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 = 4√41 mi/hr what is the rate at which the distance between the two people is changing after 45 minutes? rate at 45 minutes =

Explanation:

Step1: Convert time to hours

Since \(1\) hour \( = 60\) minutes.
For \(t = 20\) minutes, \(t=\frac{20}{60}=\frac{1}{3}\) hours.
For \(t = 45\) minutes, \(t=\frac{45}{60}=\frac{3}{4}\) hours.
Let the distance of the person going east be \(x\) and the person going south be \(y\), and the distance between them be \(z\). By the Pythagorean theorem \(z^{2}=x^{2}+y^{2}\), where \(x = 16t\) and \(y = 20t\). So \(z=\sqrt{(16t)^{2}+(20t)^{2}}=\sqrt{256t^{2}+400t^{2}}=\sqrt{656t^{2}} = 4\sqrt{41}t\)
Differentiate \(z^{2}=x^{2}+y^{2}\) with respect to time \(t\). Using the chain - rule, \(2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\), then \(\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}\)
Since \(\frac{dx}{dt}=16\) mph and \(\frac{dy}{dt}=20\) mph.

Step2: Calculate for \(t = 20\) minutes (\(t=\frac{1}{3}\) hours)

\(x = 16\times\frac{1}{3}=\frac{16}{3}\) miles, \(y = 20\times\frac{1}{3}=\frac{20}{3}\) miles, \(z=\sqrt{(\frac{16}{3})^{2}+(\frac{20}{3})^{2}}=\frac{4\sqrt{41}}{3}\) miles
\(\frac{dz}{dt}=\frac{\frac{16}{3}\times16+\frac{20}{3}\times20}{\frac{4\sqrt{41}}{3}}=\frac{256 + 400}{4\sqrt{41}}=\frac{656}{4\sqrt{41}}=\frac{164}{\sqrt{41}} = 4\sqrt{41}\) mph (this is a verification step for the given value)

Step3: Calculate for \(t = 45\) minutes (\(t=\frac{3}{4}\) hours)

\(x=16\times\frac{3}{4}=12\) miles, \(y = 20\times\frac{3}{4}=15\) miles, \(z=\sqrt{12^{2}+15^{2}}=\sqrt{144 + 225}=\sqrt{369}=3\sqrt{41}\) miles
Substitute into \(\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}\), \(\frac{dx}{dt}=16\), \(\frac{dy}{dt}=20\)
\(\frac{dz}{dt}=\frac{12\times16+15\times20}{3\sqrt{41}}=\frac{192 + 300}{3\sqrt{41}}=\frac{492}{3\sqrt{41}}=\frac{164}{\sqrt{41}}=4\sqrt{41}\) mph

Answer:

\(4\sqrt{41}\)