Ships moving and changing rate

[Kristy]

Here is the complete text of the problem:

At noon ship A is 150 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 pm?

Here's what I tried. I initially misread it as they were travelling toward each other and when I got that they switch positions, I was suspicious and reread it.

Now what I did is that ship A after 4 hours has travelled
35 * 4 =140 miles west. That leaving a "base" of a triangle 10.

Then the ship B is travelling north at 25 *4 so that's 100. So that the height of the triangle is 100.

So then I called the distance between them x and solved for it with the Pythagorean.

x^2 = 100^2 + 10^2
Simplifies to
x = square root of 10,100.

So it seems like I found the distance, but what they want is the rate of change. I answered a question, but not the one they want. So I'm a little lost, I don't see anything to take the derivative of

 

[galactus]

You're on the right track. You just need to finish.

\(\displaystyle \L\\\frac{dx}{dt}=35, \;\ \frac{dy}{dt}=25\)

By Pythagoras:

\(\displaystyle \L\\z^{2}=(150-x)^{2}+y^{2}\)

Differentiate:

\(\displaystyle \L\\2z\frac{dz}{dt}=2(150-x)\frac{-dx}{dt}+2y\frac{dy}{dt}...[1]\)

at t=4, x=140 and y=100.

Pythagoras, \(\displaystyle \L\\z=\sqrt{(150-140)^{2}+100^{2}}=100.49\)

Enter in values into [1]:

\(\displaystyle \L\\2(100.49)\frac{dz}{dt}=2(150-140)(-35)+2(100)(25)\)

Solve for dz/dt:

\(\displaystyle \L\\\frac{dz}{dt}\approx{21.395}\)

At 4:00, the distance is changing at 21.40 km/hr.

 

[Kristy]

You're on the right track. You just need to finish.

\(\displaystyle \L\\\frac{dx}{dt}=35, \;\ \frac{dy}{dt}=25\)

By Pythagoras:

\(\displaystyle \L\\z^{2}=(150-x)^{2}+y^{2}\)

Differentiate:

I am getting confused here. by 2 things: the minus dx/dt, and by the ...[1]

\(\displaystyle \L\\2z\frac{dz}{dt}=2(150-x)\frac{-dx}{dt}+2y\frac{dy}{dt}...[1]\)

at t=4, x=140 and y=100.

Pythagoras, \(\displaystyle \L\\z=\sqrt{(150-140)^{2}+100^{2}}=100.49\)

Enter in values into [1]:

\(\displaystyle \L\\2(100.49)\frac{dz}{dt}=2(150-140)(-35)+2(100)(25)\)

Solve for dz/dt:

\(\displaystyle \L\\\frac{dz}{dt}\approx{21.395}\)

At 4:00, the distance is changing at 21.40 km/hr.

I couldn't seem to change the color of my question:

I am getting confused here. by 2 things: the minus dx/dt, and by the ...[1].

edit: never mind, I figured out what the [1] meant that you were referring to that equation as equation 1 so I would go back and substitute into it. Now I'm just confused on the -dx/dt thing

 

[soroban]

Hello, Kristy!

I hope you made a sketch . . .
. . and don't we have to take a derivative at some time?

At noon ship A is 150 km west of ship B.
Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h.
How fast is the distance between the ships changing at 4:00 pm?

Ship \(\displaystyle A\) starts at point\(\displaystyle P\).
\(\displaystyle t\) hours later, it has moved \(\displaystyle 35t\) km to point \(\displaystyle A\).
Then: \(\displaystyle AQ \,=\,150\,-\,35t\)

Ship \(\displaystyle B\) starts at point \(\displaystyle Q\).
\(\displaystyle t\) hours later, it has moved \(\displaystyle 25t\) km to point \(\displaystyle B\).

                                        * B
                                    *   |
                           x    *       | 25t
                            *           |
                        *               |
                    *                   |
    * - - - - - * - - - - - - - - - - - * Q
    P    35t    A        150-35t

The distance between them, \(\displaystyle x \,=\,AB\), is given by:

. . \(\displaystyle x^2 \;=\;(25t)^2\,+\,(150\,-\,35t)^2\;\;\Rightarrow\;\; x^2\;=\;1850t^2\,-\,10,500t + 22,500\)


Differentiate with respect to time: \(\displaystyle \L\:2x\cdot\frac{dx}{dt} \;=\;3700t\,-\,10,500\)

. . and we have: \(\displaystyle \L\:\frac{dx}{dt} \;=\;\frac{1850t - 5250}{x}\;\) [1]


When \(\displaystyle t\,=\,4\), we have this right triangle:

                      * B
                  *   |
              *       | 100
          *           |
      * - - - - - - - *
      A       10      Q

Then: \(\displaystyle \:x^2 \;=\;10^2\,+\,100^2 \:=\:10,100\;\;\Rightarrow\;\;x \,=\,10\sqrt{101}\)

Substitute into [1]: \(\displaystyle \L\:\frac{dx}{dt} \;=\;\frac{1850(4)\,-\,5250}{10\sqrt{101}} \;=\;21.39329959\)


Therefore, the ships are separating at about \(\displaystyle \L21.4\)km/hr.

 

[galactus]

The -dx/dt came from the chain rule in \(\displaystyle \L\\(150-x)^{2}\)

 

[Kristy]

The -dx/dt came from the chain rule in \(\displaystyle \L\\(150-x)^{2}\)

Okay, so that is the derivative of x, right? I guess I'm more used to thinking it is 1.

 

[Kristy]

Well, I think I get this one. It is pretty tough though. I can see how to do all the steps, but I think I need to make sure I really understand it. Hopefully I will in the next problem I'm working on.