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

1. Aug 29, 2010

### pillar

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.

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 [Broken]

3.[PLAIN]http://img408.imageshack.us/img408/9339/hillo2.png [Broken]

Last edited by a moderator: May 4, 2017
2. Aug 29, 2010

### Staff: Mentor

This doesn't make any sense to me. What is a "process"? What sort of multiplication is done with AB?
Are these the parametric equations for the curve?
The radius of this sphere is not sqrt(29).

Last edited by a moderator: May 4, 2017
3. Aug 29, 2010

### pillar

AB is a Vector.

Yes those are parametric equations for the curve.

I don't know how to solve 3, I need some help.

4. Aug 30, 2010

### Chipz

When does a position function hit the z axis? When the y and x axis are 0.

5. Aug 30, 2010

### HallsofIvy

It is against forum policy to "bump" a thread and can get you banned.

6. Aug 30, 2010

### HallsofIvy

If x= x(t) and y= y(t) then the arclength is given by
$$\int\sqrt{(x'(t))^2+ (y'(t))^2} dt$$
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.

7. Aug 30, 2010

### HallsofIvy

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".

8. Aug 30, 2010

### pillar

Sorry about the bump, thanks for checking my work.