anonymous
  • anonymous
Hi guys I'm stuck in Problem Set 4, Part B, Problem 2: Consider a triangle inscribed in the unit circle in the plane, with one vertex at (1, 0) and the two other vertices given by polar angles θ1 and θ2, in that order counterclockwise. a) Express the area A of the triangle in terms of θ1 and θ2. What is the set of possible values for θ1 and θ2? Next is to find the maximum area. Need your help thanks!
MIT 18.02 Multivariable Calculus, Fall 2007
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I would start with figuring out the possible values for \theta1 and \theta2: What's the biggest possible value before it loops around? Is there such a thing?
anonymous
  • anonymous
Here it is. Problem 2 http://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/assignments/ps4.pdf I shifted to 18.02SC because there are no solutions here in 18.02 Fall 2007. :(
anonymous
  • anonymous
Ah, so it's just the first part in a multi-part problem. Hmm. Let's assume that theta1 and theta2 are separate (see drawing), rather than theta2 "containing" (and thus necessarily being larger than) theta1.|dw:1350514074023:dw|

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anonymous
  • anonymous
Of the three angles theta1, theta2, 2pi - (theta1 + theta2), at least one angle must be less than or equal to 2pi/3, and at least one other angle must be greater than or equal to 2pi/3. Without loss of generality (WLOG), we can rotate and/or flip the circle so that the smallest angle is theta1, the largest is 2pi - (theta1 + theta2), and the in-between-sized angle is theta2.
anonymous
  • anonymous
At this point, it seems like breaking it into cases is a good idea. Case 1: The inscribed triangle contains the origin in its interior (ie, all three angles are less than pi). Case 2: The inscribed triangle has a side that is a diameter-length chord through the origin. Case 3: The origin is completely outside of (exterior to) the inscribed triangle.
anonymous
  • anonymous
Case 1 is easy: Calculate each angle's area separately, then add them. |dw:1350516691235:dw| |dw:1350515228416:dw|
anonymous
  • anonymous
Just for you to know in the future - among all triangles with a given perimeter p, the biggest in area is the equilateral triangle, so theta is 2pi/3.
anonymous
  • anonymous
I can help with part of it. If you let theta1 and theta2 be both from the positive x-axis, then the three points are at: P0 = (1,0) P1 = (cos(theta1), sin(theta1)) P2 = (cos(theta2), sin(theta2)) You can then easily get the vectors P0P1 and P0P2, and the area of the triangle is half the determinant of those two vectors. As for max area, if you use the symmetry argument that max area happens when theta2 = -theta1, you can reduce the kind of complicated area formula to something simpler with only one variable: A = sin(theta) - sin(theta)cos(theta) - or - A = sin(theta) - (1/2)sin(2*theta) (You can probably see this more easily drawing the picture and going with straight 1/2*base*hieght, forget the vector stuff.) If you differentiate w/r to theta and set to zero, you get: cos(theta) = cos(2*theta) which is satisfied by theta = 2*pi/3 and it is indeed an equilateral triangle.

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