Math CentralQuandaries & Queries


Question from Russell, a student:

Consider the family of functions
f(t)= Asin3t + Acos3t +Bsin8t + Bcos8t

find exact values of parameters A and B so that f(0) = 2 and f ' (0) = 1

Hi Russell.

The expression f(0) means t = 0. When you substitute in 0 for t you get:

f(0) = A sin(3(0)) + A cos(3(0)) + B sin(8(0)) + B cos(8(0))

which simplifies to

f(0) = A + B

and s ince you are told that f(0) = 2, this means

2 = A + B.

Now take the derivative of f(t) using the chain rule for each term:

f ' (t) = 3A cos(3t) - 3A sin(3t) + 8B cos(8t) - 8B sin(8t)

This simplifies to:

f ' (0) = 3A + 8B

and since you are told that f ' (0) = 1,

1 = 3A + 8B.

Thus you have two equations involving two unknowns and you can solve this in any of the usual ways.

Stephen La Rocque.

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