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 =

Explanation:

Step1: Establish the relationship

Let \(x\) be the distance of the person going east, \(y\) be the distance of the person going south, and \(z\) be the distance between them. By the Pythagorean theorem, \(z^{2}=x^{2}+y^{2}\). Differentiating both sides with respect to time \(t\), we get \(2z\frac{dz}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}\), so \(\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}\). Given \(\frac{dx}{dt}=16\) mph and \(\frac{dy}{dt} = 20\) mph.

Step2: Calculate \(x\) and \(y\) at \(t = 20\) minutes (\(t=\frac{20}{60}=\frac{1}{3}\) hours)

\(x=\frac{dx}{dt}\times t=16\times\frac{1}{3}=\frac{16}{3}\) miles, \(y=\frac{dy}{dt}\times t=20\times\frac{1}{3}=\frac{20}{3}\) miles.

Step3: Calculate \(z\) at \(t=\frac{1}{3}\) hours

\(z=\sqrt{x^{2}+y^{2}}=\sqrt{(\frac{16}{3})^{2}+(\frac{20}{3})^{2}}=\sqrt{\frac{256 + 400}{9}}=\sqrt{\frac{656}{9}}=\frac{\sqrt{656}}{3}=\frac{4\sqrt{41}}{3}\) miles.

Step4: Calculate \(\frac{dz}{dt}\)

Substitute \(x = \frac{16}{3}\), \(y=\frac{20}{3}\), \(z=\frac{4\sqrt{41}}{3}\), \(\frac{dx}{dt}=16\), \(\frac{dy}{dt}=20\) into \(\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}\)

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Answer:

\(25.61\) mph