Apostle, chapter 14 problems: Vector Calculus

[zonk]

Homework Statement

Problem 1: Two fixed unit vectors A and B make an angle θ with each other, where 0 < θ < π. A particle moves in a space curve in such a way that its position vector r(t) and velocity v(t) are related by the equation v(t) = A x r(t). If r(0) = B, prove that the curve has constant curvature and compute this curvature in terms of θ.

Problem 2: A point moves in space according to the vector equation r(t) = 4cos(t)i + 4sin(t)j + 4cos(t)k. Show this path is an ellipse.

Homework Equations

k = | a x v |/|v|^3

The Attempt at a Solution

1: I've already figured out the motion must move in a plane because r(t) x v(t) = A and is constant, but I don't know where to go from there.The answer is k = 1/|B|*sin(θ).

2: I put x = 4cos(t), y = 4sin(t), and z = 4cos(t), and deduce xz + y^2 = 16 but this is not a ellipse. Where am I going wrong?

 

[mighty2000]

For problem 1, since r(t) and v(t) are related by the equation v(t) = A x r(t), we can take the derivative of both sides with respect to t to get:

a(t) = d/dt (A x r(t)) = d/dt (A) x r(t) + A x d/dt (r(t))

Since A is a fixed unit vector, its derivative is 0, so we can simplify this to:

a(t) = A x v(t)

Now, we know that a(t) is the acceleration vector and v(t) is the velocity vector. We also know that the acceleration is always perpendicular to the velocity in a circular motion. Therefore, we can say that a(t) is always perpendicular to v(t) and thus, a(t) is always perpendicular to r(t). This means that the motion is always in a circular path.

Now, we can use the formula for curvature, k = | a x v |/|v|^3, to compute the curvature. Since a(t) = A x v(t), we can rewrite this as:

k = | A x v(t) x v(t) |/|v(t)|^3

Since A is a unit vector, we can rewrite this as:

k = | v(t) x v(t) |/|v(t)|^3

Now, since v(t) is a vector in the direction of the motion, we can rewrite this as:

k = | v(t) |/|v(t)|^2

Since | v(t) | is just the speed of the particle, we can rewrite this as:

k = 1/|v(t)|

But we know that the speed of the particle is constant since the motion is circular. Therefore, k is also constant.

Since r(0) = B, we can say that the initial velocity, v(0), is also perpendicular to A and thus, perpendicular to B. This means that the motion starts at a right angle to the fixed unit vectors A and B. Therefore, the angle between A and v(t) will always remain constant, which means that the curvature, k, will also remain constant. And since k is constant, we can write it as:

k = 1/|B|*sin(θ)

For problem 2, you are correct in saying that the path is not an ellipse. It is in fact a helix.