Related Rates

Date: 11/26/98 at 14:44:34
From: Joe Verrico
Subject: Related Rates

1. A ladder 4m long is leaning against the vertical wall of a house. If 
the base of the ladder is pulled horizontally away from the house at 
a rate of 0.7 meters per second, how fast is the top of the ladder 
sliding down that wall at the instant when it is 2m from the ground? 

2. An airplane flies at a height of 9km in the direction of an observer 
on the ground at a speed of 800 km/hr. Find the rate of change of the 
angle of elevation of the plane from the observer at the instant when 
this angle is */3 radians.

3. Water is running out of a conical funnel at the rate of 3 cubic 
inches per second.  The funnel has a radius of 2 inches and a height of 
8 inches.  How fast is the water level dropping when it is 3 inches 
from the top?  

Show work.

Date: 11/26/98 at 20:07:24
From: Doctor Kate
Subject: Re: Related Rates

Joe:

You've sent us three related rates questions. I'll help you through 
one of them as an example and give you some hints on the others to get 
you going.

The idea behind related rates is that if you can find an equation 
relating two quantities, you can use that to find the relation not just 
of the two quantities, but of the rates of change of those quantities.  
So the general idea is:

   Figure out which quantities we're interested in.
   Find a relation between the quantities.
   Differentiate to find a relation between their rates.

This will become clear in question 1. The first question we have to ask 
is: 

a. What do we want?

   We want to know how FAST the top of the ladder is sliding. That is,  
   the rate of change of the height of the top of the ladder with 
   respect to time. (Remember that speed is a rate of change of 
   distance with respect to time; as time passes, the top of ladder 
   gets closer to the ground).

   So we want dx/dt, where x is the distance of the top of the ladder 
   from the ground.

b. What are we given?

   We are given the rate at which the base of the ladder moves. This is 
   the rate of change of the distance between the base and the house 
   with respect to time.

   So we have dy/dt, where y is the distance between the ladder and the 
   house.

In related rates problems, this is generally the setup. We are given 
one rate, and asked to determine another.

Let's look at a picture:

         wall
          |
         _|
          |\
          | \ladder
         x|  \
          |   \
         ----------- ground
          | y  |

The next step is to find a relation between x and y, the two quantities 
we care about (the distance from the ground to the top of the ladder up 
the wall, and the distance from the house to the bottom of the ladder 
along the ground).

What information will help us?

We know the ladder is 4 m long. Do you remember the Pythagorean 
Theorem?  It says the square of the hypotenuse of a triangle is the sum 
of the squares of the other two sides. Since the ladder is the 
hypotenuse, and its length is 4 m.,

   4^2 = x^2 + y^2

Now we have a relation between x and y (that is, an equation containing 
x and y). The next step is to differentiate with respect to time (t).  
Remember that the rates we are interested in are with respect to time.  
This is just implicit differentiation, so you may recall how this goes:

       0 = 2x(dx/dt) + 2y(dy/dt)

Now that we have this, what do we want to do? Well, we are looking for 
dx/dt, so let's solve for that:

   dx/dt = (-2y/2x)(dy/dt) = (-y/x)(dy/dt)

In order to determine dx/dt, we need three pieces of information:

        x
        y
        dy/dt

How can we determine x? Well, it's given in the question! It asks us to 
find dx/dt when x = 2m (that is, when the top of the ladder is 2m from 
the ground).  So

        x = 2

How can we determine y? This is only a bit trickier. Now that we know 
x, and we know the length of the ladder, our original equation should 
help us. Recall that

      4^2 = x^2 + y^2

Substituting in x = 2,

       16 = 4 + y^2

So    y^2 = 12

And, therefore, 

        y = 2*sqrt(3) = 2 times the square root of 3

So      y = 2*sqrt(3)

Now what's left? dy/dt! Do we know that yet? Yes! It's our given! In 
the question it says the base of the ladder is pulled horizontally away 
from the house at a rate of 0.7 meters per second. That means:

    dy/dt = 0.7

Plugging these in,

    dx/dt = (-2*sqrt(3)/2)*0.7 = -1.21

What does this mean?  It says "The top of the ladder is sliding down 
the house at 1.21 meters per second." (Note that 'sliding down' means 
the height is decreasing, because of the negative sign.)

So we've figured it out!  

Each step is relatively easy, but the overall attack for related rates 
problems is difficult. To review:

- Find out what rates you are interested in. They will often be with 
respect to time. You will usually be given one, and asked to find the 
other.

- Find something relating the changing quantities to each other. At 
this point, a diagram is most useful. Create an equation. This is the 
creative problem-solving step.

- Differentiate the equation with respect to the appropriate thing 
(usually time) to obtain some relation between the rates of change.  
This will yield another equation.

- Determine what you need at this point to complete the problem. That 
usually means finding numbers to plug into your equation so you can 
solve for the unknown rate.

Here are a few hints on the other two problems you sent:

For the airplane problem, the tricky part will be finding an equation 
relating your two quantities, the plane's position and the angle.  
Remember that the speed of the plane is really the rate of change of 
its position with respect to time. Draw a diagram and use your 
knowledge of trigonometry.

For the water problem, it takes a moment to figure out what quantities 
you're interested in. Here you need to consider the volume of water, 
and the height of the surface. How can we possible relate these things?  
Here, a diagram is essential. Consider perhaps what you know about 
similar triangles.

Good luck!

- Doctor Kate, The Math Forum
  http://mathforum.org/dr.math/