like sine, cosine, and their reciprocals can often relate various physical quantities, such as angles and distances, in a straightforward way.