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? rate at 20 minutes = 25.61 mi/hr

Explanation:

Step1: Set up variables

Let \(x\) be the distance of the person heading east, \(y\) be the distance of the person heading south, and \(z\) be the distance between them. By the Pythagorean theorem, \(z^{2}=x^{2}+y^{2}\). The rates are \(\frac{dx}{dt} = 16\) mph and \(\frac{dy}{dt}=20\) mph.

Step2: Convert time

20 minutes \(=\frac{20}{60}=\frac{1}{3}\) hours. Then \(x = 16\times\frac{1}{3}=\frac{16}{3}\) miles and \(y = 20\times\frac{1}{3}=\frac{20}{3}\) miles.

Step3: Differentiate the Pythagorean equation

Differentiate \(z^{2}=x^{2}+y^{2}\) with respect to \(t\): \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\), so \(\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}\).

Step4: Calculate \(z\)

\(z=\sqrt{x^{2}+y^{2}}=\sqrt{(\frac{16}{3})^{2}+(\frac{20}{3})^{2}}=\sqrt{\frac{256 + 400}{9}}=\sqrt{\frac{656}{9}}=\frac{\sqrt{656}}{3}\).

Step5: Calculate \(\frac{dz}{dt}\)

Substitute \(x=\frac{16}{3}\), \(y = \frac{20}{3}\), \(\frac{dx}{dt}=16\), \(\frac{dy}{dt}=20\), and \(z=\frac{\sqrt{656}}{3}\) into \(\frac{dz}{dt}\):

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45 minutes \(=\frac{45}{60}=\frac{3}{4}\) hours. \(x = 16\times\frac{3}{4}=12\) miles, \(y = 20\times\frac{3}{4}=15\) miles. \(z=\sqrt{12^{2}+15^{2}}=\sqrt{144 + 225}=\sqrt{369}\).

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Answer:

After 20 minutes: \(25.61\frac{mi}{hr}\). After 45 minutes: \(25.61\frac{mi}{hr}\).