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Session 17: General Parametric Equations; the Cycloid
Can someone please explain how the OC vector in the Parmetric Curves PDF was found to be 12sqrt2t<1/sqrt 2, 1/sqrt 2>? Thank you.
 one year ago
 one year ago
Session 17: General Parametric Equations; the Cycloid Can someone please explain how the OC vector in the Parmetric Curves PDF was found to be 12sqrt2t<1/sqrt 2, 1/sqrt 2>? Thank you.
 one year ago
 one year ago

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WaynexBest ResponseYou've already chosen the best response.1
The center of the circle is moving in the direction of <1,1> at a speed of \[12\sqrt{2} * t.\]We need a unit vector in the direction of motion:\[<\frac{ 1 }{ \sqrt{2} },\frac{ 1 }{ \sqrt{2} }>.\] For every second, the center of the circle moves the distance described by this unit vector multiplied by the speed.
 one year ago

JohnMBest ResponseYou've already chosen the best response.0
Thank your for the guidance. My real question is how was the vector <1/√2,1/√2> derived? I already understand that this has to be multiplied by 12√2∗t.
 one year ago

WaynexBest ResponseYou've already chosen the best response.1
Sure thing. The direction the center is going is <1,1>. A unit vector of <1,1> is <1/√2,1/√2>. What you do is take the dot product of <1,1> which is 2. Then take the square root of that, which is sqrt(2). Then divide the directional vector by that number. Think of the unit circle. All points on the circle satisfy x^2+y^2 = 1. Each point on the circle is a unit vector in the direction starting from the origin to the point in question. <1,1> is a vector that is longer than the unit circle. 1^2+1^2=2.
 one year ago

JohnMBest ResponseYou've already chosen the best response.0
OK, got it. Thanks. I forgot to think about this in terms of taking a multiple (in this case 12√2∗t) of the unit vector for <1,1>, which is <1/√2i + 1/√2j>.
 one year ago

JohnMBest ResponseYou've already chosen the best response.0
Oops. For sake of accuracy ignore latter vector notation. It should read 1/√2i + 1/√2j.
 one year ago

WaynexBest ResponseYou've already chosen the best response.1
You're welcome. Good luck in the course!
 one year ago
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