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Need help with question 13. It requires you to somehow find the value without applying any integration rules. Topic is area of a region. Source: Additional Maths 360. All help appreciated!
Date Posted: 4 years ago
Views: 226
The graph of y = √(1 - x²) can be used to form a semicircle with the x axis, for -1 ≤ x ≤ 1.
1 represents the length of the hypotenuse of a right angled triangle.
x represents the length of 1 side of it, and it's part of the x axis.
y represents the length of the third side of the triangle, and is parallel to the y axis.
Using Pythagoras theorem,
x² + y² = 1²
y² = 1 - x²
y = ± √(1 - x²)
We only consider the positive part as the negative part is another semicircle , a reflection of this semicircle along the x axis.
So the hypotenuse length is fixed at 1 but the x and y coordinates are variable.
You'll see that all the coordinates on the graph form a semicircle with radius of 1 unit.
From x = 0 to x = 1, it's a quarter circle.
Area of quarter circle = ¼π(1units)²
= ¼π units²
The line y = 1 - x forms a right angled isosceles triangle with the x and y axis, with x intercept 1 and y intercept 1.
Area of this triangle is = ½ x 1 x 1 = ½ units²
Required area
= Area of quarter circle - area of triangle
= ¼π units² - ½ units²
= ¼ (π - 2) units²
You are essentially finding a small half-leaf/segment
1 represents the length of the hypotenuse of a right angled triangle.
x represents the length of 1 side of it, and it's part of the x axis.
y represents the length of the third side of the triangle, and is parallel to the y axis.
Using Pythagoras theorem,
x² + y² = 1²
y² = 1 - x²
y = ± √(1 - x²)
We only consider the positive part as the negative part is another semicircle , a reflection of this semicircle along the x axis.
So the hypotenuse length is fixed at 1 but the x and y coordinates are variable.
You'll see that all the coordinates on the graph form a semicircle with radius of 1 unit.
From x = 0 to x = 1, it's a quarter circle.
Area of quarter circle = ¼π(1units)²
= ¼π units²
The line y = 1 - x forms a right angled isosceles triangle with the x and y axis, with x intercept 1 and y intercept 1.
Area of this triangle is = ½ x 1 x 1 = ½ units²
Required area
= Area of quarter circle - area of triangle
= ¼π units² - ½ units²
= ¼ (π - 2) units²
You are essentially finding a small half-leaf/segment
Another thing to notice :
The two semicircles form a circle with centre (0,0) and radius 1
From x² + y² = 1²
(x - 0)² + (y - 0)² = 1²
The two semicircles form a circle with centre (0,0) and radius 1
From x² + y² = 1²
(x - 0)² + (y - 0)² = 1²
I’m not sure if students are able to tell the sketch of such graphs though, since they are not taught to draw such graphs (they would not have been able to recognise that the equation is the same as the circle equation).
Annie, for this question, we will need a visual sketch of the graph. As J noted, the shape of the equation of the square root thing is that of a quarter circle (quadrant).
The straight line will form a right angled triangle with the x-axis.
From there we just calculate the integral by the sum/difference of areas.
The straight line will form a right angled triangle with the x-axis.
From there we just calculate the integral by the sum/difference of areas.
y = 1 - x would be no problem.
y = √(1 - x²) would need some inference from the Pythagoras' theorem or circle equation. They are linked.
Technically not the same as circle, but derived from it.
Anyway it's a H.O.T corner question so it's more of a higher order thinking question that goes beyond the mere usual applying of integration/expand the student's horizons/connecting the dots.
It won't come out in national exams
y = √(1 - x²) would need some inference from the Pythagoras' theorem or circle equation. They are linked.
Technically not the same as circle, but derived from it.
Anyway it's a H.O.T corner question so it's more of a higher order thinking question that goes beyond the mere usual applying of integration/expand the student's horizons/connecting the dots.
It won't come out in national exams
Thanks J and Eric!
Welcome Annie. You can go to desmos.com and plot the graphs of y = 1 - x and y = √(1 - x²) for visualisation
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Annie, here is the relevant graph obtained from Desmos which supplements J's answer in the chat box.
I have included the graph of the curve (in red), the straight line (in blue) and an additional graph below the x-axis (in green) to illustrate J's point that the equation of the curve is actually that of a (semi) circle with the origin (0, 0) as its centre and radius 1 unit.
To describe in words, the integral gives the area of the region bounded by the curve y = sqrt (1 - x^2) and the line y = 1 - x.
As we well know it, this area is simply the area of the quarter circle (quadrant) with radius 1 unit minus the area of the triangle.
Value of integral
= Area of quadrant - Area of triangle
= 1/4 x Pi x 1 x 1 - 1/2 x 1 x 1
= (Pi/4 - 1/2) units^2
I have included the graph of the curve (in red), the straight line (in blue) and an additional graph below the x-axis (in green) to illustrate J's point that the equation of the curve is actually that of a (semi) circle with the origin (0, 0) as its centre and radius 1 unit.
To describe in words, the integral gives the area of the region bounded by the curve y = sqrt (1 - x^2) and the line y = 1 - x.
As we well know it, this area is simply the area of the quarter circle (quadrant) with radius 1 unit minus the area of the triangle.
Value of integral
= Area of quadrant - Area of triangle
= 1/4 x Pi x 1 x 1 - 1/2 x 1 x 1
= (Pi/4 - 1/2) units^2
Date Posted:
4 years ago
Thanks!
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