Problem 68 - Thomas Calculus in SI Units

Question

You will explore some functions and their inverses together with their derivatives and linear approximating functions at specified points. Perform the following steps using your CAS: a. Plot the function \(y=f(x)\) together with its derivative over the given interval. Explain why you know that \(f\) is one-to-one over the interval. b. Solve the equation \(y=f(x)\) for \(x\) as a function of \(y,\) and name the resulting inverse function \(g\) c. Find the equation for the tangent line to \(f\) at the specified point \(\left(x_{0}, f\left(x_{0}\right)\right)\) d. Find the equation for the tangent line to \(g\) at the point \(\left(f\left(x_{0}\right), x_{0}\right)\) located symmetrically across the \(45^{\circ}\) line \(y=x\) (which is the graph of the identity function). Use Theorem 1 to find the slope of this tangent line. e. Plot the functions \(f\) and \(g,\) the identity, the two tangent lines, and the line segment joining the points \(\left(x_{0}, f\left(x_{0}\right)\right)\) and \(\left(f\left(x_{0}\right), x_{0}\right) .\) Discuss the symmetries you see across the main diagonal. $$y=\sin x, \quad-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}, \quad x_{0}=1$$

Step-by-Step Solution

Answer

\( f(x) = \sin x \) is one-to-one, inverse is \( g(y) = \arcsin y \); tangent lines exhibit symmetry across \( y = x \).

Step 1: Plot the Function and Derivative

Plot the function \( y = \sin x \) and its derivative \( y' = \cos x \) over the interval \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \). Because the derivative \( \cos x \) is non-zero and positive within this interval, the function \( \sin x \) is strictly increasing. Hence, \( f(x) \) is one-to-one on this interval.

Step 2: Find the Inverse Function

Solve the equation \( y = \sin x \) for \( x \). The inverse function is \( x = \arcsin(y) \), which we name \( g(y) = \arcsin(y) \).

Step 3: Find Tangent Line to \( f \)

Calculate the derivative \( f'(x) = \cos x \). The slope of the tangent line at \( x_0 = 1 \) is \( \cos(1) \). The equation of the tangent line at the point \( (1, \sin(1)) \) is \( y = \cos(1)(x - 1) + \sin(1) \).

Step 4: Find Tangent Line to \( g \)

To find the tangent line to \( g \) at \( (\sin(1), 1) \), use the inverse function theorem. If \( f'(x_0) = \cos(1) \), then the slope of the tangent to \( g \) at this point is \( \frac{1}{\cos(1)} \). Thus the equation of this tangent line is \( y = \frac{1}{\cos(1)}(x - \sin(1)) + 1 \).

Step 5: Plot and Discuss Symmetries

Plot \( f(x) = \sin x \), \( g(y) = \arcsin y \), the identity function \( y = x \), and the lines from Steps 3 and 4. Also add the segment joining \( (1, \sin(1)) \) and \( (\sin(1), 1) \). The plots show that \( f \) and \( g \) are symmetric around the line \( y = x \). Both tangent lines reflect this symmetry; the tangent line to \( g \) appears as the reflection of the tangent line to \( f \) across \( y = x \).

Key Concepts

Understanding Derivatives

When you encounter a derivative, you're looking at the rate at which a function is changing. Derivatives tell you the slope of the tangent line at any given point on a function. For the function \(y = \sin x\), its derivative is \(y' = \cos x\). This derivative tells us that the function is changing at the rate of \(\cos x\) at any point \(x\) within the interval \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\). This is crucial because the derivative provides a lot of information about the behavior of the function:

• It reveals where the function is increasing or decreasing.
• In this case, \(\cos x\) is positive, meaning \(\sin x\) is strictly increasing in this interval.
• This behavior helps us determine that \(y = \sin x\) is one-to-one here, which is necessary for an inverse to exist.

One-to-One Functions

A one-to-one function is a function where each value of the range corresponds to a unique value of the domain. For a function to have an inverse, it must be one-to-one over the interval we're examining. In the exercise, the function \(y = \sin x\) is considered over the interval \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\).

• The derivative \(y' = \cos x\) is positive in this interval, ensuring that \(y = \sin x\) is strictly increasing.
• This increasing nature signifies that no two values of \(x\) will map to the same \(y\), thus ensuring the function is one-to-one.

In simpler terms, each output \(y\) comes from one and only one input \(x\), allowing us to find an inverse function.

Tangent Line Explanation

A tangent line is a straight line that just "touches" a curve at a given point. This line represents the instantaneous rate of change of the function at that point. Finding the equation of a tangent line involves determining its slope and using the point-slope form of a line equation.

For the function \(y = \sin x\) at \(x_0 = 1\):

• The derivative \(y' = \cos x\) gives the slope of the tangent line at any \(x\).
• At \(x = 1\), the slope becomes \(\cos(1)\).
• Thus, the equation of the tangent line at the point \((1, \sin(1))\) is \(y = \cos(1)(x - 1) + \sin(1)\).

This implies that at \(x = 1\), the curve and the tangent line touch exactly, giving the best linear approximation of the function at that point.

Trigonometric Functions

Trigonometric functions are based on the relationships of angles in triangles. The sine function, \(y = \sin x\), is periodic and oscillatory, moving between -1 and 1. In this exercise, we looked at \(y = \sin x\) within the interval \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\), where it is smoothly increasing.

Key properties of \(y = \sin x\) include:

• It has an inverse over this interval, known as \(x = \arcsin(y)\), because \(\sin x\) is one-to-one within this range.
• The derivative \(y' = \cos x\) describes how the function's value changes.
• Related to these concepts, the curves and their corresponding tangent lines introduce symmetry, particularly when reflected over the line \(y = x\).

This symmetry and the trigonometric properties help explore the deep connections between a function and its inverse.