# need help on what equation to use [related-rates]

• Oct 15th 2013, 04:30 PM
Jonroberts74
need help on what equation to use [related-rates]
the question is

A ship K is sailing due north at 16km/h, and a second ship R , which is 44 km north of K, is sailing due east at 10 km/h. At what rate is the distance between K and R changing 90 minutes later? Are they approaching one another or separating at this time? explain

so the two known rates are $\displaystyle \frac{dK}{dt} = 16, \frac{dR}{dt} = 10$ and the unknown is $\displaystyle \frac{ds}{dt}$

I am a little lost of what equation to use to relate them.
can I use $\displaystyle x^2 + y^2 = s^2$ ?
• Oct 15th 2013, 05:07 PM
SlipEternal
Re: need help on what equation to use [related-rates]
You can use that formula, but you need to do a little work to make it meaningful. What are x,y, and s? Draw a picture. Think about what those distances actually mean. You may benefit from drawing multiple diagrams. Draw the initial positions of the two ships. Draw the positions of the two ships after 90 minutes. You may even want to draw a diagram of the positions of the two ships at some point in between. If you are using $\displaystyle x^2+y^2=s^2$, then you want x, y, and s to be the sides of a triangle. In the "initial" diagram, this triangle will be difficult to see because one side will have length zero.
• Oct 15th 2013, 05:20 PM
Jonroberts74
Re: need help on what equation to use [related-rates]
so in the initial diagram they would both lie on the y axis and K would be at the origin with R at the point (0,44) correct?

In the initial triangle x will be the side that equals 0 and y and s will equal 44, ?

then after 90 minutes R will be at (15, 44) and y will be at (0,24) and so s will be the distance between them still.

am I on the right track?
• Oct 15th 2013, 05:31 PM
SlipEternal
Re: need help on what equation to use [related-rates]
Yes, but consider how x and y changed. x is increasing, y is decreasing. That is the important point. So $\displaystyle \dfrac{dx}{dt} = \dfrac{dR}{dt}$, but $\displaystyle \dfrac{dy}{dt} = -\dfrac{dK}{dt}$.
• Oct 15th 2013, 07:51 PM
Jonroberts74
Re: need help on what equation to use [related-rates]
would this make the points R(15, 24) and K(0,24) and not R(15,44)?
and y = 24 or y = -24?
$\displaystyle -\frac{dK}{dt} = -16$
and does this make it
$\displaystyle 2x\frac{DR}{dt} - 2y\frac{DK}{dt} = 2s\frac{ds}{dt}{$
• Oct 15th 2013, 10:18 PM
Jonroberts74
Re: need help on what equation to use [related-rates]
so side y decreases by 24 making side y= 20 and thus the negative rate for dy/dt

so I have

$\displaystyle 15^2 + 20^2 =s^2 \Rightarrow s =25$

now I sub this information and the known rates

$\displaystyle 2(15)(10) + 2(24)(-16) = 2(25)\frac{ds}{dt}$

but that's still off.
• Oct 15th 2013, 10:29 PM
Jonroberts74
Re: need help on what equation to use [related-rates]
OH!

$\displaystyle 2(15)(10) + 2(20)(-16) = 2(25)\frac{ds}{dt} \Rightarrow \frac{ds}{dt} = -\frac{34}{5} km/h$ and they are approaching because the current rate of change is negative!!