Problem 13 - The Calculus with Analytic Geometry

Question

If the centroid of the region bounded by the parabola \(y^{2}=4 p x\) and the line \(x=a\) is to be at the point \((p, 0)\), find the value of \(a\).

Step-by-Step Solution

Answer

The value of \(a\) is \(\frac{5p}{12}\).

Step 1: Understand the Problem

Given a parabola defined by the equation \(y^2 = 4px\) and a vertical line \(x = a\), the problem asks for the value of \(a\) such that the centroid (geometric center) of the region between these two curves is at the point \((p, 0)\).

Step 2: Find the Area of the Region

First, calculate the area of the region enclosed by the parabola \(y^2 = 4px\) and the line \(x = a\). The area can be found using the integral: \[A = 2 \int_0^a \sqrt{4px}\, dx\]Integrate and simplify: \[A = 2 \int_0^a 2\sqrt{px}\, dx = 4 \sqrt{p} \int_0^a \sqrt{x}\, dx = 4\sqrt{p} \cdot \frac{2}{3} x^{3/2} \Big|_0^a = \frac{8\sqrt{p}}{3} a^{3/2}\]

Step 3: Determine the Coordinates of the Centroid

The coordinates of the centroid \((\bar{x}, \bar{y})\) for a region bounded by curves can be found as:\[\bar{x} = \frac{1}{A} \int_{-a}^a x f(x) \, dx\]Since we know the centroid is at \((p,0)\), focus on finding \(\bar{x}\):\(\bar{x} = \frac{4}{A} \int_0^a \sqrt{4px} x\, dx\).This simplifies and integrates to: \[\bar{x} = \frac{4}{\frac{8\sqrt{p}}{3} a^{3/2}} \int_0^a x\sqrt{4px}\, dx = \frac{3}{2a^{3/2}} \int_0^a 2 \sqrt{4p} x^{3/2}\, dx\]\(\bar{x} = \frac{3}{2a^{3/2}} \cdot \frac{8}{5} \sqrt{p} x^{5/2} \Big|_0^agives \bar{x} = \frac{3}{2a^{3/2}} \cdot \frac{8}{5} \sqrt{p} a^{5/2} = \frac{12}{5}a \)

Step 4: Solve for the Value of a

Since we know the centroid is located at \(\bar{x} = p\), equate and solve:\[\frac{12a}{5} = p\]Solve for \(a\):\[a = \frac{5p}{12}\]

Key Concepts

Parabola

A parabola is a symmetrical open plane curve that is created by the intersection of a cone with a plane parallel to one of its sides. The most basic form of the parabola's equation is: \[ y^2 = 4px \] Where:

• p represents the distance from the vertex to the focus.
• y and x are the coordinates on the Cartesian plane.

In our exercise, the parabola is given by \( y^2 = 4px \). This sets the foundation for finding the centroid because it defines one of the boundaries of the region we are considering. Understanding this shape is crucial because it helps us recognize how the area and centroid calculations will apply. We are also working with a vertical line \( x = a \) that serves as the other boundary for the region.

Integral Calculus

Integral calculus involves the process of integrating functions to find areas, volumes, and other quantities. In this problem, we use integration to find the area of the region enclosed by the parabola and the vertical line. The steps are as follows:

1. Set up the integral for the area of the region: \[ A = 2 \int_0^a \sqrt{4px} \, dx \].
2. Simplify the integrand and perform the integration:
\[ A = 4 \sqrt{p} \int_0^a \sqrt{x} \, dx = 4 \sqrt{p} \frac{2}{3} x^{3/2} \Big|_0^a = \frac{8\sqrt{p}}{3} a^{3/2} \].
This gives the area of the region as a function of a and p. Integral calculus is critical here because it helps us determine the exact size of the area, which is necessary for calculating the centroid.

Geometric Center

The geometric center, or centroid, is a point that represents the average position of all the points in a shape. For a region bounded by curves, the centroid's coordinates, \( \bar{x} \) and \( \bar{y} \), can be found using the area and the coordinates of the region.

• The x-coordinate of the centroid is calculated as: \[ \bar{x} = \frac{1}{A} \int_{-a}^a x f(x) \, dx \].
• Since the centroid is located at \( (p, 0) \), we focus on finding \( \bar{x} \).

For our exercise:
\[\ \bar{x} = \frac{4}{A} \int_0^a \sqrt{4px} x \, dx = \frac{4}{\frac{8\sqrt{p}}{3} a^{3/2}} \int_0^a x\sqrt{4px} \, dx \].
Simplifying further, we find that:
\[ \bar{x} = \frac{3}{2a^{3/2}} \int_0^a 2 \sqrt{4p} x^{3/2} \, dx \] Integrating:
\[ \bar{x} = \frac{3}{2a^{3/2}} \frac{8}{5} \sqrt{p} x^{5/2} \Big | _0 ^a = \frac{12}{5}a \]
The equation \( \bar{x} = p \) leads to finding the value for a: \[ a = \frac{5p}{12} \].
This calculation ensures that the centroid is at the desired point \( (p, 0) \), completing the task.