Related Rates, Boats Moving Around

[QuarkCharmer]

Homework Statement

Stewart Calculus 6E, 3.8 #14
At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h, and ship B is sailing north at 25km/h. How fast is the distance between the ships changing at 4pm?

Homework Equations

Distance Formula I think.

The Attempt at a Solution

Again, I am having trouble phrasing the question. It seems to me, that I need to find an equation in terms of distance and then implicitly differentiate. If the coordinate points of the ships are A(0,0) and B(150,0) at time 12, I thought that I could solve for the distance like this.

$$D = \sqrt{\Delta X^2+\Delta Y^2}$$

Since, I am looking for D', I would differentiate that to be:
$$D' = 1/2(\Delta X^2+\Delta Y^2)^{\frac{-1}{2}}\frac{d}{dt}(\Delta X^2+\Delta Y^2)$$

$$D' = 1/2(\Delta X^2+\Delta Y^2)^{\frac{-1}{2}}(2 \Delta X \Delta X'+2 \Delta Y \Delta Y')$$

...and that is where I realized that I am going about this all wrong.

 

[micromass]

No, first you need to find the general position of the boats at time t.

Let's begin with boat A.
What is the position of the boat at time 12?
What is the position one hour after?
Two hourds after?
t hours after?

 

[QuarkCharmer]

Boat A
What is the position of the boat at time 12?
(0,0)
What is the position one hour after?
(35,0)
Two hours after?
(70,0)
t hours after?
(35t,0)

 

[micromass]

Indeed, now do the same for boat B!

 

[QuarkCharmer]

Boat B
What is the position of the boat at time 12?
(150,0)
What is the position one hour after?
(150,25)
Two hours after?
(150,50)
t hours after?
(150,25t)

 

[micromass]

Ok, so at t, boat A is at (35t,0) and boat B is at (150,25t). What is the distance between the boats?

 

[QuarkCharmer]

Ok, so at t, boat A is at (35t,0) and boat B is at (150,25t). What is the distance between the boats?

$$D(t)=\sqrt{(35t-150)^{2}+(0-25t)^{2}}$$
$$D(t)=\sqrt{(35t-150)^{2}+(-25t)^{2}}$$
?

 

[micromass]

Correct!

Now, how do you find how fast the ship changes?

 

[QuarkCharmer]

Correct!

Now, how do you find how fast the ship changes?

You mean the distance right?
Find D'(t)
I'm working that out now...

 

[micromass]

You mean the distance right?
Find D'(t)
I'm working that out now...

Yes,sorry!

You might want to simplify D first...

 

[QuarkCharmer]

$$D(t)=\sqrt{1225t^2-10500t+22500+625t^2}$$
$$D(t)=\sqrt{1850t^2-10500t+22500}$$
$$D(t)=\sqrt{50(37t^2-210t+450)}$$
$$D(t)=5\sqrt{2(37t^2-210t+450)}$$
$$D(t)=5\sqrt{74t^2-420t+900}$$

$$D'(t)=5\frac{d}{dt}(74t^2-420t+900)^{\frac{1}{2}}$$
$$D'(t)=5\frac{1}{2}(74t^2-420t+900)^{\frac{-1}{2}}\frac{d}{dt}(74t^2-420t+900)$$
$$D'(t)=\frac{5}{2}(74t^2-420t+900)^{\frac{-1}{2}}(150t-420)$$

Then for when t=4;
$$D'(4)=\frac{5}{2}(74(4)^2-420(4)+900)^{\frac{-1}{2}}(150(4)-420)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$
$$D'(4)=132605.769105$$

?? That can't be right! (It should be 21.3932995895, so says my calculator)

 

[micromass]

$$D(t)=\sqrt{1225t^2-10500t+22500+625t^2}$$
$$D(t)=\sqrt{1850t^2-10500t+22500}$$
$$D(t)=\sqrt{50(37t^2-210t+450)}$$
$$D(t)=5\sqrt{2(37t^2-210t+450)}$$
$$D(t)=5\sqrt{74t^2-420t+900}$$

$$D'(t)=5\frac{d}{dt}(74t^2-420t+900)^{\frac{1}{2}}$$
$$D'(t)=5\frac{1}{2}(74t^2-420t+900)^{\frac{-1}{2}}\frac{d}{dt}(74t^2-420t+900)$$
$$D'(t)=\frac{5}{2}(74t^2-420t+900)^{\frac{-1}{2}}(150t-420)$$

In the last term I get 148t-420

Then for when t=4;
$$D'(4)=\frac{5}{2}(74(4)^2-420(4)+900)^{\frac{-1}{2}}(150(4)-420)$$
$$D'(4)=\frac{5}{2}(86836)^{\frac{-1}{2}}(180)$$

Wow, this step can't be correct! The 86836 is way to big! Can you calculate this again, I get something very different.

 

[QuarkCharmer]

Haha, yeah, I figured it out. I was just unsure how to set up the problem. Thanks for the help.