# Insane, out of control boat problem

I had this problem on my Calc 1 exam today and found it to be a bit difficult. I'll walk you through the problem and my attempt at solving it, hopefully you guys will be able to help me!

At noon, boat A is 20 miles west of boat B. Boat A is traveling north at 6 mph and boat B is traveling east at 4 mph. At 5:00 PM, what is the rate at which the distance between the two boats is changing?

I sketched up this beautiful diagram:

https://i.imgur.com/orwcKH7.jpg (Sorry, I can't embed images yet)

And I then tried to use the distance formula to solve this problem $$d = \sqrt{(y_2-y_1)^2 + (x_2-x_1)^2}$$ $$d = \sqrt{(0-6t)^2 + (4t + 20)^2}$$ $$= \sqrt{52 t^2+160 t+400}$$ But when I try to derive $\frac{dd}{dt}$ to get the rate of change in distance with respect to time, sh*t starts getiting funky af. Can someone tell me where I'm going wrongwith this? Thank you!

• My bad, I forgot to include that! Starts at noon. – Defacto Nov 15 '16 at 3:43
• I don't see what gets weird. The derivative $\frac{dd}{dt}$ is pretty easy to compute, then you just want to plug in $t = 5$. – Morgan Rodgers Nov 15 '16 at 3:46
• "Insane" and "out of control" may be overstating the situation just a bit. – Tpofofn Nov 15 '16 at 3:50
• Oops, look like I had it right all along. Sorry fellas – Defacto Nov 15 '16 at 3:51

The distance function between $A$ and $B$ at time $t$ is given by $$L(t) = \sqrt{(6t)^2 + (4t + 20)^2} = 2\sqrt{13t^2 + 40t + 100}.$$ It follows that $$\frac{dL}{dt} = \frac{26t + 40}{\sqrt{13t^2+40t+100}}.$$ When $t = 5$ hours, this is simply $$L'(5) = \frac{34}{5} = 6.8 \text{ mi/hr}.$$

I would recommend drawing a picture of where the ships are at noon and where they are and at 5:00 PM. Your second picture should be a right triangle with side lengths:

$$a=6 \ \frac{mi}{h}*5 \ h = 30 \ mi,$$ $$b=20 \ mi + \bigg(4 \ \frac{mi}{h} * 5 \ h\bigg) = 40 \ mi.$$

Then use Pythagorean Theorem to find the hypotenuse.

$$c^2=a^2+b^2$$ $$c^2=30^2+40^2$$ $$c^2=2,500$$ $$c=50 \ mi$$

You also know that $$\frac{da}{dt}=6\frac{mi}{h}$$ because ship A is moving $$6\frac{mi}{h}$$ and making side $$a$$ larger as it moves. Similarly, $$\frac{db}{dt}=4\frac{mi}{h}$$ because side $$b$$ of your triangle is growing at that rate.

Now all you need to do is differentiate the Pythagorean Theorem with respect to time, solve for $$\frac{dc}{dt}$$, and plug in the other values.

$$c^2=a^2+b^2$$ $$2c\frac{dc}{dt}=2a\frac{da}{dt}+2b\frac{db}{dt}$$ $$c\frac{dc}{dt}=a\frac{da}{dt}+b\frac{db}{dt}$$ $$\frac{dc}{dt}=\frac{a\frac{da}{dt}+b\frac{db}{dt}}{c}$$ $$\frac{dc}{dt}=\frac{30*6+40*4}{50}=6.8 \ \frac{mi}{h}$$

Here's a link to a site with a much more detailed explanation of how to solve a problem like this. This site has a good explanation of how to draw your picture and go through the problem.

https://jakesmathlessons.com/derivatives/related-rates/