Problem 1 - Calculus of a Single Variable

Question

Finding Distance Using Two Methods In Exercises 1 and \(2,\) find the distance between the points using (a) the Distance Formula and (b) integration. $$ (0,0), \quad(8,15) $$

Step-by-Step Solution

Answer

The distance between the points (0,0) and (8,15), calculated both using the Distance Formula and Integration, is 17 units.

Step 1: Part(a): Using the Distance Formula

The distance formula in the plane is derived from Pythagoras' theorem and is generally given by \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\). Plug in the coordinates of the given points into this formula: \(d = \sqrt{(8-0)^2 + (15-0)^2}\).

Step 2: Calculation of Distance using the Distance Formula

Calculate the square root of sum of squares of differences of x-coordinates and y-coordinates: \(d = \sqrt{(8)^2 + (15)^2}\) = \(\sqrt{64+225}\) = \(\sqrt{289}\). So, the distance using the Distance Formula is 17 units.

Step 3: Part(b): Integration

Identify the line passing through the points (0,0) and (8,15) using the point-slope form y - y1 = m(x - x1). The slope of the line, m, between the two points is (15 - 0) / (8 - 0) = 15 / 8. So, the equation of the line segment is y = (15/8) x. The length of an arc from (0,0) to (8,15) in this function is given by the integral from 0 to 8 of the square root of [1 + ((dy/dx)²)] dx.

Step 4: Calculation of Distance using Integration

Now calculate the integral: the derivative of the line segment function, (dy/dx), is 15/8. So, the integral becomes \(\int_{0}^{8} \sqrt{1+(15/8)^2} dx = \int_{0}^{8} \sqrt{1+(225/64)} dx = \int_{0}^{8} \sqrt{289/64} dx = 17\int_{0}^{8} dx = 17 * x \Big|_{0}^{8} = 17 * 8 \). Thus, the distance using integration is also 17 units.

Key Concepts

Distance Formula

The Distance Formula is a simple mathematical formula used to calculate the distance between two points in a plane. This formula is derived from the Pythagorean theorem. When visualizing two points on a coordinate plane, you can form a right-angled triangle by drawing horizontal and vertical lines. The hypotenuse of this triangle represents the distance between the two points.
To find this distance, the Distance Formula is expressed as:

• d = \( \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)

Here,

• \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.

By plugging these coordinates into the formula, we calculate the hypotenuse length, which is the distance. In our example exercise, we find the distance between the points (0, 0) and (8, 15). By substituting, you can calculate the distance:

• \( \sqrt{(8-0)^2 + (15-0)^2} = \sqrt{64 + 225} = \sqrt{289} \)

This results in a distance of 17 units between the points.

integration

Integration is a concept from calculus used to find the area under a curve, among other applications, such as determining the length of a curve. In this exercise, integration is used to find the arc length between two points forming a straight line.
It is an alternative method to the Distance Formula in determining how far apart two points are in a plane. The main idea behind using integration to find distance involves setting up an integral of the form:

• \( \int_{a}^{b} \sqrt{1+(dy/dx)^2} dx \)

For our line described by points (0, 0) and (8, 15), we first find the derivative \( dy/dx \), which is the slope of the line, 15/8. Then the integral becomes determining the arc length by computing:

• \( \int_{0}^{8} \sqrt{(1 + (15/8)^2)} dx = \int_{0}^{8} \sqrt{(289/64)} dx \)

Calculating this integral, we again arrive at 17 units as the distance.

arc length

Arc length is the term used to describe the distance along a curved path or the edge of a curve. In contexts like our exercise, it can also be applied to linear paths. To find the arc length between two points on a curve, integration is used to sum the infinitely small pieces making up the length.
If you have a function curve described by \( y = f(x) \) and differentiable over an interval \([a, b]\), its arc length \(L\) is computed by:

• \( L = \int_{a}^{b} \sqrt{1+(f'(x))^2} dx \)

In the scenario of our linear example with straight line segments, this simplifies since the rate of change is constant.Calculating via integration still leads us to the real-world related dimension or distance, being entirely consistent--span of 17 units.

Pythagorean theorem

The Pythagorean theorem is a fundamental principle of geometry that relates to right triangles. It states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
Mathematically expressed as:

• \( c^2 = a^2 + b^2 \)

Where \(c\) is the length of the hypotenuse and \(a\) and \(b\) are the lengths of the other two sides. This theorem is foundational for the Distance Formula. When we solve for \( c \), we use the square root to find the direct distance or hypotenuse:

• \( c = \sqrt{a^2 + b^2} \)

In the case of two specific points on a plane, \(a\) and \(b\) are the differences in x and y coordinates respectively. Therefore, the Pythagorean theorem is effectively baked into the computations of the distance formula, embodying how these spatial distances connect with ancient geometric truths. It confirms the calculated 17 units in the distance method as accurately valid.