Rotating Light vs Wall: Related-Rates Analysis

Problem Setup

  • Scenario: A beacon/light rotates in a horizontal circle and projects a spot onto a vertical wall.

  • Fixed distance from light to wall: 12 ft12\text{ ft} (adjacent side in a right triangle).

  • The light completes one full revolution in 4 s4\text{ s}.

  • At the instant of interest, the beam makes a 10∘10^{\circ} angle ((\theta)) with the perpendicular (normal) to the wall.

  • Unknown required: the speed at which the light spot moves along the wall, i.e. dwdt\frac{dw}{dt} where (w) = horizontal distance the spot has travelled from the foot of the perpendicular.

  • Expected unit for the answer: ft/s\text{ft/s}.

Key Variables

  • (\theta): instantaneous angle between the beam and the perpendicular to the wall (NOT the wall itself). (Perpendicular means a line extending straight out from the wall.)

  • (w): length of the segment along the wall from the perpendicular foot to the projected light spot (opposite side of the right triangle).

  • Time variable: tt (seconds).

Known Rates & Conversions

1 revolution =2π rad= 2\pi\text{ rad}

  • Rotation period =4 s=4\text{ s} (\Rightarrow) angular speed
    dθdt=2π rad4 s=π2  rad/s\frac{d\theta}{dt}=\frac{2\pi \text{ rad}}{4\text{ s}} = \frac{\pi}{2}\; \text{rad/s}.

Geometric Relation

Right triangle with

  • opposite side = ww

  • adjacent side = 1212

Tangent function:
tan⁡θ=w12.\tan\theta = \frac{w}{12}.

Differentiate Implicitly (Related-Rates)

Differentiate both sides with respect to time tt:

  1. Left side: derivative of tan⁡θ\tan\theta is sec⁡2θ\sec^{2}\theta, followed by chain rule factor dθdt\frac{d\theta}{dt}.

  2. Right side: ddt(w12)=112dwdt.\frac{d}{dt}\left(\frac{w}{12}\right)=\frac{1}{12}\frac{dw}{dt}.

Putting it together:
sec⁡2θ dθdt=112dwdt.\sec^{2}\theta\,\frac{d\theta}{dt}=\frac{1}{12}\frac{dw}{dt}.

Solve for dwdt\frac{dw}{dt}:
dwdt=12 sec⁡2θ dθdt.\frac{dw}{dt}=12\,\sec^{2}\theta\,\frac{d\theta}{dt}.

Plug In Numerical Values

At the instant considered, (\theta = 10^{\circ}) (convert? We may leave in degrees but the calculator must be in degree mode) and dθdt=π2 rad/s\frac{d\theta}{dt}=\frac{\pi}{2}\text{ rad/s}.

Compute symbolically first:
dwdt=12⋅sec⁡2(10∘)⋅π2<br>=6π sec⁡2(10∘).\frac{dw}{dt}=12 \cdot \sec^{2}(10^{\circ}) \cdot \frac{\pi}{2}<br>= 6\pi\, \sec^{2}(10^{\circ}).

Because most calculators lack a direct secant key, rewrite using cosine:
sec⁡2(10∘)=(1cos⁡10∘)2\sec^{2}(10^{\circ}) = \left(\frac{1}{\cos 10^{\circ}}\right)^{2}
dwdt=6π(1cos⁡10∘)2.\frac{dw}{dt}=6\pi\left(\frac{1}{\cos 10^{\circ}}\right)^{2}.

Numerical Approximation

Ensure calculator is in DEGREE mode.
Evaluate:

  1. cos⁡10∘≈0.984807753.\cos 10^{\circ}\approx 0.984807753.

  2. Reciprocal squared: (10.984807753)2≈1.0301537.\left(\frac{1}{0.984807753}\right)^{2}\approx 1.0301537.

  3. Multiply by 6π  (≈18.84955592)6\pi \;(\approx 18.84955592):
    18.84955592×1.0301537≈19.4356.18.84955592\times1.0301537\approx19.4356.

Rounded result:
dwdt ≈ 19.44 ft/s\boxed{\frac{dw}{dt}\,\approx\,19.44\ \text{ft/s}}

Interpretation & Contextual Notes

  • The spot races fastest when the beam is nearly parallel to the wall (larger (\theta)). At (\theta=0^{\circ}) (directly perpendicular), the speed would be zero.

  • Changing either the angular velocity ((d\theta/dt)) or the distance from the light to the wall alters the linear speed proportionally.

  • Under the given parameters, every four seconds the beam resets; whenever it reaches a small (10^{\circ}) deviation from the normal, the spot is already sliding across the wall at almost 20 ft/s20\text{ ft/s}.

Mathematical / Calculus Connections

  • Classic "related-rates" problem: convert rotational/angular motion into linear motion using trigonometric relations.

  • Utilizes chain rule for differentiating composite functions ((\tan\theta) where (\theta=\theta(t))).

  • Implicit differentiation because (\theta) and (w) are both functions of time, not explicit functions of one another.

  • Reinforces reciprocal-trig identity: sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}.

  • Highlights importance of unit consistency: radians for derivatives of trig functions, degrees must be converted or calculator set appropriately.

Practical Implications

  • This style of reasoning models lighthouse beams, radar sweeps, rotating lasers in industry, or security scans.

  • Safety calculations (e.g.
    laser scanning speed across human vision) rely on similar conversions from angular speed to linear wall/ground speed.

Summary of Final Answer

At (\theta=10^{\circ}) and rotation period 4 s4\text{ s},
dwdt≈19.44 ft/s.\frac{dw}{dt}\approx19.44\text{ ft/s}.