To solve this problem, we'll need to set up a relationship between the length of the shadow, the height of the building, and the angle $\theta$ the sun makes with the ground.

Let's define:

• $L$ as the length of the shadow (in feet),
• $h = 85$ ft as the height of the building,
• $\theta$ as the angle the sun makes with the ground.

From basic trigonometry, we know that: $\tan(\theta) = \frac{h}{L}$ Differentiating both sides of this equation with respect to time $t$, we get: $\sec^2(\theta) \cdot \frac{d\theta}{dt} = -\frac{h}{L^2} \cdot \frac{dL}{dt}$ We need to solve for $\frac{dL}{dt}$, which is the rate at which the shadow is changing.

Step 1: Convert the angle rate to radians

The rate of change of $\theta$ is given in degrees per minute, so we need to convert it to radians per minute. We know that: $1^\circ = \frac{\pi}{180} \text{ radians}$ Thus, $\frac{d\theta}{dt} = 0.28^\circ/\text{min}$ is equivalent to: $\frac{d\theta}{dt} = 0.28 \times \frac{\pi}{180} \approx 0.00489 \text{ radians/min}$

Step 2: Use the current values

At the moment in question, the height of the building is 85 ft, and the shadow is 204 ft long. So, we can compute $\theta$ using the inverse tangent: $\tan(\theta) = \frac{85}{204} \quad \Rightarrow \quad \theta = \tan^{-1}\left(\frac{85}{204}\right) \approx 0.3948 \text{ radians}$

Now, compute $\sec^2(\theta)$: $\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{85}{204}\right)^2 \approx 1 + 0.1734 \approx 1.1734$

Step 3: Solve for $\frac{dL}{dt}$

Substituting these values into the differentiated equation: $1.1734 \cdot 0.00489 = -\frac{85}{204^2} \cdot \frac{dL}{dt}$ Simplify the right-hand side: $\frac{85}{204^2} = \frac{85}{41616} \approx 0.00204$ Now, solve for $\frac{dL}{dt}$: $1.1734 \cdot 0.00489 = -0.00204 \cdot \frac{dL}{dt}$ $\frac{dL}{dt} = -\frac{1.1734 \cdot 0.00489}{0.00204} \approx -2.814 \text{ ft/min}$

Step 4: Convert to inches per minute

Since there are 12 inches in a foot, we convert the rate to inches per minute: $\frac{dL}{dt} = -2.814 \times 12 \approx -33.8 \text{ inches/min}$

Thus, the shadow is decreasing at approximately 33.8 inches per minute.

Would you like further clarification or details on any step?

5 Related Questions:

1. How does the rate of change of the angle affect the shadow length?
2. What happens to the shadow length if the angle rate increases?
3. How would you find the time at which the shadow reaches a specific length?
4. What is the significance of using radians in these calculations?
5. Can the process be simplified using linear approximations for small angles?

Tip:

Always ensure that angles are converted to radians when differentiating trigonometric functions in calculus.