# Thread: Related rates problem

1. ## Related rates problem

Hey, I have this problem that I am halfway done, but I do not know how to solve the second portion.

Question
Ship S is 41 Km due North of ship T. Ship S sails south at a rate of 9 km/h. Ship T sails west at 10 km/h.
a) At what rate are they approaching or separating one hour later?
b) When do they cease to approach one another and how far apart are they at this time?

Solution
a)
I figured out the distance after 1 hour: 33.52km
I then figured out the rate of change of the seperation/approach: 0.57kmh (approaching)

b) I do not know how to solve this part though. I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables. Help?

2. I think you need to find out when the hypotenuse changes from minimizing the distance as t moves west and s moves south to when the hypotenuse starts increasing the distance and then find that distance of the hypotenuse at that moment.

3. Originally Posted by dwsmith
I think you need to find out when the hypotenuse changes from minimizing the distance as t moves west and s moves south to when the hypotenuse starts increasing the distance and then find that distance of the hypotenuse at that moment.
Okay. How do I do that?

Thank you for the response!

4. I haven't done related rates in awhile but when you are minimizing or maximizing you need two equations. One will be $a^2+b^2=c^2$ since we want to minimize c. The other might be the area of triangle not 100% on that though.
What we know:

$\dot x =10$

$\dot y =-9$

From this, I think you will have to implicit differentiate the Pythagorean identity but after that I am a little hazy on what needs to be done.

5. Originally Posted by dwsmith
I haven't done related rates in awhile but when you are minimizing or maximizing you need two equations. One will be $a^2+b^2=c^2$ since we want to minimize c. The other might be the area of triangle not 100% on that though.
What we know:

$\dot x =10$

$\dot y =-9$

From this, I think you will have to implicit differentiate the Pythagorean identity but after that I am a little hazy on what needs to be done.
Thanks for trying!

6. Anyone know what I need to do? I think I just need to find when they are at a minimum distance from each other, that is the same as when the rate of change of distance apart would be zero.

7. Originally Posted by Kakariki
I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables.
Yes it does - at least, you don't need to worry about hypotenuse z because it goes to zero with dz/dt (if it's on the dz/dt side of the equals sign) or else is merely the denominator of a zero-valued fraction (if it's on that side). And you know that for y you can substitute 41 - 9/10 x.

8. Originally Posted by Kakariki
Hey, I have this problem that I am halfway done, but I do not know how to solve the second portion.

Question
Ship S is 41 Km due North of ship T. Ship S sails south at a rate of 9 km/h. Ship T sails west at 10 km/h.
a) At what rate are they approaching or separating one hour later?
b) When do they cease to approach one another and how far apart are they at this time?

Solution
a)
I figured out the distance after 1 hour: 33.52km
I then figured out the rate of change of the seperation/approach: 0.57kmh (approaching)

b) I do not know how to solve this part though. I could use the equation I used to find the rate of change, but setting that to zero really doesn't help in finding 3 other variables. Help?

A is the starting position of ship S

B is the starting position of ship T

AB = 41

$x=10t$

$y=41-9t$

$z=\sqrt{(10t)^2+(41-9t)^2}$

$\frac{dx}{dt}=10$ and ${dy}{dt}=-9$

$x^2+y^2=z^2$

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=2z\frac{dz}{dt}$

$\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}} {z}$

Set $\frac{dz}{dt}=0$

$x\frac{dx}{dt}+y\frac{dy}{dt}=0$

$(10t)(10)+(41-9t)(-9)=0$

Solve for t and plug into the formula for z