Math Problem Statement

How do you determine from just an interval where it is concave up or down for sinusoidal functions

Solution

To determine where a sinusoidal function (like y=sin(x)y = \sin(x) or y=cos(x)y = \cos(x)) is concave up or concave down over an interval, you need to analyze the second derivative of the function. Here's how you can do it step-by-step:

1. Find the Second Derivative:

  • For y=sin(x)y = \sin(x), the first derivative is y=cos(x)y' = \cos(x), and the second derivative is: y=sin(x)y'' = -\sin(x)
  • For y=cos(x)y = \cos(x), the first derivative is y=sin(x)y' = -\sin(x), and the second derivative is: y=cos(x)y'' = -\cos(x)

2. Determine Concavity:

  • A function is concave up where its second derivative is positive: y>0y'' > 0.
  • A function is concave down where its second derivative is negative: y<0y'' < 0.

3. Apply This to Sinusoidal Functions:

  • For y=sin(x)y = \sin(x):

    • Since y=sin(x)y'' = -\sin(x), concavity depends on the sign of sin(x)-\sin(x).
    • Concave Up: sin(x)>0-\sin(x) > 0sin(x)<0\sin(x) < 0, which occurs on intervals like (π,2π)( \pi, 2\pi ), (3π,4π)(3\pi, 4\pi), etc.
    • Concave Down: sin(x)<0-\sin(x) < 0sin(x)>0\sin(x) > 0, which occurs on intervals like (0,π)(0, \pi), (2π,3π)(2\pi, 3\pi), etc.
  • For y=cos(x)y = \cos(x):

    • Since y=cos(x)y'' = -\cos(x), concavity depends on the sign of cos(x)-\cos(x).
    • Concave Up: cos(x)>0-\cos(x) > 0cos(x)<0\cos(x) < 0, which occurs on intervals like (π2,3π2)\left( \frac{\pi}{2}, \frac{3\pi}{2} \right), (5π2,7π2)\left( \frac{5\pi}{2}, \frac{7\pi}{2} \right), etc.
    • Concave Down: cos(x)<0-\cos(x) < 0cos(x)>0\cos(x) > 0, which occurs on intervals like (0,π2)(0, \frac{\pi}{2}), (3π2,5π2)\left( \frac{3\pi}{2}, \frac{5\pi}{2} \right), etc.

4. Analyze the Interval:

  • Given a specific interval, check where sin(x)\sin(x) or cos(x)\cos(x) is positive or negative within that interval to determine concavity.

Example:

For y=sin(x)y = \sin(x) on the interval [0,2π][0, 2\pi]:

  • From 00 to π\pi, sin(x)>0\sin(x) > 0, so y=sin(x)y = \sin(x) is concave down.
  • From π\pi to 2π2\pi, sin(x)<0\sin(x) < 0, so y=sin(x)y = \sin(x) is concave up.

Would you like more detailed steps on any part of this?


Here are 5 questions to explore the topic further:

  1. What is the concavity of y=cos(x)y = \cos(x) on the interval [0,2π][0, 2\pi]?
  2. How would you determine the concavity of y=sin(x)+2cos(x)y = \sin(x) + 2\cos(x)?
  3. What are the points of inflection for y=sin(x)y = \sin(x) in the interval [0,2π][0, 2\pi]?
  4. How do concavity and points of inflection relate to the graph's curvature?
  5. How does changing the amplitude of y=Asin(x)y = A\sin(x) affect its concavity?

Tip: The second derivative test is a powerful tool to determine not just concavity but also to test for local maxima or minima of functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Sinusoidal Functions
Concavity
Second Derivative

Formulas

y'' = -sin(x) for y = sin(x)
y'' = -cos(x) for y = cos(x)

Theorems

Second Derivative Test

Suitable Grade Level

Grades 11-12 or Introductory College Level