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

To determine where a sinusoidal function (like \( y = \sin(x) \) or \( y = \cos(x) \)) 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 \( y = \sin(x) \), the first derivative is \( y' = \cos(x) \), and the second derivative is:
     \[
     y'' = -\sin(x)
     \]
   - For \( y = \cos(x) \), the first derivative is \( y' = -\sin(x) \), and the second derivative is:
     \[
     y'' = -\cos(x)
     \]

### 2. **Determine Concavity:**
   - A function is **concave up** where its second derivative is **positive**: \( y'' > 0 \).
   - A function is **concave down** where its second derivative is **negative**: \( y'' < 0 \).

### 3. **Apply This to Sinusoidal Functions:**

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

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

### 4. **Analyze the Interval:**
   - Given a specific interval, check where \( \sin(x) \) or \( \cos(x) \) is positive or negative within that interval to determine concavity.

### Example:
For \( y = \sin(x) \) on the interval \( [0, 2\pi] \):
- From \( 0 \) to \( \pi \), \( \sin(x) > 0 \), so \( y = \sin(x) \) is concave down.
- From \( \pi \) to \( 2\pi \), \( \sin(x) < 0 \), so \( y = \sin(x) \) 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 \( y = \cos(x) \) on the interval \( [0, 2\pi] \)?
2. How would you determine the concavity of \( y = \sin(x) + 2\cos(x) \)?
3. What are the points of inflection for \( y = \sin(x) \) in the interval \( [0, 2\pi] \)?
4. How do concavity and points of inflection relate to the graph's curvature?
5. How does changing the amplitude of \( y = A\sin(x) \) 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.

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

Learn how to determine where sinusoidal functions like sin(x) or cos(x) are concave up or down using the second derivative. This guide provides a step-by-step approach to analyzing the concavity over a given interval, utilizing key calculus concepts. Ideal for students in Grades 11-12 or introductory college calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation