Maximum of a Trigonometric Function (Calculus)

Question

For the function h(t)=e^-tsin(t), state the first maximum and when it occurs.

Answer 1

To find the maximum, we need to first find the critical points by taking the derivative and finding where it is either 0 or undefined. Using the product rule, we know that h'(t) = -e^-tsin(t) + e^-tcos(t) = e^-t(cos(t) - sin(t)). Now this cannot be undefined, for e^-t is defined for all reals and so is sin and cos, so let e^-t(cos(t) - sin(t)) = 0. Now since e^-t can never be 0 (ln(0) is undefined), we know that cos(t) - sin(t) must = 0, so cos(t) = sin(t) and thus tan(t) = 1 when we divide both sides by cos(t). Thus our critical points are in the form π/4 + nπ where n is an integer. Now we can use the second derivative test to find which ones are minimums and which ones are maximums. The second derivative is h''(t) = -e^-t(cos(t) - sin(t) + e^-t(-sin(t) - cos(t)) = -2e^-tcos(t). We need to find the sign of this second derivative to know whether it is a min or max. If it is positive, it is concave up and therefore a min, but if it is negative, it is concave down and therefore a max. Now 2e^-t is always positive, so it does not affect the sign of the second derivative, so we can ignore it. Thus the sign only depends on -cos(t). This is negative whenever the critical point is of the form π/4 + 2nπ, and positive when it is of the form 5π/4 + 2πn (using the unit circle and the fact that cos is positive in quadrant I and negative in quadrant III). This means that all the x values of the form π/4 + 2nπ. Now since this means maximums exist indefinitely to the left (negative with large magnitude), I assume the first maximum implies we are looking for the first positive one, which occurs at π/4. Plugging this into our original function gives a y value of e^-pi/4 *((√2)/2). Thus, the first maximum is (π/4, e^-pi/4 *((√2)/2)).