Karl's Calculus Tutor - Solution to Exercise 8.3-1

Solution to Exercise 8.3-1

The problem was: Find l and w if the function,

   f(t)  =  e^lt cos(wt)

has a zero crossing at t = p/4 and has a critical point at t = p/6.

Step 1: What can you learn from the zero crossing? Solve what you can using the zero crossing first. You do that by setting t to p/4 and f(t) to zero.

   0  =  e^lt cos(w × p/4)

You can divide the e^lt out of both sides and you get:

   0  =  cos(w × p/4)

Step 2: Use trig to solve for w. If cos(q) is zero, then q = p/2 or q = 3p/2 or q = 5p/2 etc. Here q = w × p/4. You can see from this that w can be 2, 6, 10, or any value in the form of 2 + 4n, where n is an integer. But the problem asks for the smallest positive value of w, and that is w = 2.

Step 3: Now that you've solved for w, find f'(t) so you can solve for l. Using the product rule, you find that

   f'(t)  =  le^lt cos(wt)  -  we^lt sin(wt)

Step 4: Put the values that you know into f'(t). Put in zero for f(t) (critical points always occur where the derivative is equal to zero), put in 2 for w (we just figured out that that is what w is equal to), and put in p/6 for t (because the problem says the critical point is at that value of t).

   0  =  le^lp/6 cos(p/3)  -  2e^lp/6 sin(p/3)

Step 5: Solve for l. You can divide out the e^lp/6. You also know from trig that

               1
  cos(p/3)  =   
               2
                _
               Ö3
  sin(p/3)  =    
                2

So with that knowledge the equation becomes

         l    _
   0  =    - Ö3
         2

           _
   l  =  2Ö3

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