Karl's Calculus Tutor - Solution to Exercise 8.3-2

Solution to Exercise 8.3-2

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

   f(t)  =  e^lt (sin(t) + A cos(t))

has a zero crossing at t = p/8 and a critical point at t = p/12.

Step 1: Set up the zero crossing equation. That is, put p/8 in for t and zero in for f(t).

   0  =  e^lp/8 (sin(p/8) + A cos(p/8))

Step 2: Solve for A. You can divide out the e^lp/8. You find that

           sin(p/8)                        _
   A  =  -           =  -tan(p/8)  =  1 - Ö2  =  -0.4142136
           cos(p/8)

If you didn't know that -tan(p/8) = 1 - Ö2 you can be excused for using your calculator for finding its numerical value.

Step 3: Find f'(t). Using the product rule you find

   f'(t)  =  le^lt(sin(t) + A cos(t))  +  e^lt(cos(t) - A sin(t) )

          =  e^lt( (l - A)sin(t) + (lA + 1)cos(t) )

Step 4: Put in what you know at about the critical point. You know that at the critical point f'(t) = 0. And you know that the critical point happens at t = p/12.

   0  =  e^lp/12( (l - A)sin(p/12) + (lA + 1)cos(p/12) )

Step 5: Solve for l. You can divide out the e^lp/12. Use your calculator to find

   sin(p/12)  =  0.2588190

   cos(p/12)  =  0.9659258

Putting all that in (and using your calculator a lot) you have

   0  =  0.2588190(l - A)  +  0.9659258(lA + 1)

   0.2588190A - 0.9659258  =  (0.2588190 + 0.9659258A)l

   A  =  -0.4142136


         0.2588190A - 0.9659258
   l  =                          =  7.595752
         0.2588190 + 0.9659258A

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