Solution to Exercise 8.3-1

The problem was: Find λ and ω if the function,

   f(t)  =  e^λt cos(ωt)

has a zero crossing at t = π/4 and has a critical point at t = π/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 π/4 and f(t) to zero.

   0  =  e^λt cos(ω × π/4)

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

   0  =  cos(ω × π/4)

Step 2: Use trig to solve for ω. If cos(θ) is zero, then θ = π/2 or θ = 3π/2 or θ = 5π/2 etc. Here θ = ω × π/4. You can see from this that ω 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 ω, and that is ω = 2.

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

   f'(t)  =  λe^λt cos(ωt)  -  ωe^λt sin(ωt)

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 ω (we just figured out that that is what ω is equal to), and put in π/6 for t (because the problem says the critical point is at that value of t).

   0  =  λe^λπ/6 cos(π/3)  -  2e^λπ/6 sin(π/3)

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

               1
  cos(π/3)  =   
               2
                _
               √3
  sin(π/3)  =    
                2

So with that knowledge the equation becomes

         λ    _
   0  =    - √3
         2

           _
   λ  =  2√3

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