Problem 49 - Essential Calculus

Question

Sketch the graph of \(f(x)=2-2 \sin x\) on the interval \([0, \pi / 2]\) (a) Find the distance from the origin to the \(y\) -intercept and the distance from the origin to the \(x\) -intercept. (b) Write the distance \(d\) from the origin to a point on the graph of \(f\) as a function of \(x\). Use a graphing utility to graph \(d\) and find the minimum distance. (c) Use calculus and the zero or root feature of a graphing utility to find the value of \(x\) that minimizes the function \(d\) on the interval \([0, \pi / 2]\). What is the minimum distance? (Submitted by Tim Chapell, Penn Valley Community College, Kansas City, MO.)

Step-by-Step Solution

Answer

The distances from the origin to the y-intercept and the x-intercept are respectively 2 units and \(\frac{\pi}{2}\) units. The function for the distance from the origin to a point on the graph of \(f\) is \(d=\sqrt{x^2 + (2-2 \sin x)^2}\). Using calculus and a graphing utility, the value of \(x\) that minimizes this distance, and the minimum distance can be found.

Step 1: Sketch the graph of the function

Begin by plotting \(f(x)=2-2 \sin x\) on the x-y plane. Note the intercepts: the y-intercept is \(f(0)=2\) and the x-intercept is the root of the equation \(2-2 \sin x=0\), which gives \(x=\frac{\pi}{2}\). Draw the curve connecting these points.

Step 2: Calculate the distance from the origin to the intercepts

Apply the distance formula \(d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\) where (x1, y1) is the origin and (x2, y2) the coordinates of a point. For the y-intercept (0, 2), this gives \(d=\sqrt{0^2 + 2^2} = 2\). For the x-intercept \(\left(\frac{\pi}{2}, 0\right)\), this gives \(d=\sqrt{(\frac{\pi}{2}-0)^2 + (0-0)^2} = \frac{\pi}{2}\).

Step 3: Write the distance function

The distance \(d\) from the origin to a general point \((x, y) = (x, f(x))\) on the graph of \(f\) is \(d=\sqrt{x^2 + (2-2 \sin x)^2}\). To see the graph of \(d\) and find its minimum value, use a graphing utility.

Step 4: Find the minimizing value of \(x\) and the minimum distance

To find the value of \(x\) that minimizes \(d\), apply calculus: find the derivative of \(d\) with respect to \(x\), set equal to zero and solve for \(x\). Then verify this is a minimum by checking that the second derivative at this point is greater than 0. Cross-check your result using the zero or root feature of a graphing utility. This will give you the minimizing \(x\) and hence the minimum distance, by substituting this \(x\) into \(d\).

Key Concepts

Calculus Optimization

Calculus optimization is a critical topic in mathematics, particularly in calculus, that deals with finding the maximum or minimum values of functions within certain constraints. By using derivatives, we can identify these extreme values—a process that is crucial in many real-world applications, from engineering to economics.

To start with, one calculates the first derivative of the function and sets it equal to zero. This step results in potential points, called critical points, where maxima or minima can occur. To confirm whether each critical point is a maximum or minimum, the second derivative test can be used. If the second derivative is positive at a critical point, the function has a local minimum there; if negative, a local maximum. In the specific exercise of graphing the function and minimizing the distance from the origin to a point on the graph, the derivative helps us find the exact point on the curve closest to the origin.

Distance Formula

The distance formula is essential in geometry and helps calculate the distance between two points in a coordinate plane. This formula is derived from the Pythagorean theorem and is written as \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\), where \((x_1, y_1)\) and \((x_2, y_2)\) represent the coordinates of the two points. In the context of our exercise, the distance formula allows us to measure the distance from the origin (0,0) to any point \((x, f(x))\) on the graph. For points that lie on the x- or y-intercept, the formula simplifies, as one of the coordinates will be zero, making calculations even easier.

Graphing Utility

A graphing utility is an invaluable tool in visualizing mathematical functions and their properties. It enables users to quickly plot graphs for complex equations and observe behaviors that might be difficult to identify purely through calculation.

For trigonometric functions, graphing utilities can show periodicity, amplitude, and phase shifts. When combined with its features like the zero or root function, one can easily find points where the graph intersects the axis. In the context of our exercise, using a graphing utility facilitates both the plotting of the trigonometric function \(f(x) = 2 - 2 \sin x\) and the subsequent visualization of the distance function \(d\) to determine the minimum distance to the origin.

Minimizing Distance

Minimizing distance in a mathematical sense often means finding the shortest path between two points. This concept is not only important in math but in everyday life scenarios such as routing GPS systems or optimizing the layout of a building. In our exercise, minimizing distance translates to finding the point on the graph of the function \(f(x) = 2 - 2 \sin x\) which is closest to the origin.

To achieve this, we use the distance formula for any given point on the graph and apply calculus optimization techniques to find the value of \(x\) that yields the smallest distance. The combination of these mathematical strategies leads us to the most efficient path, in this case, the shortest distance from the origin to the curve.