Problem 60 - Calculus for Biology and Medicine

Question

Find the points on the curve \(y=\cos ^{2} x\) that have a horizontal tangent.

Step-by-Step Solution

Answer

The points with horizontal tangents are \((\frac{\pi}{2} + k\pi, 0)\) and \((k\pi, 1)\) for integer \(k\).

Step 1: Understand the Problem

We need to determine the points on the curve defined by the function \( y = \cos^2(x) \) where the tangent is horizontal. This means we need to find the values of \( x \) for which the derivative \( \frac{dy}{dx} = 0 \).

Step 2: Differentiate the Function

To find when the tangent is horizontal, we first need the derivative of \( y = \cos^2(x) \). Using the chain rule, we differentiate: \( y = (\cos(x))^2 \). Let \( u = \cos(x) \), so \( y = u^2 \). The derivative, \( \frac{dy}{du} = 2u \) and \( \frac{du}{dx} = -\sin(x) \). Thus, \( \frac{dy}{dx} = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x) \).

Step 3: Set the Derivative to Zero

For a horizontal tangent, the derivative must be zero: \( -2\cos(x)\sin(x) = 0 \). This equation can be simplified to \( \cos(x)\sin(x) = 0 \).

Step 4: Solve the Zero-Product Equation

From \( \cos(x)\sin(x) = 0 \), we find the solutions by separately setting \( \cos(x) = 0 \) and \( \sin(x) = 0 \). \( \cos(x) = 0 \) when \( x = \frac{\pi}{2} + k\pi \), and \( \sin(x) = 0 \) when \( x = k\pi \), where \( k \) is any integer.

Step 5: Calculate the Corresponding y-values

For \( x = \frac{\pi}{2} + k\pi \), \( y = \cos^2\left(\frac{\pi}{2} + k\pi\right) = 0 \). For \( x = k\pi \), \( y = \cos^2(k\pi) = 1 \).

Step 6: Present the Points With Horizontal Tangents

Thus, the points where the tangents are horizontal are \( (\frac{\pi}{2} + k\pi, 0) \) and \( (k\pi, 1) \), where \( k \) is any integer.

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which helps us understand how a function changes at any given point. If you think of a function as describing a curve, differentiation allows us to find the slope of the tangent line to the curve at any point. This tangent line represents the rate of change of the function's value with respect to changes in the input.

To differentiate a function, we use rules and formulas that correspond to types of functions we are working with. For example, the basic rule of differentiation, the power rule, tells us how to differentiate functions of the form \(x^n\).

In our problem with the function \(y = \cos^2(x)\), differentiation helps us obtain the expression \(-2\cos(x)\sin(x)\), which represents the tangent's slope at any point \(x\).

• Finding the derivative tells us where the slope is zero: Horizontal tangents occur where the derivative equals zero.
• By setting the derivative to zero, we can locate points where the curve "flattens out." This is essential for identifying horizontal tangents.

Trigonometric Functions

Trigonometric functions, such as sine \(\sin(x)\) and cosine \(\cos(x)\), are based on the relationships between the angles and sides of triangles. They are fundamental in understanding periodic phenomena.

A particular feature of these functions is that they oscillate between -1 and 1, creating a wave-like pattern.

• For instance, \(\cos(x)\) is 0 at \(\frac{\pi}{2} + k\pi\) for integers \(k\), and 1 or -1 at multiples of \(\pi\).
• Similarly, \(\sin(x)\) is zero at multiples of \(\pi\).

In our problem, when differentiating \(\cos^2(x)\), these zero points play a crucial role in identifying where the tangent is horizontal. The equation \(\cos(x)\sin(x) = 0\) arises from the trigonometric relationships, and solving this gives us the x-values where horizontal tangents occur.

Chain Rule

The chain rule is a key technique in differentiation used when dealing with composite functions. This rule helps find the derivative of a function that is the composition (or combination) of two or more functions.

Consider the function \(y = \cos^2(x)\). We can view it as a composition: let \(u = \cos(x)\), then \(y = u^2\). This requires us to first differentiate \(u^2\) with respect to \(u\), then \(u\) with respect to \(x\).

• The derivative \(\frac{dy}{du} = 2u\) results in multiplying by the derivative \(\frac{du}{dx} = -\sin(x)\).
• The complete derivative \(\frac{dy}{dx} = 2\cos(x)\cdot (-\sin(x))\) provides the slope function we're interested in.

This method is powerful because it breaks down complex expressions into manageable parts, allowing us to solve problems involving multiple layers of functions. In our scenario, the chain rule simplifies handling \(\cos^2(x)\), letting us achieve the necessary derivative to test for horizontal tangents.