Related Rates-Ships Probelm > A sailboat 4 miles west of Port Stanza sails toward Port Stanza at 5 miles per hour. At the same moment, a motorboat 3 miles north of Port Stanza travels due south, toward Port Stanza, at 2 miles per hour. I set up a traingle with the origin being Port Stanza, and set the first sail boat 4 units to the left of the origin, and the other sail boat 3 units above the origin. Which leads to my questions $$\frac{a(\frac{da}{dt})+b(\frac{db}{dt})}{c}=\frac{dc}{dt}$$ Should $a$, and $\frac{db}{dt}$ here be substituted as a negative number, since $a$ is west of the origin, and $\frac{db}{dt}$ since the second sail boat is going south to the origin? $$a=4mi$$ $$\frac{da}{dt}=5\frac{mi}{h}$$ $$b=3mi$$ $$\frac{db}{dt}=2\frac{mi}{h}$$ $$c=5mi$$ The answer I got with it all being plugged in postive is $\frac{26}{5}\frac{mi}{h}$

the sailboat is at $a$ and the motor-boat is at $b$.

you are using $c$ for distance.

$c = \sqrt {a^2 + b^2}\\\ \frac {dc}{dt} = \frac {a \frac {da}{dt} + b\frac{db}{dt}}{c}$

$a<0$ (to the left of the origin) and $\frac {da}{dt} > 0$ (headed toward the origin)

$b>0$ (above of the origin) and $\frac {db}{dt} < 0$ (headed toward the origin).

$\frac {dc}{dt} < 0$ since the boats are getting closer together $c$ is getting smaller. (and both terms in the numerator are $<0$ while the denominator is strictly positive)