Problem 15 - Calculus for Biology and Medicine

Question

Find } c \text { so that } f^{\prime}(c)=0 . $$ $$ f(x)=\sin \left(\frac{\pi}{2} x\right) $$

Step-by-Step Solution

Answer

The critical points are at odd integers: \( c = 1, 3, 5, \ldots \ (or \ c = 2n+1) \).

Step 1: Differentiate the Function

To find where the derivative is zero, we first need to find the derivative of the given function. The function is \( f(x) = \sin\left(\frac{\pi}{2}x\right) \). Applying the chain rule, the derivative is \( f'(x) = \cos\left(\frac{\pi}{2}x\right) \cdot \frac{\pi}{2} \).

Step 2: Set the Derivative Equal to Zero

Next, we need to set the derivative equal to zero to find the critical points. This means solving \( \cos\left(\frac{\pi}{2}x\right) \cdot \frac{\pi}{2} = 0 \). Since \( \frac{\pi}{2} \) is a constant and non-zero, we can divide it out to simplify the equation to \( \cos\left(\frac{\pi}{2}x\right) = 0 \).

Step 3: Solve for \( x \)

The cosine function equals zero at odd multiples of \( \frac{\pi}{2} \), i.e., \( \frac{\pi}{2}x = \frac{\pi}{2} + n\pi \), where \( n \) is any integer. Solving for \( x \) gives us \( \frac{\pi}{2}x = (2n+1)\frac{\pi}{2} \). Simplifying this equation, \( x = (2n+1) \).

Step 4: Write General Solution for \( c \)

The solution for where \( f'(c) = 0 \) in terms of \( c \) is \( c = (2n+1) \), where \( n \) is any integer. This means \( c \) could be any odd integer.

Key Concepts

Chain Rule

The Chain Rule is an essential concept in calculus that helps us differentiate composite functions, which are functions within functions. It's like peeling an onion: you start from the outer layer and work your way inward. Suppose we have two functions, an outer function, and an inner function. The Chain Rule tells us how to find the derivative of these layered functions.

Here's how it works:

• Identify the outer function and differentiate it, treating the inside function as a single variable.
• Multiply the derivative of the outer function by the derivative of the inner function.

In the exercise, our function is composed of the sine function, a trigonometric function, and a linear function inside. We use the Chain Rule to differentiate: first the outer \(\sin\) and then the inner \(\frac{\pi}{2}x\). The derivative becomes the cosine of the inner function multiplied by the derivative of the inner linear function, giving us \(f'(x) = \cos\left(\frac{\pi}{2}x\right) \cdot \frac{\pi}{2}\).

This process allows us to break down complex differentiations into simpler, manageable parts.

Critical Points

Critical points of a function are where the function’s derivative is either zero or undefined. These points are significant because they can indicate local maxima, minima, or points of inflection. Understanding critical points allows us to analyze the behavior of a function in depth.

To find a critical point, follow these steps:

• First, compute the derivative of the function.
• Next, set the derivative equal to zero and solve for the variable. This gives possible critical points.
• Finally, solve for cases where the derivative is undefined (if any) to find additional critical points.

In our example, after taking the derivative of \(f(x) = \sin\left(\frac{\pi}{2}x\right)\), we set \(f'(x)\) to zero, resulting in the equation: \(\cos\left(\frac{\pi}{2}x\right) = 0\). Solving this equation helps us identify the places where the slope of the tangent to the function is horizontal, which is key to identifying critical points.

Derivative

A derivative represents the rate at which a function is changing at any given point and is a fundamental concept in calculus. The derivative can be thought of as the slope of the tangent line to the curve of the function at a particular point.

Key points about derivatives include:

• They provide information on the function's increasing or decreasing nature.
• They help in finding maxima and minima of functions.
• They are central in understanding motion and rate changes.

In the given problem, the function \(f(x) = \sin\left(\frac{\pi}{2}x\right)\) was differentiated to find its slope, represented as \(f'(x) = \cos\left(\frac{\pi}{2}x\right) \cdot \frac{\pi}{2}\). The derivative helps us to set up the equation needed to find where the function's rate of change is zero, identifying the critical points.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are cyclic and periodic, forming the backbone for modeling wave-like phenomena. They have particular properties and behaviors that make them very important in calculus and applied mathematics.

Key aspects of trigonometric functions:

• The sine and cosine functions have ranges from -1 to 1 and are continuous.
• Sine functions start from the origin (0,0), while cosine functions start from the maximum point (1,0).
• They exhibit periodic behavior, meaning they repeat their values in regular intervals; for sine and cosine, the period is \(2\pi\).

In this exercise, the function \(f(x) = \sin\left(\frac{\pi}{2}x\right)\) uses these properties to solve for critical points. By understanding where the \cos\ function becomes zero, we find the critical points where the derivative equals zero. This generalized approach is vital in analyzing wave patterns and oscillating systems in science and engineering.