Related rates, question 24

Topic: Related rates
Difficulty: Challenging
Suggested time: 7 min
Marks: 4
Theory: UBC Math 100 §5 CLP-1 §3.2Key results

Late in the afternoon the sun’s angle of elevation, $\theta$ , is falling at 0.2 radians per hour. How fast is the shadow of a 25 m building on level ground lengthening at the moment the sun’s elevation is $\dfrac{\pi}{6}$ ?

Write your solution in full, with units and the sign of each rate, then compare it against the model solution.

Your answer

Theory UBC Math 100 §5 CLP-1 §3.2Key results

Worked solution

Strategy

The height of the building is fixed and the angle changes, so a trigonometric ratio links the shadow to the angle at every instant. Differentiate it with respect to time, put in the signed rate of the angle, and only then the angle of this instant.

Working

Let s be the length of the shadow in metres and θ the sun’s elevation, both functions of the time t in hours. The 25 m building is the side opposite θ, and the shadow is the side next to it.
$s = 25\cot\theta$
From tan θ = 25/s; this holds at every instant.
$\frac{ds}{dt} = -25\csc^{2}\theta\,\frac{d\theta}{dt}$
Differentiate with respect to t; the chain rule brings in dθ/dt.
$\csc\frac{\pi}{6} = 2 \implies \csc^{2}\frac{\pi}{6} = 4, \qquad \frac{d\theta}{dt} = -0.2$
The elevation is falling, so its rate is negative.
$\frac{ds}{dt} = -25(4)(-0.2) = 20$
Positive: as the sun sinks, the shadow grows.

Answer

\frac{ds}{dt} = 20 \text{ m/h}

What earns the marks

Names the variables: s for the length of the shadow and θ for the elevation, both changing with time, with the 25 m height fixed.
Writes a relation that holds at every instant, such as tan θ = 25/s or s = 25 cot θ.
Differentiates with respect to time before substituting: ds/dt = −25 csc²θ · dθ/dt, or the same from tan θ = 25/s.
Uses dθ/dt = −0.2, negative because the elevation is falling, and csc²(π/6) = 4.
Concludes that ds/dt = 20 m/h: the shadow is lengthening at 20 metres per hour.

Check

In 0.01 h the elevation falls by 0.002 rad, and the shadow 25 cot θ grows from 25√3 ≈ 43.301 m to about 43.502 m, 0.2 m longer.

Where this goes wrong

Entering dθ/dt as +0.2, which gives −20 m/h, a shadow that shrinks as the sun goes down. The elevation is falling, so dθ/dt = −0.2, and the two minus signs make the rate of the shadow positive.

Leans on

Derivatives of the trigonometric functions, from Exponential and trigonometric functions

Try a similar question

A camera on a tripod stands 0.4 km from a straight road. A car drives along the road at 1.5 km/min, moving away from the point on the road nearest the camera, and the camera turns to keep the car in frame. How fast is the camera rotating, in radians per minute, at the instant the car is 0.5 km from the camera?

My working

Saved in this browser only.

Related rates practice question 24