Problem 43 - Single Variable Calculus

Question

In Exercises 43 and 44, use integration to find the area of the triangle with the given vertices. $$ (0,0),(1,6),(4,2) $$

Step-by-Step Solution

Answer

The area of the triangle with vertices (0,0), (1,6), and (4,2) is \(9\frac{1}{3}\).

Step 1: Find the equations of the lines passing through the vertices

To find the coordinates of the lines passing through the given vertices, we will use the slope-point form of a line equation, which is given by \(y-y_1=m(x-x_1)\), where m is the slope of the line and \((x_1,y_1)\) is a point on the line. 1. Line passing through (0,0) and (1,6): The slope of the line is given by \(m=\frac{y_2-y_1}{x_2-x_1}\) So, \(m=\frac{6-0}{1-0} = 6\) Plugging this in the slope-point form with the point (0,0), we get: \(y=6x\) 2. Line passing through (1,6) and (4,2): The slope of the line is given by \(m=\frac{2-6}{4-1} = -\frac{4}{3}\) Plugging this in the slope-point form with the point (1,6), we get: \(y=-\frac{4}{3}(x-1)+6\) 3. Line passing through (4,2) and (0,0): The slope of the line = \(\frac{2-0}{4-0}= \frac{1}{2}\) Plugging this in the slope-point form with the point (4,2), we get: \(y=\frac{1}{2}(x-4)+2\)

Step 2: Set up the integral to find the area of the triangle

We know that the triangle starts from x = 0 and ends at x = 4. The area can be found by taking the difference between the upper and lower curve integrals. For x between 0 and 1, the upper curve is created by the line passing through (1,6) and (4,2) while the lower curve is created by the line passing through (0,0) and (1,6). For x between 1 and 4, the upper curve is created by the line passing through (1,6) and (4,2) and the lower curve is created by the line passing through (4,2) and (0,0). The area of the triangle can be calculated as the sum of these two integrals: Area = \(\int_{0}^{1} [(-\frac{4}{3}(x-1)+6) - (6x)]dx\) + \(\int_{1}^{4} [(-\frac{4}{3}(x-1)+6) - (\frac{1}{2}(x-4)+2)]dx\)

Step 3: Calculate the integrals

Let's calculate the two integrals individually: First integral = \(\int_{0}^{1} [-\frac{4}{3}(x-1)+6-6x]dx\) = \(\int_{0}^{1} [-\frac{4}{3}x+2]dx\) Now, we integrate: = \(-\frac{4}{3}\int_{0}^{1}xdx + 2\int_{0}^{1}dx\) = \(-\frac{4}{3}[\frac{x^2}{2}]_{0}^{1}+2[x]_{0}^{1}\) = \(-\frac{4}{3}[\frac{1^2}{2}-\frac{0^2}{2}]+ 2(1-0)\) = \(-\frac{4}{6}+2\) = \(\frac{4}{3}\) Second integral = \(\int_{1}^{4} [(-\frac{4}{3}(x-1)+6) - (\frac{1}{2}(x-4)+2)]dx\) Simplifying the expression inside the integral: = \(\int_{1}^{4} [-\frac{4}{3}x+\frac{8}{3}-\frac{1}{2}x+2]dx\) Now, we integrate: = \(-\frac{4}{3}\int_{1}^{4}xdx + \frac{8}{3}\int_{1}^{4}dx - \frac{1}{2}\int_{1}^{4}xdx + 2\int_{1}^{4}dx\) = \(-\frac{4}{3}[\frac{x^2}{2}]_{1}^{4}+\frac{8}{3}[x]_{1}^{4}-\frac{1}{2}[\frac{x^2}{2}]_{1}^{4}+2[x]_{1}^{4}\) = \(-\frac{4}{3}[\frac{4^2}{2}-\frac{1^2}{2}]+\frac{8}{3}(4-1)-\frac{1}{2}[\frac{4^2}{2}-\frac{1^2}{2}]+2(4-1)\) = \(8-\frac{10}{2}+6\) = \(14-5\) = \(9\) Now, let's find the sum of the two integrals: Area = \(\frac{4}{3} + 9\) Area = \(9\frac{1}{3}\) So, the area of the triangle with the given vertices is \(9\frac{1}{3}\).

Key Concepts

Area of a Triangle

To find the area of a triangle, especially when you are given its vertices in coordinate form, you can use multiple methods. One popular method involves setting up definite integrals. When you know a triangle's three vertices on a coordinate plane, you can use those coordinates to establish lines forming a closed figure.

• Method of Integration: In this exercise, you find the area by integrating over a given interval and subtracting the integral of the lower line from the upper line. This involves calculating the area between two curves (the lines here) over a certain range of x-values. • Determinant Formula: Alternatively, the area of a triangle with vertices \((x_1, y_1), (x_2, y_2), (x_3, y_3)\) can also be computed directly using the formula: \[Area = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\]This method uses determinants and gives a scalar value representing the enclosed area.

Both methods are accurate and can be used depending on the given problem context.

Slope-Point Form of a Line

The slope-point form of a line is a fundamental concept in understanding how lines behave on a coordinate plane. This form is particularly useful for deriving the equation of a line when you know a point on the line and its slope. Given by the formula \(y-y_1=m(x-x_1)\), it rests on two critical features:

• Point: The known point on the line, \((x_1, y_1)\).
• Slope: The steepness or incline of the line, denoted as \(m\).

When solving problems, this form is instrumental for determining equations of the lines connecting triangle vertices, as seen in our exercise. You start by calculating the slope between two points using \(m =\frac{y_2-y_1}{x_2-x_1}\). Once the slope is known, insert a point from the line as \((x_1, y_1)\) into the formula to express the line equation.

Definite Integral

The definite integral is a crucial tool for calculating areas under curves or between curves on a graph. Using calculus, it provides a way to sum up infinitely small sections under a curve across an interval \([a, b]\). For triangle-related problems:

• Boundaries: The upper and lower bounds define where the integration begins and ends on the x-axis, e.g., from x = 0 to x = 4.
• Upper and Lower Functions: The functions that outline the top and bottom boundaries over the specified interval.

In the context of our exercise, definite integrals were used to find the area between lines—representing part of the triangle. You set up an integral for the upper function, subtract the lower function, and evaluate it over the given x-interval. This subtractive process is what accurately measures the shaded area between the curves.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is the study of geometry using a coordinate system. It merges algebra to the spatial visualization of geometric figures. The foundation is built on:

• Coordinates: Points represented by \((x, y)\) on the Cartesian plane.
• Line Equations: Mathematical expressions that describe a series of points forming a straight path. It involves formulas like the slope-point form and the linear equation form \(y = mx + c\).

This field enables solving complex geometric problems by transforming them into algebraic equations, much like how we derived the equations of the triangle sides in the exercise. It allows for precise and formulaic handling of shapes, areas, distances, and other geometric properties using algebra.