Calc3 vectors, lenght of curve, equation of the sphere.

pillar
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1.Consider the process A[1,1] and B [-1,4]; let Vector U=AB

1a.Write u using components. Write u using the standard basis.
1b.Draw u as a position vector. Find u.
1c.If C [1,1] and D [5,y], then find the value of y so that AB is parallel to CD.

2. Find the length of the curve 2t2 3t2 , 0<t<[3/4]
3. Find the equation of the sphere with center [2,3,4] and is tangent to thee xyplane. At what points does the sphere intersect the z axis?
2.distance formula

3.

My Answers

1a. -1-1=-2 4-1=3 u=<-2,3> u=<1,1> <-1,4>
1b. u=-22+32[1/2] =131/2
1c. [5-1/2],[y-1/2] [2,y-1/2]

2. [PLAIN]http://img255.imageshack.us/img255/9178/hillo1.png

3.[PLAIN]http://img408.imageshack.us/img408/9339/hillo2.png
 
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pillar said:
1.Consider the process A[1,1] and B [-1,4]; let Vector U=AB
This doesn't make any sense to me. What is a "process"? What sort of multiplication is done with AB?
pillar said:
1a.Write u using components. Write u using the standard basis.
1b.Draw u as a position vector. Find u.
1c.If C [1,1] and D [5,y], then find the value of y so that AB is parallel to CD.

2. Find the length of the curve 2t2 3t2 , 0<t<[3/4]
Are these the parametric equations for the curve?
pillar said:
3. Find the equation of the sphere with center [2,3,4] and is tangent to thee xyplane. At what points does the sphere intersect the z axis?
2.distance formula

3.

My Answers

1a. -1-1=-2 4-1=3 u=<-2,3> u=<1,1> <-1,4>
1b. u=-22+32[1/2] =131/2
1c. [5-1/2],[y-1/2] [2,y-1/2]

2. [PLAIN]http://img255.imageshack.us/img255/9178/hillo1.png

3.[PLAIN]http://img408.imageshack.us/img408/9339/hillo2.png
The radius of this sphere is not sqrt(29).
 
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AB is a Vector.

Yes those are parametric equations for the curve.

I don't know how to solve 3, I need some help.
 
When does a position function hit the z axis? When the y and x-axis are 0.
 
pillar said:
bump
It is against forum policy to "bump" a thread and can get you banned.
 
If x= x(t) and y= y(t) then the arclength is given by
\int\sqrt{(x&#039;(t))^2+ (y&#039;(t))^2} dt
not
\int\sqrt{(x(t))^2+ (y(t))^2} dt
which is what you seem to be trying.

For x= 2t^2, y= 3t^2 that would be
\int\sqrt{4t^2+ 36t^2}dt= 2\sqrt{10}\int t dt
not what you have.
 
3: If a sphere has center (a, b, c) and is "tangent to the xy-plane" then the radius to that plane is perpendicular to it. That means that the radius of the sphere is "c".
 
Sorry about the bump, thanks for checking my work.
 
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