One lap is 2Ï€. What ratio of a circle must 2Ï€3 be?
A
a
No, it's not a function.
B
b
Domain: - 10≤ x ≤ 10
Range: - 10≤ y ≤ 10
C
c
About 209.44 units^2
Practice makes perfect
a
Examining the equation, we see that it contains a squared x and a squared y. Therefore, the equation must be describing a circle. To draw its graph, we have to identify the center and radius. We can do that if we rewrite it in the following format.
(x-a)^2+(y-b)^2=r^2
In this format, (a,b) is the center and r is the radius.
As we can see, the graph is a circle with its center at the origin and a radius of 10. Let's graph it.
The definition of a function is a graph where one input corresponds to only one output. We can check if this is the case by performing a Vertical Line Test. If we draw a vertical line anywhere in the diagram, it should never hit the graph more than once if it's a function. Let's put our circle to the test.
As we can see, the graph failed the vertical line test, and therefore it is not a function.
b
The domain tells us the x-values for which the graph is defined, and the range is the y-values for which the graph is defined. Examining the diagram, we can identify the range and domain.
Now we can write the domain and range.
Domain:& - 10≤ x ≤ 10
Range:& - 10≤ y ≤ 10
c
Let's first calculate the area of the circle. From Part A, we know it has a radius of 10 units. With this information, we can calculate the area of the circle.
A_C=Ï€( 10)^2=100Ï€
A full lap corresponds to 2Ï€. Therefore, 2Ï€3 must correspond to an arc that is a third of a circle.
If we multiply the circle's area with the ratio of the central angle to 2Ï€, we can find the sector's area.
A_S=100Ï€(2Ï€/3/2Ï€)=100Ï€/3
Finally, to determine how much area remains after we remove the wedge, we should subtract the sector's area from the circle's area.
A_C- A_S= 100π- 100π/3≈ 209.44 units^2
