Problem 74 - Calculus Single Variable

Question

Approximate \(f^{\prime}(c)\) for the given \(f\) and \(c\) in the following way: find a small viewing window with \(P=\) \((c, f(c))\) near the center. The window should be small enough so that the graph of \(f\) appears to be a straight line. Let \(Q\) and \(R\) be the endpoints of the graph of \(f\) as it exits this window. Use the slope of \(Q R\) as the approximation to \(f^{\prime}(c)\). $$ f(x)=\tan \left(\frac{\pi}{4} \sin (\pi x / 2)\right), c=1 $$

Step-by-Step Solution

Answer

The approximate derivative \(f'(1)\) is 0.

Step 1: Evaluate f at c

First, substitute the value of \(c\) into the function \(f\). Since \(f(x) = \tan\left(\frac{\pi}{4} \sin\left(\frac{\pi x}{2}\right)\right)\), we find \(f(1) = \tan\left(\frac{\pi}{4} \sin\left(\frac{\pi \cdot 1}{2}\right)\right)\). Simplify inside the function: \(\frac{\pi}{2} \times 1 = \frac{\pi}{2}\), and \(\sin\left(\frac{\pi}{2}\right) = 1\). Therefore, \(f(1) = \tan\left(\frac{\pi}{4} \times 1\right) = \tan\left(\frac{\pi}{4}\right) = 1\). This gives us point \(P = (1, 1)\).

Step 2: Define Viewing Window

To approximate the derivative, define a small window around \(x = 1\). Let's choose the window to be from \(x = 0.9\) to \(x = 1.1\) to ensure we capture the linear behavior of the function. Evaluate \(f(x)\) at these boundaries to find points \(Q\) and \(R\).

Step 3: Find Q and R Points

For \(x = 0.9\), calculate \(f(0.9) = \tan\left(\frac{\pi}{4} \sin\left(\frac{\pi \cdot 0.9}{2}\right)\right)\). Evaluate: \(\frac{\pi \cdot 0.9}{2} = 0.45\pi\), \(\sin(0.45\pi) = \sin(\frac{9\pi}{20})\). Approximating using a calculator, \(\sin(0.45\pi) \approx 0.809\). Therefore, \(f(0.9) \approx \tan\left(\frac{\pi}{4} \cdot 0.809\right)\) which gives us approximately 0.577 after calculating. Thus, \(Q = (0.9, 0.577)\).For \(x = 1.1\), calculate \(f(1.1) = \tan\left(\frac{\pi}{4} \sin\left(\frac{\pi \cdot 1.1}{2}\right)\right)\). Evaluate: \(\frac{1.1\pi}{2} = 0.55\pi\), \(\sin(0.55\pi) = \sin(\frac{11\pi}{20})\). Approximating, \(\sin(0.55\pi) \approx 0.809\). Therefore, \(f(1.1) \approx \tan\left(\frac{\pi}{4} \cdot 0.809\right)\) which gives us approximately 0.577. Thus, \(R = (1.1, 0.577)\).

Step 4: Calculate Slope of QR

The slope of a line passing through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by the formula \(m = \frac{y_2 - y_1}{x_2 - x_1}\). Use points \(Q = (0.9, 0.577)\) and \(R = (1.1, 0.577)\). Calculate: \(m = \frac{0.577 - 0.577}{1.1 - 0.9} = \frac{0}{0.2} = 0\). Thus, the slope of \(QR\) is approximately 0.

Key Concepts

Calculus

Calculus is an essential subject in understanding changes and motion. It mainly deals with two fundamental concepts: differentiation and integration. In simpler terms, differentiation is about finding how a quantity changes with respect to another. Integration is about finding the total size or value from smaller parts. When we look at the motion of a car, differentiation could help us find its speed or rate of change of position, whereas integration could provide the total distance traveled.

In the given exercise, we focus on the differentiation part of Calculus. Specifically, we are interested in finding the derivative of a function at a specific point. The derivative tells us the rate at which the function's output is changing at that particular point. Think of it as the slope of the tangent line at that point on the curve of the function, which leads us to our next concept.

Tangent Line

The tangent line to a curve at a given point is the straight line that touches the curve exactly at that point without crossing it. Imagine a hillside and you're standing at one point; the tangent line is like a plank that's perfectly flat against the slope of the hill at your feet. This line shows the immediate direction that a small object would move if set to roll.

In mathematical terms, when we say we want to find the tangent line to the graph of a function at a point, we are seeking to find the line that minimally touches the graph at that point and shares the same slope as the function itself there. The slope of this tangent line is actually the derivative of the function at that point.

Slope Calculation

Calculating the slope of a line involves understanding how steep that line is. The slope gives us the rate of change of the vertical direction compared to the horizontal direction between two points. Numerically, it's calculated using the formula:

• \( m = \frac{y_2 - y_1}{x_2 - x_1} \)

Here,

• \( y_2 \) and \( y_1 \) are the output or 'y-values' for two different 'x-values' (\(x_2\) and \(x_1\)).
• This difference quotient essentially means change in 'y' over change in 'x'.

In the exercise, the slope of the line joining points \(Q\) and \(R\) is used to approximate the derivative at point \(P\). It's like taking two slightly different points on either side of our main point and seeing how the function behaves linearly there.

Trigonometric Functions

Trigonometric functions are a core part of geometry and calculus. Functions like sine and tangent help us describe the angles and lengths in right-angled triangles. In our exercise, the function involves the tangent of a sine function. This presents a more complex function where angles are 'transformed' by another angle-related function.

The sine function, abbreviated as \( \sin \), varies from -1 to 1. In our specific function, \( \tan \left( \frac{\pi}{4} \sin \left( \frac{\pi x}{2} \right) \right) \), we calculate sine first, which gets multiplied by a factor, followed by applying the tangent operation on it. This combination can sound intricate. Therefore, we approximate its derivative using a visual method like finding the slope between two very nearby points \(Q\) and \(R\).

Understanding these reactions to changes near our chosen point is crucial for deriving where the function leads as it moves forward.