Answer :
The spotlight is rotating at approximately -0.16 radians per second.
Step 1: Understand the Setup
- The model is walking on a runway at a speed of 1.5 meters per second.
- The spotlight is 6 meters away from the closest point on the runway to the spotlight.
- When the model is 4.5 meters from the closest point, we want to find the rate at which the angle of the spotlight (let's call it [tex]\theta[/tex]) is changing.
Step 2: Define the Relationship
We can visualize the scenario as a right triangle:
- One leg of the triangle is the distance from the runway to the spotlight (6 meters).
- The other leg is the horizontal distance from the closest point on the runway to the model (which is changing as the model walks).
- We define the horizontal distance as [tex]x[/tex], which at this moment is 4.5 meters.
Using trigonometry, we have:
[tex]\tan(\theta) = \frac{x}{6}[/tex]
From this, we can derive the angle as follows:
[tex]\theta = \tan^{-1}(\frac{x}{6})[/tex]
Step 3: Differentiate with Respect to Time
- To find the rate of change of the angle [tex]\frac{d\theta}{dt}[/tex], we need to use implicit differentiation:
[tex]\frac{d\theta}{dt} = \frac{1}{1 + (\frac{x}{6})^2} \cdot \frac{d}{dt}(\frac{x}{6})[/tex]
Step 4: Calculate [tex]\frac{dx}{dt}[/tex]
- The model is moving towards the spotlight at a speed of 1.5 m/s, so:
[tex]\frac{dx}{dt} = -1.5 \, \text{m/s}[/tex]
(Negative because x is decreasing as the model approaches the spotlight.)
Step 5: Substitute and Solve
- At this moment, when [tex]x = 4.5[/tex]:
Calculate [tex]\frac{d\theta}{dt}[/tex]:
[tex]\frac{d\theta}{dt} = \frac{1}{1 + (\frac{4.5}{6})^2} \cdot \frac{-1.5}{6}[/tex]
First find [tex](\frac{4.5}{6})^2[/tex]:
[tex](\frac{4.5}{6})^2 = \frac{20.25}{36} = 0.5625[/tex]
Thus:
[tex]\frac{d\theta}{dt} = \frac{1}{1 + 0.5625} \cdot \frac{-1.5}{6}[/tex]
[tex]= \frac{1}{1.5625} \cdot \frac{-1.5}{6}[/tex]
[tex]= \frac{-1.5}{9.375} \approx -0.16 \, \text{radians/second}[/tex]
