College

A model walks a straight runway at a speed of 1.5 meters per second. A spotlight is located on the same plane as the runway, 6 meters from the runway, and is kept focused on the model. At what rate is the spotlight rotating when the model is 4.5 meters from the point on the runway closest to the spotlight?

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]