Problem 16 - Calculus for Biology and Medicine

Question

Use the first derivative test and the second derivative test to determine where each function is increasing, decreasing, concave up, and concave down. You do not need to use a graphing calculator for these exercises. $$ y=\sin \left(\pi x^{2}\right), 0 \leq x \leq 1 $$

Step-by-Step Solution

Answer

Increasing on \([0, \frac{1}{\sqrt{2}})\) and decreasing on \((\frac{1}{\sqrt{2}}, 1]\). Concave up on \([0, \frac{1}{\sqrt{2}})\) and concave down on \((\frac{1}{\sqrt{2}}, 1]\).

Step 1: Find the First Derivative

The function is given as \( y = \sin(\pi x^2) \). First, find the derivative \( y' \) with respect to \( x \). Using the chain rule: \( y' = \cos(\pi x^2) \cdot 2\pi x \).

Step 2: Determine Critical Points

Set the first derivative \( y' \) equal to zero to find critical points. \[ \cos(\pi x^2) \cdot 2\pi x = 0. \] This gives \( 2\pi x = 0 \) or \( \cos(\pi x^2) = 0 \). Therefore, \( x = 0 \) or \( \pi x^2 = (2n+1)\frac{\pi}{2} \), where \( n \) is an integer. Within the interval \( 0 \leq x \leq 1 \), this simplifies to \( x = 0 \) and \( x = \frac{1}{\sqrt{2}} \).

Step 3: First Derivative Test for Increasing/Decreasing

Evaluate the sign of \( y' \) on intervals \( [0, \frac{1}{\sqrt{2}}) \) and \( (\frac{1}{\sqrt{2}}, 1] \). For \( 0 < x < \frac{1}{\sqrt{2}} \), \( y' > 0 \), so \( y \) is increasing. For \( x > \frac{1}{\sqrt{2}} \), \( y' < 0 \), so \( y \) is decreasing.

Step 4: Find the Second Derivative

Find the second derivative \( y'' \) of \( y \). Again using the chain rule and product rule: \( y'' = -\sin(\pi x^2) \cdot (2\pi x)^2 + \cos(\pi x^2) \cdot 2\pi \). This simplifies to \( y'' = -4\pi^2 x^2 \sin(\pi x^2) + 2\pi \cos(\pi x^2) \).

Step 5: Second Derivative Test for Concavity

Evaluate the sign of \( y'' \) on the same intervals. For \( 0 < x < \frac{1}{\sqrt{2}} \), since \( \cos(\pi x^2) > 0 \), \( y'' > 0 \), meaning the function is concave up. For \( x > \frac{1}{\sqrt{2}} \), \( y'' < 0 \) because both terms contribute negatively, indicating the function is concave down.

Key Concepts

First Derivative Test

The First Derivative Test is a handy technique in calculus used to evaluate the behavior of a function to determine its increasing or decreasing nature over specific intervals. This test primarily involves finding and analyzing critical points, which occur where the derivative of a function equals zero or is undefined.

For the function we examined, which is given by \( y = \sin(\pi x^2) \), we determined the first derivative to be \( y' = \cos(\pi x^2) \cdot 2\pi x \). Our task was to set this derivative equal to zero to find potential critical points:
\[ \cos(\pi x^2) \cdot 2\pi x = 0 \]
From this, we found critical points at \( x = 0 \) and \( x = \frac{1}{\sqrt{2}} \).

Then, by testing intervals between these critical points, we can infer that:

• On the interval \( 0 < x < \frac{1}{\sqrt{2}} \), the derivative is positive so \( y \) is increasing.
• On the interval \( \frac{1}{\sqrt{2}} < x \leq 1 \), the derivative is negative, indicating that \( y \) is decreasing.

This behavior is crucial in sketching a function's graph and understanding its behavior without relying on a calculator.

Second Derivative Test

The Second Derivative Test helps to determine the concavity of a function, as well as to confirm if a critical point is a local maximum or minimum. By examining the sign of the second derivative, we can find out where the function is concave up or down.

For our function, we derived the second derivative as:
\[ y'' = -4\pi^2 x^2 \sin(\pi x^2) + 2\pi \cos(\pi x^2) \]
Evaluating this over our intervals:

• For \( 0 < x < \frac{1}{\sqrt{2}} \), because \( \cos(\pi x^2) \) is positive and dominates, \( y'' > 0 \) indicates the function is concave up.
• For \( \frac{1}{\sqrt{2}} < x \leq 1 \), both terms of the second derivative contribute negatively, resulting in \( y'' < 0 \) which indicates the function is concave down.

This alternation of concavity confirms the nature of the points and helps refine our understanding of the function's shape.

Concavity

Concavity describes how a function curves and tells us much about its shape. Specifically, understanding concavity can help in identifying inflection points—points where the function changes curvature from concave up to concave down or vice versa.

A function is said to be:

• Concave up when its second derivative, \( y'' \), is positive. This describes a U-shape curve.
• Concave down when \( y'' \) is negative, describing an n-shape curve.

In our analysis of \( y = \sin(\pi x^2) \), we noticed regions of varying concavity:

• From \( 0 < x < \frac{1}{\sqrt{2}} \), the concavity is upwards since \( y'' > 0 \).
• From \( \frac{1}{\sqrt{2}} < x \leq 1 \), concavity shifts downwards as here \( y'' < 0 \).

These details provide us with a deeper comprehension of the function's behavior across its domain, enabling us to illustrate it more accurately.