Unit-1 Abir Bro - asxcasc PDF

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The equation x² + (y - 2)² and the relation "(x, y) R (0, 2)", where R is read as "has distance 1 of". This relation can also be read as “the point (x, y) is on the circle of radius 1 with center (0, 2)” and “(x, y) satisfies this equation, if and only if, (x, y) R (0, 2)”. The equation of a circle of center (h, k) and radius r is ( x – h)2 +( y – k ) 2=r 2 From here we can see the center of the circle is (0, 2) h = 0 and k = 2 and the radius r = 1 The equation originally = x ²+( y −2)²=1 The equation simplified =

(x−0)²+( y−2)² =1²

Therefore, the circle equation shows that there is a relation between x and y.

The variables can be easily manipulated to be shown as a function of either. Objective: Show x as a function of y e.g. x=g ( y ) By taking the original equation: x ²+( y −2)²=1 2

x 2=1− ( y −2 )

x=± √ 1−( y−2)

2

[Shown]

Objective: Show y as a function of x e.g.

y=h ( x )

Again by taking the original equation: x ²+( y −2)²=1 ( y−2)²=1−x ² y=± √ 1−x 2+ 2

[Shown]

The domain of these two functions can be shown as: First taking the function:

y=± √ 1−x 2+ 2

we can find the domain of x

√ 1−x 2 ≥ 0 x2 ≤ 1

√ x2 ≤ √ 1 1≥ x ≥−1 Therefore, the domain of x is: 1≥ x ≥ 1 Secondly, by taking the function: x=± √ 1−( y−2)2

√ 1−( y−2) ≥ 0 2

2 1≥ ( y−2)

√ 1≥ √ ( y−2)

2

1 ≥ y− 2 ≥−1

we can find the domain of y

1 + 2 ≥ y ≥−1+2 Therefore, the domain of y is: 3 ≥ y ≥ 1

What are the graphs of these two functions? Answer: The graphs of these two function are half circle with center (0, 2)

Are there points of the coordinate axes that relate to (0, 2) by means of R? Answer: Every point on the circle that has a distance 1 from point (0, 2) relate to R...