Question

Show, in Figure 4.4, that for a point like ${\mathbf{P}}_{\mathbf{3}}$,$\frac{{\mathbf{x}}_{\mathbf{3}}}{{\mathbf{y}}_{\mathbf{3}}}\mathbf{>}\frac{\mathbf{\pi }}{\mathbf{2}}$and for${\mathbf{P}}_{\mathbf{2}}$,$\frac{{\mathbf{x}}_{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{2}}$.

Short answer

It is proved that for ${\mathrm{P}}_3$and ${\mathrm{P}}_2$, $\frac{{\mathrm{x}}_3}{{\mathrm{y}}_3}>\frac{\mathrm{\pi }}{2}$and $\frac{{\mathrm{x}}_2}{{\mathrm{y}}_2}=\frac{\mathrm{\pi }}{2}$, respectively, in figure 4.4,

Step-by-step solution

Given Information

The half-circle in figure 4.4 and points${\mathrm{P}}_3$ and${\mathrm{P}}_2$ are given.

Definition of Calculus

In the same way that geometry is the study of shape and algebra is the study of generalisations of arithmetic operations, calculus, sometimes known as infinitesimal calculus or "the calculus of infinitesimals," is the mathematical study of continuous change.. Differentiation and integration are the two main branches.

Draw the circle

Redraw the half-circle given in figure 4.4 with appropriate points.

Points travelled more than half the circumference

In figure 4.4, in previous step, observe point ${\mathrm{P}}_3$that has travelled a distance greater than half of the circumference, so the coordinate ${\mathrm{x}}_3$is greater than half of the circumference.

$$\begin{gathered} {\mathrm{x}}_3>\frac{1}{2}\times 2\mathrm{\pi a} \\ {\mathrm{x}}_3>\mathrm{\pi a} \end{gathered}$$

The $y_3$coordinate has less height than the circle.

$$\begin{gathered} y_3<2a \\ \frac{x_3}{y_3}>\frac{\mathrm{\pi }}{2} \end{gathered}$$

Points travelled equal to half the circumference

In figure 4.4, observe point ${\mathrm{P}}_2$that has travelled a distance equal to half of the circumference, so the coordinate $x_2$is equal to half of the circumference.

$$\begin{gathered} x_2=\frac{1}{2}\times 2\mathrm{\pi a} \\ {\mathrm{x}}_2=\mathrm{\pi a} \end{gathered}$$

The $y_2$coordinate has height exactlysame as that of the of the circle.

$$\begin{gathered} y_2=2a \\ \frac{x_2}{y_2}=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Therefore, it is proved that for points ${\mathrm{P}}_3$and ${\mathrm{P}}_2$, in figure 4.4,

$\frac{{\mathrm{x}}_3}{{\mathrm{y}}_3}>\frac{\mathrm{\pi }}{2}\mathrm{and}\frac{{\mathrm{x}}_2}{{\mathrm{y}}_2}=\frac{\mathrm{\pi }}{2},\mathrm{respectively}.$