How is the area under the arch of y=sinx related to the unit circle?

  • Thread starter Thread starter Trav44
  • Start date Start date
  • Tags Tags
    Area
AI Thread Summary
The discussion centers on the relationship between the area under the sine curve and the unit circle, particularly in the context of a year 12 math investigation. The original poster, Travis, seeks clarification on calculating areas using triangles and trapeziums, noting that the area under the sine curve should logically equal 2. Forum members emphasize the importance of showing attempts at problem-solving to receive help and guide Travis in using auxiliary lines drawn in the provided diagrams. Travis shares his calculations for areas of triangles and trapeziums, but struggles to understand how these relate to the formula pi/4(1+(√2))^3 and the concept of dividing pi into increments. The conversation highlights the need for a deeper understanding of the mathematical principles involved in these calculations.
Trav44
Messages
4
Reaction score
0
Hi guys I'm new to the forum but with year 12 maths coming up next year I assume I'll be coming here a lot.

This question is part of my end of year investigation.

I have scanned the questions and attached them as jpegs.

I was thinking something along the lines of using the triangle to determine the area? Or if the fact that this curve is related to half of a unit circle had something to do with it?

Any help would be much appreciated :)

Thank you, Travis
 

Attachments

  • 1.jpg
    1.jpg
    33.6 KB · Views: 936
  • 2.jpg
    2.jpg
    13.8 KB · Views: 653
  • 3.jpg
    3.jpg
    23.4 KB · Views: 911
Physics news on Phys.org
You've got to show an attempt at solving the problem in order to get some help.
 
Hi Trav, welcome to the forums. I do hope you'll be coming here - and learning - a lot.
As SteamKing said, please show us your attempt or problem. The questions are leading you in a specific direction. As you've noticed, there is a pretty big hint in the picture in the form of the auxiliary lines that have been drawn, creating for example a triangle. If you show us how far you got with that, we may be able to help you complete it.
 
Sorry I was unaware that an attempt must be shown, I will be sure to include them in every question from here on out :)
My attempts for each question as follows are:
1. The area of the outer rectangle is pi as pi x 1 = pi
The area for the inner triangle is 1/2bh so 1/2 x pi x 1= 1/2pi or pi/2
this shows that the area under the sine arch must fit the range of pi./2 < x < pi

2.
Triangle area= 1/2 x b x h
1/2 x pi/4 x sin(pi/4)= 0.2776

Trapezium area = (a + b)/2 x h
(sin (pi/4) + sin(pi/2))/2 x pi/4=0.6704

total area = 0.2776x2 + 0.6704x2= 1.896

which is the correct area

I get that sin of (pi/4 or pi/2) is the height at those given points but could someone please give me an explanation as to why this is?

I have found the area for both triangles and trapeziums and multiplied them by 2 to give me the overall area and this is right how ever I cannot make the connection between this and the equation pi/4(1+(root 2))3. I can use the formula to add up all the areas and I get an answer close to 2 which seems logical as the area under a sine curve is meant to equal 2.
I do not understand how this formula has been derived though?
And for pi to be split into increments of 0.1 I thought the start of the equation should be 1/31.4 as this is pi/0.1?
1/20 would give increments close to 0.15 wouldn't it?
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top