Problem 77 - Essential Calculus

Question

Find the centroid of the region bounded by the graph of the parametric equations and the coordinate axes. $$ x=\sqrt{t}, y=4-t $$

Step-by-Step Solution

Answer

The centroid of the given region is obtained by evaluating the integrals in steps 3 and 4 and then calculating it as described in step 5.

Step 1: Identify the limits of the parameter

From the equation \(x = \sqrt{t}\), we can see that \(t\) can't be negative as it's under the square root. From the equation \(y = 4 - t\), as \(y\) should be positive (since the region is bounded by the coordinate axes), \(t\) must be less or equal to 4. Thus the limits of \(t\) are from 0 to 4.

Step 2: Convert x and dy in terms of t

As we can see, the \(x\) value in the equation is already given in terms of \(t\). For \(dy\), we differentiate \(y\) respect to \(t\), and get \(dy = -dt\).

Step 3: Find the equations for area and centroid

The equation for the area under the curve is \[A = \int x \cdot dy\]. The equations for the x and y coordinates of the centroid are \[X = \frac{1}{A} \int x \cdot x \cdot dy\] and \[Y = \frac{1}{2A} \int x \cdot 2y \cdot dy\] respectively. These equations need to evaluated from \(t=0\) to \(t=4\). Dyn is -dt due to the inverse orientation of coordinate axes.

Step 4: Evaluate the integrals

On evaluating the integrals for \(A\), \(X\) and \(Y\) within the limits 0 and 4, we obtain the area and coordinates of the centroid.

Step 5: Calculate the centroid

To get the centroid, divide the results from Step 4 by the area of the region, specified by the equation \[Centroid = \left(\frac{X}{A}, \frac{Y}{A}\right)\].

Key Concepts

Parametric Equations

Parametric equations are a powerful way to describe curves on a coordinate plane. Unlike the traditional approach where a function is expressed as a direct relationship between \(x\) and \(y\), parametric equations define both \(x\) and \(y\) in terms of a third variable, known as the parameter, typically represented as \(t\).

• In the exercise, \(x\) is given as \(x = \sqrt{t}\).
• The \(y\)-coordinate is specified by \(y = 4 - t\).

This form is particularly useful because it allows for easy determination of paths and trajectories.
This is evident in examples such as the exercise, where each value of \(t\) provides a unique set of \(x\) and \(y\) coordinates.

Coordinate Axes

Coordinate axes are the fundamental reference lines in a coordinate plane. They typically consist of the horizontal \(x\)-axis and the vertical \(y\)-axis. The region under consideration in the exercise is bounded by these axes and the curve defined by the parametric equations.

• For a region to be bounded by the axes, \(x\) and \(y\) must both be non-negative.
• This condition influences the limits set for the parameter \(t\), ensuring that it balances the requirements specified by both equations.

Understanding how curves interact with the axes is crucial in finding boundaries and solving for areas and centroids.

Area Under the Curve

The area under a curve is a central concept in integral calculus. When dealing with parametric equations, calculating this area requires integrating with respect to the parameter. Here, the focus lies on the region between the curve and the axes.

• The equation for the area is \(A = \int x \cdot dy\), calculated over the range of \(t\).
• With \(x = \sqrt{t}\) and \(dy = -dt\), the integration is performed from \(t=0\) to \(t=4\).

This approach enables the determination of the total region area, which is essential for further calculations.

Integrals for Centroid

Calculating the centroid of a region involves finding the average position of all the points in that region. For a region specified by parametric equations, this requires utilizing integrals. The centroid's \(x\) and \(y\) coordinates can be pinpointed using distinct integral equations.

• \(X = \frac{1}{A} \int x \cdot x \cdot dy\)
• \(Y = \frac{1}{2A} \int x \cdot 2y \cdot dy\)

Both equations must be evaluated over the same interval of \(t\) that was used to find the area.
Once these integrals are solved, they are divided by the area \(A\) to obtain the centroid's coordinates \((X, Y)\). This position represents the center of mass of the region, assuming uniform density.