Q. 11 - Calculus

Question

Explain why the inequalities $r>0$and $0<\theta <\frac{\pi }{2}$ together describe the points in the first quadrant. Use similar inequalities to describe the points in the third quadrant.

Step-by-Step Solution

Answer

The answer is $r>0$ and $\pi <\theta <\frac{3\pi }{2}$

Step 1: Given information

The inequalities $r>0$ and $0<\theta <\frac{\pi }{2}$

Step 2: Simplification

Consider the inequalities $r>0$ and $0<\theta <\frac{\pi }{2}$.

The objective is to write the inequality which is in the third quadrant.

In the first quadrant, all the values are more than $0°$ and less than $\frac{\pi }{2}$.

All the angles are positive in the first quadrant.

Every point $(r,\theta )$ lies on the ray since $r>0$.and the $\theta$ values are in between $0^{\text{ο}} to \frac{\pi }{2}$.

In the third quadrant $r>0$ and all the $\theta$ values are in between $\pi$ to $\frac{3\pi }{2}$.

Every point $(r,\theta )$ lies on the ray and $\theta$ values are in between $\pi$ to $\frac{3\pi }{2}$.

Thus, representation of angles by using inequality is as follows,

$r>0$ and $\pi <\theta <\frac{3\pi }{2}$

Therefore, the answer is $r>0$ and $\pi <\theta <\frac{3\pi }{2}$