Q. 71 - Calculus

Question

Recall that a median of a triangle is a segment connecting a vertex of a triangle to the midpoint of the opposite side. Let T be the triangle with vertices $(0, 0), (a, 0), and (c, d).$In Exercises 70–72, prove the given statements.

Use the integral definition for the centroid to show that the centroid of T is point P from Exercise 70.

Step-by-Step Solution

Answer

The Center of mass of a triangle is The centroid of the triangle.

So centroid of the triangle is $p(\overline{x},\overline{y})=\left(\frac{a+c}{3},\frac{d}{3}\right).$

Step 1. Given information.

 Vertices of given triangle T are $(0, 0), (a, 0), and (c, d).$

Step 2. The centroid of the triangle.

The Center of mass of a triangle is The centroid of the triangle.

x coordinate of the center of mass is following.

$$\begin{gathered} \overline{x}=\frac{\int {\int }_{\Omega }x dA}{\int {\int }_{\Omega }dA} \\ \overline{x}=\frac{{\int }_0^c{\int }_0^{{\displaystyle \frac{d}{cx}}}x dydx+{\int }_0^c{\int }_0^{d/c-a(x-a)}x dydx}{{\int }_0^c{\int }_0^{d/cx}dydx} \\ \overline{x}=\frac{a+c}{3} \end{gathered}$$

y coordinate of the center of mass is following.

$$\begin{gathered} \overline{y}=\frac{\int {\int }_{\Omega }y dA}{\int {\int }_{\Omega }dA} \\ \overline{y}=\frac{{\int }_0^c{\int }_0^{{\displaystyle \frac{d}{cx}}}y dydx+{\int }_0^c{\int }_0^{d/c-a(x-a)}y dydx}{{\int }_0^c{\int }_0^{d/cx}dydx} \\ \overline{y}=\frac{d}{3} \end{gathered}$$

So centroid is $p(\overline{x},\overline{y})=\left(\frac{a+c}{3},\frac{d}{3}\right).$