Solving for v(t) and a(t): Calculating B

[TranscendArcu]

Homework Statement

A particle is moving along the path c(t) = <cos(t),sin(t),2t>. Suppose its acceleration a(t) is given by a(t) = v(t) x B where v(t) is the velocity and B(x,y,z) is a constant vector (i.e. independent of (x,y,z)).

Find v(t) and a(t)
Find B?

Homework Equations

The Attempt at a Solution

Finding v(t) is relatively simple: v(t) = <-sint(t),cos(t),2>. Finding a(t) is more tricky and I don't know how to begin. I would be appreciative of at least the first step.

 

[Dick]

You found the velocity v(t) by taking the first derivative of c(t), v(t)=c'(t) right? The acceleration is the derivative of the velocity. Why is that trickier?

 

[SammyS]

Well, how did you find v(t) ?

a(t) has the same relationship to v(t) as v(t) has with c(t).

So, if finding v(t) from c(t) was easy, finding a(t) from v(t) should be equally easy, maybe easier.

 

[TranscendArcu]

I don't understand a(t) as simply being the derivative of v(t). That would be easy. Instead, I see a(t) as being defined as the cross product of v(t) with some vector B. Given this definition, how can I assume that a(t) = v'(t)?

 

[Dick]

I don't understand a(t) as simply being the derivative of v(t). That would be easy. Instead, I see a(t) as being defined as the cross product of v(t) with some vector B. Given this definition, how can I assume that a(t) = v'(t)?

v(t) and a(t) you find the usual way. a(t)=v(t) x B isn't a definition of a(t), it's a definition of B. Try and find B from that.

 

[TranscendArcu]

Okay, that makes sense. Thank you!

But another question: Suppose we have the surface, S, which is defined by x^2 + y^2 - z^2 + 1 = 0, and the plane, H, x - 3z = 0. I want to find all the points on S where the tangent plane is parallel to H. Here's how I tried to approach this:

I know the vector normal to H is <1,0,-3>. Therefore, are points on S which have gradients that point in the same direction as <1,0,-3> should be parallel. So, finding the gradient of S gives

graf f = <2x,2y,-2z> = <1,0,-3>, so,

x = 1/2
y = 0
z = 3/2, so, the point (.5,0,1.5) should satisfy the condition, right? Except then I realize that this point doesn't lie on the surface S. So I don't know how to proceed.

 

[Dick]

Okay, that makes sense. Thank you!

But another question: Suppose we have the surface, S, which is defined by x^2 + y^2 - z^2 + 1 = 0, and the plane, H, x - 3z = 0. I want to find all the points on S where the tangent plane is parallel to H. Here's how I tried to approach this:

I know the vector normal to H is <1,0,-3>. Therefore, are points on S which have gradients that point in the same direction as <1,0,-3> should be parallel. So, finding the gradient of S gives

graf f = <2x,2y,-2z> = <1,0,-3>, so,

x = 1/2
y = 0
z = 3/2, so, the point (.5,0,1.5) should satisfy the condition, right? Except then I realize that this point doesn't lie on the surface S. So I don't know how to proceed.

Two vectors can point in the same direction without having the same components. c*<1,0,-3> for any constant c is also parallel to <1,0,-3>, right?