Expert: Paul Klarreich - 11/28/2007

Question
Hi. the I got stuck with the following question:

A ship K, is sailing due north at 16 km/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 each other or separating at this time? Explain.

Could you please help me?

Thank you.
Ruedi

Answer
Questioner:   Ruedi
Category:  Calculus
Private:  No
 
Subject:  Simple Related Rates problem
Question:  Hi. the I got stuck with the following question:

A ship K, is sailing due north at 16 km/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 each other or separating at this time? Explain.

Could you please help me?

Thank you.
Ruedi
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Hi, Ruedi,

Suggestion: This site allows you to read its archive of questions.  Look for questions with subject line Related Rates.  You will probably find every standard example in the archive.

Is this your first attempt at Related-Rate problems?  If so, the scheme is something like this:

1. Identify the variables in the problem -- the things that change.  Give them names.
2. Write their rates of change as derivatives WITH RESPECT TO time.  Note which are known and which is to be found.
3. Determine a relationship (yes, it is called 'related rates' for a reason) between the variables.  Use a diagram, use your life experience, your general knowledge and brilliance, whatever you have to.  This may be the hard part.
4. Differentiate implicitly, THEN substitute the known quantities and rates, and solve for the unknown rate.


Your problem looks like this:
      x  
A+-------R-->
 |
 |
 |
 |
44|44 - y
 |
 |
 |
 |
 |
 ^K, going up
 |
 |
y|
 |
 +






In this case, your variables:

x = horizontal position of R.= AR
y = vertical position of K. = 44 - y
r = distance between R,K.

dx/dt = speed of R, given as 10
dy/dt = speed of K, given as 16
dr/dt = rate of change of distance between them, TO FIND.

At t = 0,  R is at (0,44), and  K is at (0,0).

As they move,  (44 - y)^2 + x^2 = r^2

Now diff:

2(44 - y)(-dy/dt) + 2x dx/dt = 2r dr/dt.
(44 - y)(-dy/dt) + x dx/dt = r dr/dt.

Now where is everyone after 90 min? or 1.5 hrs?

K has moved up 24 miles, so y = 24
R .........right 15 miles, so x = 15

We need an r, so:

(44 - y)^2 + x^2 = r^2, and:
(44 - 24)^2 + (15)^2 = r^2

(20)^2 + (15)^2 = r^2

-->  r = 25

Now subst:
(44 - 24)(-16) + 15(10) = (25) dr/dt.

(20)(-16) + 15(10) = (25) dr/dt.

(4)(-16) + 3(10) = (5) dr/dt.

-64 + 30  = (5) dr/dt.

-34  = (5) dr/dt.

dr/dt = -34/5;  decreasing.