Question

What part of the area of a square is cut off by the parabola passing through two adjacent vertices of the square and touching the midpoint of one of its sides?

Short answer

Answer: The ratio of the area cut off by the parabola to the area of the square is \(\frac{13}{24}\).

Step-by-step solution

Setting up the problem

Let's consider a square of side length 1. Place this square in the coordinate plane with vertices A(0,0), B(1,0), C(1,1), and D(0,1). We'll say that the parabola passes through vertices A and B (0,0) and (1,0) respectively, and touches the midpoint of side BC at (1,0.5).

Find the equation of the parabola

Let the equation of the parabola be y = ax^2 + bx + c. Since it passes through vertices A and B, let's plug in the coordinates (0,0) and (1,0) into the equation of the parabola to find the values of a, b, and c: When x=0, y=0: 0 = a(0)^2 + b(0) + c, which implies that c = 0. When x=1, y=0: 0 = a(1)^2 + b(1) + 0, which implies that a + b = 0. Now, let's use the fact that the parabola touches the midpoint of side BC at (1,0.5): When x=1, y=0.5: 0.5 = a(1)^2 + b(1) + 0 = a + b, which implies that a + b = 0.5. From the previous conditions, we know a + b = 0, so a = -b. Substituting this into a + b = 0.5, we get -b + b = 0.5, leading to a = 0.25 and b = -0.25. Thus, the equation of the parabola is y = 0.25x^2 - 0.25x.

Determine area cut off by the parabola

Split square into two triangles. The area cut off by the parabola includes a left triangle with vertices at A(0,0), D(0,1), and E(1,1) and a curving region below the parabola from x=0 to x=1. To find the area of this curving region, we'll need to compute the integral of y = 0.25x^2 - 0.25x from x=0 to x=1. Area of left triangle = 0.5 * base * height = 0.5 * 1 * 1 = 0.5. Area under the parabola (curved region) = \int_{0}^{1} (0.25x^2 - 0.25x)dx. Compute the integral: = [\frac{1}{12}x^3 - \frac{1}{8}x^2]_{0}^{1} = (\frac{1}{12}(1)^3 - \frac{1}{8}(1)^2) - (\frac{1}{12}(0)^3 - \frac{1}{8}(0)^2) = \frac{1}{12} - \frac{1}{8} = -\frac{1}{24}. Since the area cannot be negative, we take the absolute value: |-\frac{1}{24}| = \frac{1}{24}. Total area cut off = Area of left triangle + Area under the parabola = 0.5 + \frac{1}{24} = \frac{13}{24}.

Find the ratio of the area cut off to the area of the square

The area of the square is 1*1 = 1. The area cut off by the parabola is \frac{13}{24}. Therefore, the ratio of the area cut off to the area of the square is \frac{\frac{13}{24}}{1} = \frac{13}{24}. So, the part of the area of the square cut off by the parabola is \(\frac{13}{24}\).

Key concepts

Integral Calculus

Integral calculus is an essential branch of mathematics focused on finding the total size or value, such as area or volume, when it is not possible to calculate it through simple geometry. It deals with the concept of an integral, which provides a way to calculate accumulations, like areas under curves and along surfaces. This is crucial when dealing with irregular shapes where traditional geometry falls short.

For example, finding the area under a parabola or any other curve on a graph is a typical problem solved using integral calculus. The parabola, which does not have straight lines and uniform shape, requires this approach for accuracy. An integral is evaluated from a starting point to an ending point, known as the limits of integration, and it results in the net area between the curve and the x-axis within those limits. The definite integral, which we will discuss later on, is specifically used for calculating the exact numerical value of this area.

Parabola and Geometry

The parabola is a symmetrical, open plane curve formed by the intersection of a cone with a plane parallel to its side. It's a type of conic section and has noteworthy geometric properties. In a Cartesian coordinate system, a parabolic shape can be represented by a quadratic function, typically in the form of \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants.

Parabolas have a unique geometric feature where any point on the curve is equidistant to a fixed point (the focus) and a line (the directrix). This property is harnessed in various applications ranging from satellite dishes to architecture. When combining parabolas with geometry, it's possible to determine intersections with other shapes, such as squares or lines, to compute areas of regions bounded by these intersections as in the given exercise.

Definite Integration

Definite integration is the process of calculating the numerical value of an integral with specific boundaries, called the limits of integration. It finds the net area under a curve between two points, which can be quite useful in practical applications like physics for computing distances and areas, among others.

Returning to our problem with the parabola and the square, the definite integral can be used to find the exact area underneath the curve of the parabola within the square. By integrating the function that describes the parabola from the left edge of the square to the right edge, which are the limits of integration, we determine the size of the 'cut-off' region under the curve. The definite integral, represented as \( \[\int_{a}^{b} f(x)dx\] \), where \( a \) and \( b \) are the limits, is evaluated to provide a precise area — a concept wonderfully utilized in answering our textbook problem.