Problem 59 - Calculus for Biology and Medicine

Question

Find the points on the curve \(y=\sin \left(\frac{\pi}{3} x\right)\) that have a horizontal tangent.

Step-by-Step Solution

Answer

The points are \( \left( \frac{3}{2} + 3k, (-1)^k \right) \) for integer \( k \).

Step 1: Understand Horizontal Tangents

A function has a horizontal tangent at points where its derivative is equal to zero. This means we need to calculate the derivative of the given function and set it equal to zero to find these points.

Step 2: Find the Derivative

The given function is \( y = \sin \left( \frac{\pi}{3} x \right) \). Using the chain rule, the derivative \( \frac{dy}{dx} \) is \( \cos \left( \frac{\pi}{3} x \right) \cdot \frac{\pi}{3} \). Hence, \( \frac{dy}{dx} = \frac{\pi}{3} \cos \left( \frac{\pi}{3} x \right) \).

Step 3: Set Derivative to Zero

We set the derivative equal to zero to find points of horizontal tangency: \( \frac{\pi}{3} \cos \left( \frac{\pi}{3} x \right) = 0 \). This simplifies to \( \cos \left( \frac{\pi}{3} x \right) = 0 \).

Step 4: Solve for x

The equation \( \cos \theta = 0 \) at \( \theta = \frac{\pi}{2} + k\pi \) for integer \( k \). Thus, we set \( \frac{\pi}{3} x = \frac{\pi}{2} + k\pi \). Solving for \( x \), we find \( x = \frac{3}{2} + 3k \).

Step 5: Find Corresponding y-values

Substituting back into the original function, when \( x = \frac{3}{2} + 3k \), \( y = \sin \left( \frac{\pi}{3} \cdot \left( \frac{3}{2} + 3k \right) \right) = \sin \left( \frac{\pi}{2} + k\pi \right) = \sin \left( \frac{\pi}{2} \right) (-1)^k = (-1)^k \).

Step 6: State the Points

The points on the curve that have a horizontal tangent are \( \left( \frac{3}{2} + 3k, (-1)^k \right) \) where \( k \) is any integer.

Key Concepts

Derivative

In calculus, a derivative represents the rate at which a function is changing at any given point on its curve. If you picture the graph of a function, the derivative at a particular point is the slope of the tangent line at that point.
For the function given in the exercise, the expression is quite similar: \( y = \sin \left(\frac{\pi}{3} x\right) \). To find the derivative, we make use of different rules in calculus, such as the chain rule.
Knowing how to differentiate basic trigonometric functions is essential, and in this process:

• The derivative of \( \sin(u) \) with respect to \( x \) is \( \cos(u) \cdot u' \), where \( u \) is a function of \( x \).
• This derivative reflects how the sine function grows or decreases as \( x \) changes.

Applying this to our function, the derivative becomes \( \frac{dy}{dx} = \cos \left( \frac{\pi}{3} x \right) \cdot \frac{\pi}{3} \). This expression will guide us as we explore conditions for horizontal tangents.

Horizontal Tangent

When dealing with curves and graphs, identifying horizontal tangents is crucial for understanding the behavior of the function. A tangent is horizontal when its slope equals zero.
This happens when the derivative is set to zero: \( \frac{dy}{dx} = 0 \).
In our specific case, this means setting the derivative we computed, \( \frac{\pi}{3} \cos \left( \frac{\pi}{3} x \right) \), equal to zero. This simplifies to solving \( \cos \left( \frac{\pi}{3} x \right) = 0 \).

• Horizontal tangents reveal turning points or plateaus in the function.
• The x-values satisfying this condition give us specific points where the curve changes direction.

Finding these points helps illustrate key "milestones" on the graph where the function neither increases nor decreases.

Chain Rule

The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. A composite function is a function inside another function, similar to nesting.
For example, in our exercise, we have \( y = \sin \left( \frac{\pi}{3} x \right) \). The chain rule applies here because \( \frac{\pi}{3} x \) is itself a function of \( x \).
Applying the chain rule involves:

• Identifying the "inner" function - here, \( \.\frac{\pi}{3} x \).
• Deriving the outer function, \( \sin(u) \), to get \( \cos(u) \).
• Finally, multiplying by the derivative of the inner function.

This allows us to find \( \frac{dy}{dx} \) in a systematic way: \( \frac{dy}{dx} = \cos \left( \frac{\pi}{3} x \right) \cdot \frac{\pi}{3} \). This conveys the combined rate of change, taking into account both function levels, ensuring an accurate reflection of the entire process.