Question

Walkers \(A\) and \(B\) are walking on straight streets that meet at right angles. \(A\) approaches the intersection at 2 \(\mathrm{m} / \mathrm{sec}\) and \(B\) moves away from the intersection at 1 \(\mathrm{m} / \mathrm{sec}\) as shown in the figure. At what rate is the angle \(\theta\) changing when \(A\) is 10 \(\mathrm{m}\)from the intersection and \(B\) is 20 \(\mathrm{m}\) from the intersection? Express your answer in degrees per second to the nearest degree.

Short answer

The angle \(\theta\) is changing at about 573 degrees per second.

Step-by-step solution

Establish the relationship

The situation forms a right triangle where the angle \(\theta\) is opposite the side that corresponds to \(B\)'s distance from the intersection, and adjacent to the side that corresponds to \(A\)'s distance. Hence, \(\tan(\theta) = \frac{B}{A}\)

Differentiate with respect to time

Differentiate the relationship established in the previous step with respect to time \(t\), using the chain rule: \(\frac{d}{dt}\tan(\theta) = \frac{d}{dt} \left(\frac{B}{A}\right)\). This gives: \(\sec^2(\theta) \frac{d\theta}{dt} = \frac{A \frac{dB}{dt} - B \frac{dA}{dt}}{A^2}\)

Substitute given values

We know that \(A = 10 \, \mathrm{m}\), \(B = 20 \, \mathrm{m}\), \(\frac{dA}{dt} = -2 \, \mathrm{m/sec}\) (because \(A\) is getting closer to the intersection), and \(\frac{dB}{dt} = 1 \, \mathrm{m/sec}\). Now we can calculate \(\theta\) at the moment in question: \(\theta = \tan^{-1} \left(\frac{B}{A}\right)= \tan^{-1}\left(\frac{20}{10}\right)= \tan^{-1}(2)\). Substitute all these values into the differentiation equation.

Calculate \(\frac{d\theta}{dt}\)

\(\frac{d\theta}{dt} = \frac{A \frac{dB}{dt} - B \frac{dA}{dt}}{\sec^2(\theta) A^2} = \frac {10*1 - 20*(-2)}{\sec^2(\tan^{-1}(2)) * 100} = \frac{50}{5}\) which gives \(\frac{d\theta}{dt} = 10\, \mathrm{rad/sec}\). Be careful, though, as the exercise requires the answer in degrees per second.

Convert to degrees

To convert from radians per second to degrees per second, use the fact that \(180^{\circ} = \pi \, \mathrm{rad}\). Hence, the rate of change in \(\theta\) is \(\frac{d\theta}{dt} = 10 * \frac{180}{\pi} \approx 573^{\circ}/\mathrm{sec}\)