Question

two people start biking from the same point. one heads east at 16 mph, the other south at 20 mph. what is the rate at which the distance between the two people is changing after 20 minutes and after 45 minutes? what is the rate at which the distance between the two people is changing after 20 minutes?

Explanation:

Step1: Convert time to hours

Since \(1\) hour \( = 60\) minutes, \(t\) (in hours) for \(20\) minutes is \(t=\frac{20}{60}=\frac{1}{3}\) hours and for \(45\) minutes is \(t = \frac{45}{60}=\frac{3}{4}\) hours.
Let the distance of the first - person (east - bound) be \(x = 16t\) and the distance of the second - person (south - bound) be \(y=20t\). By the Pythagorean theorem, the distance \(z\) between them is \(z=\sqrt{x^{2}+y^{2}}=\sqrt{(16t)^{2}+(20t)^{2}}=\sqrt{256t^{2}+400t^{2}}=\sqrt{656t^{2}} = 4\sqrt{41}t\).
Differentiate \(z\) with respect to \(t\) using the chain rule. If \(z=\sqrt{x^{2}+y^{2}}\), then \(\frac{dz}{dt}=\frac{2x\frac{dx}{dt}+2y\frac{dy}{dt}}{2\sqrt{x^{2}+y^{2}}}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{\sqrt{x^{2}+y^{2}}}\).
We know that \(\frac{dx}{dt}=16\) and \(\frac{dy}{dt}=20\).

Step2: Calculate at \(t=\frac{1}{3}\)

When \(t = \frac{1}{3}\), \(x = 16\times\frac{1}{3}=\frac{16}{3}\) and \(y=20\times\frac{1}{3}=\frac{20}{3}\).
Substitute into \(\frac{dz}{dt}\): \(\frac{dz}{dt}=\frac{\frac{16}{3}\times16+\frac{20}{3}\times20}{\sqrt{(\frac{16}{3})^{2}+(\frac{20}{3})^{2}}}=\frac{\frac{256 + 400}{3}}{\sqrt{\frac{256+400}{9}}}=\frac{\frac{656}{3}}{\frac{\sqrt{656}}{3}}=\sqrt{656}\approx 25.61\)

Step3: Calculate at \(t=\frac{3}{4}\)

When \(t=\frac{3}{4}\), \(x = 16\times\frac{3}{4}=12\) and \(y=20\times\frac{3}{4}=15\).
Substitute into \(\frac{dz}{dt}\): \(\frac{dz}{dt}=\frac{12\times16 + 15\times20}{\sqrt{12^{2}+15^{2}}}=\frac{192+300}{\sqrt{144 + 225}}=\frac{492}{\sqrt{369}}=\frac{492}{3\sqrt{41}}=\frac{164}{\sqrt{41}}\approx25.61\)

Answer:

The rate at which the distance between the two people is changing after \(20\) minutes and after \(45\) minutes is approximately \(25.6\ \frac{mi}{hr}\)