Problem 66 - Thomas Calculus

Question

Slopes on sine curves a. Find equations for the tangents to the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) at the origin \((m \text { a constant } \neq 0) ?\) Give reasons for your answer. c. For a given \(m,\) what are the largest values the slopes of the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) can ever have? Give reasons for your answer. d. The function \(y=\sin x\) completes one period on the interval \([0,2 \pi],\) the function \(y=\sin 2 x\) completes two periods, the function \(y=\sin (x / 2)\) completes half a period, and so on. Is there any relation between the number of periods \(y=\sin m x\) completes on \([0,2 \pi]\) and the slope of the curve \(y=\sin m x\) at the origin? Give reasons for your answer.

Step-by-Step Solution

Answer

Tangents for both curves are perpendicular. Slopes at origin for \(y=\sin(mx)\) and \(-\sin(x/m)\) are perpendicular. Largest slopes: \(|m|\) and \(1/m\). Period proportional to slope for \(y=\sin(mx)\).

Step 1: Find the derivative of the function

To find the slope of the tangent to the curve at the origin, we need to calculate the derivative of each given function with respect to \(x\). For \(y = \sin 2x\), the derivative \(y'\) is \(2 \cos 2x\). Similarly, for \(y = -\sin(x/2)\), the derivative \(y'\) is \(-\frac{1}{2} \cos(x/2)\).

Step 2: Evaluate the derivative at the origin

Substitute \(x=0\) into each derivative to find the slope at the origin. For \(y = \sin 2x\), \(y'(0) = 2 \cos 0 = 2\). For \(y = -\sin(x/2)\), \(y'(0) = -\frac{1}{2} \cos 0 = -\frac{1}{2}\).

Step 3: Determine tangent equations at the origin

Use the point-slope form of a line \(y - y_1 = m(x - x_1)\) where \((x_1, y_1)\) is the origin \((0,0)\) and \(m\) is the slope. The tangent to \(y = \sin 2x\) at the origin is \(y = 2x\). The tangent to \(y = -\sin(x/2)\) at the origin is \(y = -\frac{1}{2}x\).

Step 4: Analyze special relationships between the tangents

The slopes of the tangents \(2\) and \(-\frac{1}{2}\) are negative reciprocals of each other, indicating that the tangents are perpendicular.

Step 5: Generalize for functions \(y=\sin(mx)\) and \(y=-\sin(x/m)\)

For \(y=\sin(mx)\), the derivative at the origin is \(m\). For \(y=-\sin(x/m)\), the derivative is \(-\frac{1}{m}\). The slopes \(m\) and \(-\frac{1}{m}\) are negative reciprocals, so the tangents are always perpendicular.

Step 6: Find the maximum slopes of the curves

The maximum derivative for \(y = \sin(mx)\) is \(|m|\), occurring at \(x = \frac{\pi}{2m} + k\pi/m\), \(k\in\mathbb{Z}\). For \(y = -\sin(x/m)\), the maximum is \(\frac{1}{m}\), occurring at \(x = \frac{\pi}{2}\).

Step 7: Relate periods and slopes for \(y=\sin(mx)\)

The function \(y=\sin(mx)\) completes \(m\) periods over \([0, 2\pi]\). The slope of the tangent at the origin is \(m\), indicating a proportional relationship between the number of periods and the initial slope.

Key Concepts

Derivatives

Derivatives are fundamental in understanding the behavior of curves and their tangents. They represent the rate at which a function changes as the input changes. In simple terms, derivatives allow us to calculate the slope of a curve at any given point. For the sine functions explored in the exercise, the derivatives are critical for determining the tangent slopes at the origin.
To find the derivative of a sine function, such as \(y = \sin 2x\), we apply the chain rule. This results in the derivative \(y' = 2 \cos 2x\). Similarly, for the function \(y = -\sin(x/2)\), the derivative is \(y' = -\frac{1}{2} \cos(x/2)\). These derivatives indicate how steep the tangent to the curve is at different points, particularly at the origin here.
By evaluating these derivatives at \(x = 0\), we find the slopes of the tangents to the curves at the origin: \2\ for \(y = \sin 2x\) and \ -\frac{1}{2}\ for \(y = -\sin(x/2)\). This illustrates the power of derivatives in analyzing curve behavior.

Sine Functions

Sine functions are a type of periodic function, which means they repeat their shape at regular intervals. The general form of a sine function is \(y = \sin(kx)\), where \(k\) is a constant that determines how stretched or compressed the wave is.
Sine curves have specific characteristics such as amplitude (the height from the center to the peak), period (the distance required for the function to complete one full cycle), and the phase shift (the horizontal shift from the usual position).
In the exercise, \(y = \sin 2x\) and \(y = -\sin(x/2)\) are examples of modified sine functions. The \(2x\) and \(x/2\) in these functions stretch or compress the wave horizontally. The parameter \(2x\) indicates a compression, allowing the sine wave to complete twice as many cycles in the same interval, while \(x/2\) indicates a stretch, reducing the completed cycles.
The sine functions can be visualized as waves propagating along the x-axis, with their periodic nature playing a key role in their analysis. By exploring these functions, we gain insight into how functions change shape and frequency.

Slopes of Curves

The slope of a curve is a measure of how steep or flat the curve is at a particular point. It is directly tied to the concept of derivatives. A positive slope indicates the curve is rising, while a negative slope shows it is falling. At sine curve peaks or troughs, the slope is zero, indicating a horizontal tangent line.
In the context of the given exercise, we determined the slopes of the curves \(y = \sin 2x\) and \(y = -\sin(x/2)\) at the origin. By using the derivatives, we found these slopes to be \2\ and \ -\frac{1}{2}\, respectively.
These slopes also tell us about the geometric relationship between the tangents at the origin. Specifically, a slope of \2\ corresponds to a steeper incline compared to the gentler decline of \ -\frac{1}{2}\. Interestingly, because one slope is the negative reciprocal of the other, the tangents are perpendicular at the origin.

• The positive slope \2\ points upwards from the origin.
• The negative slope \ -\frac{1}{2}\ points downwards from the origin.

This demonstration shows how the direction and steepness of a curve at a point can be captured using the slope.

Periodicity of Functions

Periodicity refers to the repeating pattern of a function over a given interval. For sine functions like \(y = \sin mx\), the periodicity is adjusted by the constant \(m\). A sine wave completes a certain number of cycles within a standard interval, often \[0, 2\pi\].
By adjusting \(m\), we change how many cycles fit into that interval. For instance, \(y = \sin 2x\) fits twice the number of cycles into \ [0, 2\pi]\, while \(y = -\sin(x/2)\) fits only half a cycle.
This concept of periodicity impacts the tangent slopes at specific points like the origin because it directly influences how rapidly the sine function ascends and descends. As such, when we compute the derivatives at the origin, we observe that the slopes are directly proportional to \(m\), the cyclic frequency factor. The more periods or cycles created, the steeper the slope at the origin.
Understanding periodicity allows us to anticipate not just how often a function repeats, but also how behavior like slope changes with differing frequencies. It's a pivotal concept in comprehending the dynamics of oscillatory solutions, including many trigonometric problems involving sine and other wave functions.