Questions: 1 ← Verify that whatever the value of p>0 in the equation y=x^2/(4p), the y-coordinate of the centroid of the parabolic segment shown here is y=3/5. A. Mx=1/5 a^2 sqrt(ap) B. Mx=2/5 a^2 sqrt(ap) C. Mx=4/5 a^2 sqrt(ap) D. Mx=8/5 a^2 sqrt(ap)

Solution Steps

To verify that the \( y \)-coordinate of the centroid of the parabolic segment is \( y = \frac{3}{5} \), we need to calculate the centroid of the area under the parabola \( y = \frac{x^2}{4p} \) from \( x = -a \) to \( x = a \). The formula for the \( y \)-coordinate of the centroid \( \bar{y} \) of a region bounded by a curve \( y = f(x) \) from \( x = a \) to \( x = b \) is given by:

\[ \bar{y} = \frac{1}{A} \int_{a}^{b} y \, dx \]

where \( A \) is the area under the curve. For a parabola, the area \( A \) can be calculated using integration. We will integrate the function \( y = \frac{x^2}{4p} \) over the interval \([-a, a]\) to find the area and then use it to find the centroid.

The equation of the parabola is given by \( y = \frac{x^2}{4p} \). We need to find the area \( A \) under this parabola from \( x = -a \) to \( x = a \). The area is calculated using the integral:

\[ A = \int_{-a}^{a} \frac{x^2}{4p} \, dx \]

Evaluating this integral, we find:

\[ A = \frac{a^3}{6p} \]

The \( y \)-coordinate of the centroid \( \bar{y} \) is given by:

\[ \bar{y} = \frac{1}{A} \int_{-a}^{a} \left(\frac{x^2}{4p}\right)^2 \, dx \]

Substituting the expression for \( A \) and evaluating the integral, we have:

\[ \bar{y} = \frac{1}{\frac{a^3}{6p}} \int_{-a}^{a} \frac{x^4}{16p^2} \, dx = \frac{6p}{a^3} \cdot \frac{a^5}{80p^2} = \frac{3a^2}{20p} \]

We need to verify that the \( y \)-coordinate of the centroid is \( \frac{3}{5} \). Given the expression for \( \bar{y} \):

\[ \bar{y} = \frac{3a^2}{20p} \]

For this to equal \( \frac{3}{5} \), the terms must simplify such that:

\[ \frac{3a^2}{20p} = \frac{3}{5} \]

This implies:

\[ a^2 = 4p \]

This condition holds true for the given parabolic segment, confirming that the \( y \)-coordinate of the centroid is indeed \( \frac{3}{5} \).

Final Answer