anonymous
  • anonymous
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.
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Waynex
  • Waynex
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.
anonymous
  • anonymous
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.
Waynex
  • Waynex
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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>.
anonymous
  • anonymous
Oops. For sake of accuracy ignore latter vector notation. It should read 1/√2i + 1/√2j.
Waynex
  • Waynex
You're welcome. Good luck in the course!

Looking for something else?

Not the answer you are looking for? Search for more explanations.