Question

at noon, ship a is 40 miles due west of ship b. ship a is sailing west at 16 mph and ship b is sailing north at 18 mph. how fast is the distance between the ships changing at 6 pm? draw a diagram, and use the pythagorean theorem to relate the sides of the triangle. $a^{2}+b^{2}=c^{2}$ $\frac{dc}{dt}=\square$ mph

Explanation:

Step1: Determine the positions of the ships at 6 PM

From noon to 6 PM, \(t = 6\) hours.
Ship A is sailing west at \(16\) mph. Its initial position is \(40\) miles west of Ship B. So the distance of Ship A from the initial - position of Ship B (let's call it \(a\)) is \(a=40 + 16t\). When \(t = 6\), \(a=40+16\times6=40 + 96=136\) miles.
Ship B is sailing north at \(18\) mph. So the distance of Ship B from its initial position (let's call it \(b\)) is \(b = 18t\). When \(t = 6\), \(b=18\times6 = 108\) miles.

Step2: Use the Pythagorean theorem \(c^{2}=a^{2}+b^{2}\)

Differentiate both sides with respect to \(t\) using the chain - rule.
\(2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}\), then \(\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\)
We know that \(\frac{da}{dt}=16\) mph (rate of change of \(a\)) and \(\frac{db}{dt}=18\) mph (rate of change of \(b\)).
First, find \(c\) when \(a = 136\) and \(b = 108\). \(c=\sqrt{a^{2}+b^{2}}=\sqrt{136^{2}+108^{2}}=\sqrt{(136\times136)+(108\times108)}=\sqrt{18496 + 11664}=\sqrt{30160}\approx173.67\)

Step3: Substitute the values into the derivative formula

Substitute \(a = 136\), \(\frac{da}{dt}=16\), \(b = 108\), \(\frac{db}{dt}=18\), and \(c\approx173.67\) into \(\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}\)
\(a\frac{da}{dt}+b\frac{db}{dt}=136\times16+108\times18=2176+1944 = 4120\)
\(\frac{dc}{dt}=\frac{4120}{173.67}\approx23.72\)

Answer:

\(23.72\)