Question

Find all points on the graph of \(y=\sin ^{2} x\) where the tangent line has slope \(1 .\)

Short answer

The points are \(\left(\frac{\pi}{4} + k\pi, \frac{1}{2}\right)\) where \(k\) is any integer.

Step-by-step solution

Find the Derivative

To find the points where the tangent line has a slope of 1, we first need to find the derivative of the function. The derivative of a function gives us the slope of the tangent line at any point. The given function is \(y = \sin^2 x\). To find the derivative, we use the chain rule. If \(y = u\) and \(u = \sin x\), then \(y = u^2\). The derivative \(\frac{dy}{dx}\) is \(2u \cdot \frac{du}{dx}\), where \(\frac{du}{dx} = \cos x\). Thus, \(\frac{dy}{dx} = 2 \sin x \cos x = \sin(2x)\) using the double angle identity.

Set the Derivative Equal to 1

To find the x-values where the tangent slope is 1, set the derivative equal to 1: \(\sin(2x) = 1\). Solve for \(2x\) to find where this equation holds.

Solve the Equation

For \(\sin(2x) = 1\), \(2x\) must be \(\frac{\pi}{2} + 2k\pi\), where \(k\) is an integer. This arises because the sine function reaches 1 at these values. Divide by 2 to solve for \(x\): \(x = \frac{\pi}{4} + k\pi\).

Find Corresponding y-Values

To find the corresponding y-values, substitute these x-values back into the original function: \(y = \sin^2 x\). For \(x = \frac{\pi}{4}\), \(y = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}\). Therefore, for \(x = \frac{\pi}{4} + k\pi\), \(y\) will always be \(\frac{1}{2}\) because the sine function is periodic with a period of \(\pi\).

Conclusion and Summary

The points where the tangent line has a slope of 1 on the function \(y = \sin^2 x\) are \(\left(\frac{\pi}{4} + k\pi, \frac{1}{2}\right)\) for any integer \(k\).

Key concepts

Derivative

The derivative is a fundamental concept in calculus that describes how a function changes at any given point. To put it simply, it represents the slope of the tangent line to the curve at that particular point. This slope is extremely useful, as it helps us understand the behavior of the function, including where it increases or decreases.

When dealing with the function \(y = \sin^2 x\), we use differentiation techniques to find its derivative. For this example, we apply the chain rule, a method used to differentiate compositions of functions. The chain rule states that if a function \(y\) is dependent on another function \(u\), which is in turn dependent on \(x\), we can find the derivative \(\frac{dy}{dx}\) by multiplying \(\frac{dy}{du}\) by \(\frac{du}{dx}\).

Given that \(u = \sin x\) and \(y = u^2\), the derivative becomes \(2u \cdot \frac{du}{dx}\). Since \(\frac{du}{dx} = \cos x\), substituting back gives us \(\frac{dy}{dx} = 2 \sin x \cos x = \sin(2x)\). This expression now represents the slope of the tangent to our original function at any point \(x\).

Trigonometric Functions

Trigonometric functions are essential components of mathematics, deeply connected to geometry and periodic phenomena. Functions like sine, cosine, and tangent are cyclic, meaning they repeat values over regular intervals.

In our scenario, the function \(y = \sin^2 x\) involves squaring the sine function. This transformation retains some properties of \(\sin x\); namely, periodicity and the range that restricts \(y\) to values between 0 and 1.

The trigonometric expression \(\sin(2x)\) arises from the derivative. It is obtained by employing the double angle identity, a powerful formula that provides a way to simplify expressions involving angles that are multiples of a base angle. The identity tells us that \(2 \sin x \cos x = \sin(2x)\), helping to replace more complex products with simpler single angle equivalents.

This representation is particularly useful when solving equations involving these trigonometric functions, such as finding values of \(x\) where \(\sin(2x) = 1\).

Tangent Lines

Tangent lines are straight lines that touch a curve at only one point without crossing it. They provide crucial information about the behavior of curves at specific points by displaying the curve's direction and rate of change at that spot.

For a function like \(y = \sin^2 x\), determining where tangent lines have a particular slope (in this case, slope 1) is achieved through finding the derivative. Setting \(\sin(2x) = 1\), which we derived as the expression for the slope, tells us the `x` values where the tangent lines meet this slope condition.

When the expression is solved, we find that \(2x\) is equal to \(\frac{\pi}{2} + 2k\pi\). Dividing through by 2 determines the specific points \(x = \frac{\pi}{4} + k\pi\), indicating the x-coordinates of where these tangent lines are located. For these x-values, the tangent line will mirror the curve's direction instantly, having the slope of 1.

• Understanding tangent lines helps visualize how a function behaves at various points and is key to graphing and predicting function changes.

Substituting these \(x\) values back into the function helps us find the corresponding \(y\) values, completing the points where \(y = \sin^2 x\) has a tangent with a slope of 1.