Calc Diff. Word problem: At noon, Ship A is 100 km west of

[Stealmylilhart]

I am having trouble setting up this problem.

At noon, Ship A is 100 km west of Ship B; Ship A is sailing at 35 km/hr and Ship B is sailing 25 km/hr. Ship A is going south and B is going north. How fast is the distance between the ships changing at 4:00 pm?

I see that the picture is going to form two different triangles with 100 km as their common side. I know how to differentiate, but I don't know how to set this up. I know that the 35 is my dx/dt and 25 is my dy/dt, but that's about all I can get.

 

[skeeter]

in what directions are the two ships sailing?

 

[stapel]

At noon, Ship A is 100 km west of Ship B; Ship A is sailing at 35 km/hr and Ship B is sailing 25 km/hr. How fast is the distance between the ships changing at 4:00 pm?

If all you have is what you posted, then there is no way to solve this, since we are given no information regarding the directions of the two ships. Sorry.

Eliz.

 

[Johnwill]

one of them is west ?

 

[Stealmylilhart]

Sorry A is going south and B is going North.

 

[Johnwill]

i did something similar to this question last week,

Here's a suggestion:

Draw it

 

[Stealmylilhart]

A is going South B is going north

At noon, Ship A is 100 km west of Ship B; Ship A is sailing at 35 km/hr and Ship B is sailing 25 km/hr. How fast is the distance between the ships changing at 4:00 pm?

If all you have is what you posted, then there is no way to solve this, since we are given no information regarding the directions of the two ships. Sorry.

Eliz.

 

[Stealmylilhart]

A is going south b is going north. A is west of B

one of them is west ?

 

[galactus]

I hope I have things labeled correctly.

We want \(\displaystyle \L\\\frac{dD}{dt}\) given \(\displaystyle \L\\\frac{dx}{dt}=35\) and

\(\displaystyle \L\\\frac{dy}{dt}=25\)

Let's try it this way.

Let x=distance traveled by ship A

Let y=distance traveled by ship B

By the diagram, the distance between them is given by

\(\displaystyle \L\\D=\sqrt{(x+y)^{2}+100^{2}}\)

Now differentiate, using some chain rule.

\(\displaystyle \L\\\frac{dD}{dt}=\frac{2(x+y)(\frac{dx}{dt}+\frac{dy}{dt})}{2\sqrt{(x+y)^{2}+100^{2}}}\)

\(\displaystyle \L\\\frac{dD}{dt}=\frac{2(140+100)(35+25)}{2\sqrt{(140+100)^{2}+100^{2}}}\)

\(\displaystyle \H\\\frac{dD}{dt}=\frac{28800}{520}=\frac{720}{13}\approx{55.38} \;\ km/h\)


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Here's another way that dawned on me just as I finished the first method.

Just let x=the time traveled by both ships.

\(\displaystyle \L\\D^{2}=(35x+25x)^{2}+100^{2}\)

\(\displaystyle \L\\D^{2}=(60x)^{2}+100^{2}\)

\(\displaystyle \L\\D^{2}=3600x^{2}+100^{2}\)

Differentiate:

\(\displaystyle \L\\2D\frac{dD}{dt}=7200x\)

After 4 hours B has travelled 100 km and A has travelled 140 km

By Pythagoras: \(\displaystyle \sqrt{240^{2}+100^{2}}=260\)

\(\displaystyle \L\\2(260)\frac{dD}{dt}=7200(4)\)

\(\displaystyle \L\\540\frac{dD}{dt}=28800\)

\(\displaystyle \H\\\frac{dD}{dt}=\frac{28800}{520}=\frac{720}{13}\approx{55.38}\;\ km/h\)

Check it out. See if you or anyone else concurs.

 

[soroban]

Re: Calc Diff. Word problem: At noon, Ship A is 100 km west

Hello, Stealmylilhart!

At noon, Ship A is 100 km west of Ship B.
Ship A is sailing south at 35 km/hr and Ship B is sailing north at 25 km/hr.
How fast is the distance between the ships changing at 4:00 pm?

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

Ships A and B start at P and Q respectively.
. . Distance \(\displaystyle PQ\,=\,100\) km.

Ship A sails south at 35 kph.
In \(\displaystyle t\) hours, it has sailed \(\displaystyle 35t\) km to \(\displaystyle A.\)

Ship \(\displaystyle B\) sails north at 25 kph.
In \(\displaystyle t\) hours, it has sailed \(\displaystyle 25t\) km to \(\displaystyle B.\)



The diagram can be modified as follows:

                                          B
                                          *
                                       *  |
                                    *     | 25t
                                 *        |
                              *           |
                      x    *              +
                        *                 |
                     *                    |
                  *                       | 35t
               *                          |
            *                             |
         *                                |
      * - - - - - - - - - - - - - - - - - *
      A               100

We have a right triangle with sides \(\displaystyle 60t\) and \(\displaystyle 100\), and hypotenuse \(\displaystyle x.\)

Hence, we have: \(\displaystyle \:x^2\;=\;(60t)^2\,+\,100^2 \;=\;3600t^2\,+\,100^2\)

Then: \(\displaystyle \:2x\cdot\frac{dx}{dt}\;=\;7200t\;\;\Rightarrow\;\;\L\frac{dx}{dt}\;=\;\frac{3600t}{x}\)


At 4 p.m. \(\displaystyle (t\,=\,4)\), the right triangle has sides \(\displaystyle 100\) and \(\displaystyle 240.\)
. . hence, its hypotenuse is: \(\displaystyle \,x\,=\,260\)


So we have: \(\displaystyle \L\:\frac{dx}{dt}\;=\;\frac{(3600)(4)}{260} \;\approx\;55.38\) km/hr.


[The same answer as Galactus, of course ... but without radicals.]

 

[Stealmylilhart]

Thanks

I am sorry this thanks is late...but taht really helped a lot....all the different ways to do it was interesting to see!!! You guys rock!!! Sorry this is late again I have a lot of credit this semester.