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January 15th, 2015, 12:59 PM   #1
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Parameterization curves

Sketch and parameterize curves for each of the following:
1. From the point (-2,5) to (3,10) along a parabola passing through (0,1).

2. From the point (3,4) to (-4,3) along a circular path

3. From the north pole of the sphere centered at (7,-2,3) of radius 5 to the south pole of that sphere, with your curve lying on that sphere.

4. From the point (0,2,5) to the point (2,0,-3) lying on a vertical cylinder of radius 2.

5. The intersection of the plane 2x+4y=8 and the cylinder x^2 + z^2 = 4

Parameterize the tangent line to your answer to part 5 of the second question above at the point (-2,3,0).
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January 15th, 2015, 02:37 PM   #2
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Again, you show none of your own attempt to do these so we have no way of knowing what you already know or what help you need.

1) You should know that a parabola can be written as x= t, y= at^2+ bt+ c. Knowing that (-2,5) is on the parabola tells you that x= t= -2, y= at^2+ bt+ c= 4a- 2b+c= 5. (3,10) being on the parabola tells you that x= t= 3, y= at^2+ bt+ c= 9a+ 3b+ c= 10. Knowing (0,1) is on the grapy tells you that x= t= 0, y= at^2+ bt+ c= c= 1. Solve those equations for a, b, and c. You will want to restrict t to be from -2 to 3 so that the curve goes from (-2, 5) to (3, 10).

2) There are an infinite number of circles passing through two give points. You should know that a circle with center (a, b) and radius r can be written x= a+ rcos(t), y= b+ rsin(t). In order that the circle pass through (3, 4), you must have 3= a+ r cos(t), 4= b+ rsin(t). In order that the circle pass through (-4, 3) you must have -4= a+ r cos(t), 3= b+ r sin(t). That gives you two equations to solve for a, b, and r which will leave one of them undetermined.

3) A sphere with center at (a, b, c), with radius r, can be written x= a+ r cos(s)sin(t), y= b+ r sin(s)sin(t), z= c+ r cos(t). Here, the center is (7, -2, 3) and the radius is 5 so the sphere is given by x= 7+ 5cos(s)sin(t), y= -2+ 5sin(s)sin(t), z= 3+ 5cos(t). The "north pole" is given by t= 0, (7, -2, 8) and the south pole by t= pi, (7, -2, -2). But, again, there are an infinite number of circles that pass through those two points, depending on what value of syou choose. For example with s= 0, in the y= -2 plane, x= 7+ 5sin(t), y= -2, = 3+ 5cos(t). With s= pi/2, in the x= 7 plane, x= 7, y= -2+ 5sin(t), z= 3+ 5cos(t).

4) A vertical cylinder with radius 2 can be written as x= a+ 2cos(t), y= b+ 2sin(t), z= s.
The fact that (0, 2, 5) and (2, 0, 3) lie on that cylinder mean that x= a+ 2 cos(t)= 0, y= b+ 2sin(t)= 2, z= s= 5, and x= a+ 2cos(t)= 2, y= b+ 2sin(t)= 0, z= s= 3. The equations for x and y can be used to determine a, b, and t. Again there are many different curves, lying on the cylinder that lie on that cylinder.

5) From the equation of the plane, 2x+ 4y= 8, we can get y= 2- x/2. For any y, the cylinder x^2+ z^2= 4 is a circle with with center at (0, 0) and radius 2: x= 2 cos(t), z= 2 sin(t). From y= 2- x/2, y=2- cos(t). The parametric equations of the intersection are x= 2cos(t), y= cos(t), z= 2 sin(t).

A tangent vector to that curve is <-2 sin(t), -sin(t), 2 cos(t)>. The point (-2, 3, 0) corresponds to the t satisfying -2= 2 cos(t), 3= 2- cos(t), 0= 2 sin(t). All three of those are satisfied by t= pi so the tangent vector at (-2, 3, 0) is <0, 0, -2> and the equation of the tangent line there is 0(x+ 2)+ 0(y- 3)- 2(z- 0)= -2z= 0 or z= 0.

The tangent line to that curve at (-2, 3, 0) is
Parameterize the tangent line to your answer to part 5 of the second question above at the point (-2,3,0).

Last edited by Country Boy; January 15th, 2015 at 02:47 PM.
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January 15th, 2015, 02:42 PM   #3
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1. It helps to notice that the parabola has equation y = x² + 1.

2. Which circular path would you choose?

Can you post your attempt for each question?
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