Expert: Paul Klarreich - 12/7/2006

Question
I am taking a class that's a mixture of precalc and calc right now.  I've tried this problem already by using (tan @)' = (y/x)' and then taking (y/x)' by using the derivative formula of division.  But it doesn't work.

Question:

Let @ (in radians) be an acute angle in a right triangle, and let x and y, respectively, be the lengths of the sides adjacent to and opposite @.  At a certain instant, x= 2 units and is increasing at 1 unit/sec, while y = 2 units and is decreasing at 1/4 unit/sec.  How fast is @ decreasing or increasing at that instant?  

Answer
Hi, Joyce,

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, use your general knowledge and brilliance, do 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 example, your variables are:

@ is the angle in the triangle.
x is one leg.
y is the other leg.

The rates are:

d@/dt is the rate of increase of the angle, TO BE FOUND.
dx/dt is the rate of increase of the first leg, given as  1.
dy/dt is the rate of increase of the second leg, given as -1/4

The relation is (as you noted)

tan @ = y/x

Differentiate with respect to time:(yes, you will use the quotient rule)
               x dy/dt - y dx/dt
sec^2 @ d@/dt = -----------------
                      x^2

Now substitute known values:

x = 2,  y = 2.
dx/dt = 1,  dy/dt = -1/4

What about @?  

sec^2 @ = 1 + tan^2 @ = 1 + (y/x)^2 = 1 + 2^2 = 5

We're in business:

               x dy/dt - y dx/dt
sec^2 @ d@/dt = -----------------  (repeated)
                      x^2


           2 (-1/4) - 2(1)
(5) d@/dt = -----------------
                 2^2

         -1/2 - 2
5 d@/dt = --------
             4

         -1 - 4
5 d@/dt = --------
            8

         -5
5 d@/dt = ---
          8


       -1
d@/dt = --- radians/sec
        8