Expert: Paul Klarreich - 10/2/2006

Question
THis helped so much, but i just have one more question. At the
end ofthe problem you said deivide by the  sqroot of 2. How
does 2k/(sqroot of 2) =  k*Sqroot2?  Wouldn't it just be left as
is? Maybe I am not seeing it correctly. Thank you so much
foryour help, finding the variables is now so much easier. O and
to asnwer your question, this is not my first time doing  a
related rate problem. Thanks again.

-kels
-------------------------

Followup To

Question -
I have been working on this related rate problem forever,a nd I
don't seem to be getting anywhere. This is how it goes.

Two trucks, one traveling west and the other traveling south, are
approaching an intersecton, If both trucks are traveling at the
rate of k km/hr, show that they are approaching each other at
the rate of k(2)^(1/2) ( k* the square root of 2) km/hr when they
are each m kilometers from the intersection.

I need help setting upa question, I don't know if I am supposed
to come up with easy numbers to plug into the equation or fi I
can prove it using variables. Thanks, and hope it all works!

-kels

Answer -
kelsey Brannan Asks in Category Calculus ...
 
Subject:  an annoying related rate
Private:  no
 
Question:  I have been working on this related rate problem
forever, and I don't seem to be getting anywhere. This is how it
goes.

Two trucks, one traveling west and the other traveling south, are
approaching an intersection,

-- BETTER GET THE HELL OUT OF THERE!  And I recently had an
accident just like this.  Fortunately in my case 'k' was a small
number.   Now I'm going to have those nightmares all over
again.
--
If both trucks are traveling at the rate of k km/hr, show that they
are approaching each other at the rate of k(2)^(1/2) ( k* the
square root of 2) km/hr when they are each m kilometers from
the intersection.

I need help setting up a question, I don't know if I am supposed
to come up with easy numbers to plug into the equation or if I
can prove it using variables. Thanks, and hope it all works!

-kels

-------------------------------------
OK, Kels,

Is this your first attempt at R-R 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 is the key step.

4. Now differentiate implicitly, then substitute the known
quantities and rates, and solve for the unknown rate.

In this case, how about setting it up like this:

Variables:  Let:

x = the distance from the first truck (going W) to the crash
point, er..., sorry, intersection.

y = the distance from the second truck (going S) to the
intersection.

z = the distance between the trucks.

Rates:

dx/dt = the speed of the first truck, GIVEN as  -k  km/hr  
(minus, because the distance to the crash point is decreasing.

dy/dt = the speed of the second truck, also given as  -k  km/hr.
dz/dt = the rate at which the distance between them is
decreasing.  TO BE FOUND.

Relationship:

Your diagram will show a right triangle with  x,y as the legs and
z as the hypotenuse.

x^2 + y^2 = z^2

Differentiate implicitly:

2x dx/dt + 2y dy/dt = 2z dz/dt  or

x dx/dt + y dy/dt = z dz/dt

Put in known values.  We have  x = m and y = m,

Difficulty: We don't have a value for z.
Resolution: Compute it from the diagram and relationship.

z^2 = x^2 + y^2
   = m^2 + m^2
   = 2m^2
z = m sqrt(2)
--------------

m (k) + m (k) = m sqrt(2) dz/dt

The m divides out --
k + k = sqrt(2) dz/dt
2k = sqrt(2) dz/dt

Divide by  sqrt(2):

dz/dt = k sqrt(2)

So there is no problem using the symbolic values  k, m in the
problem.  You don't have to pick numbers or anything, just deal
with those symbolic values.  Yes, you have to be a little more
careful doing that, and it might feel uncomfortable, but it works.


Answer
kelsey Brannan Asks in Category Calculus ...
 
Subject:  an annoying related rate
Private:  no
 
Question:  This helped so much, but I just have one more question. At the
end ofthe problem you said divide by the  sqroot of 2. How
does 2k/(sqroot of 2) =  k*Sqroot2?  Wouldn't it just be left as is? Maybe I am not seeing it correctly. Thank you so much for your help, finding the variables is now so much easier. O and to answer your question, this is not my first time doing  a related rate problem. Thanks again.

-kels
................................
At the end, we had:

2k = sqrt(2) dz/dt

Divide by  sqrt(2):
         2k  
dz/dt = --------
        sqrt(2)

Now it is customary to 'rationalize the denominator.'
         2k    sqrt(2)
dz/dt = -------- -------
        sqrt(2) sqrt(2)
 2k sqrt(2)
= ----------
     2

= k sqrt(2)

.........................
And in response to YOUR note, I didn't say it WAS your first time doing these, but it was your first time asking ME about them, and I always send out that same routine the first time.  Believe me, you are not the first to ask me about R-R problems.