Suppose r(t)=cos(πt)i+sin(πt)j+2tk
represents the position of a particle on a helix, where z
is the height of the particle.
(a) What is t when the particle has height 6
(b) What is the velocity of the particle when its height is 6
(c) When the particle has height 6
, it leaves the helix and moves along the tangent line at the constant velocity found in part (b). Find a vector parametric equation for the position of the particle (in terms of the original parameter t) as it moves along this tangent line.
L(t)=
(a) To find t when the particle has height 6, we set the z-coordinate equal to 6:
2t = 6
Solving for t, we get:
t = 3
(b) The velocity of the particle is the derivative of r(t):
v(t) = -πsin(πt)i + πcos(πt)j + 2k
To find the velocity when the height is 6, we substitute t = 3 into the velocity function:
v(3) = -πsin(π(3))i + πcos(π(3))j + 2k
= -πsin(3π)i + πcos(3π)j + 2k
= 0i - πj + 2k
= -πj + 2k
So, the velocity of the particle when its height is 6 is -πj + 2k.
(c) To find a vector parametric equation for the position of the particle as it moves along the tangent line, we use the position equation of the particle and the direction vector of the tangent line (velocity):
L(t) = r(t) + tv(t)
L(t) = (cos(πt)i + sin(πt)j + 2tk) + t(-πj + 2k)
Simplifying, we get:
L(t) = cos(πt)i + (sin(πt) - πt)j + (2t - tπ)k
So, a vector parametric equation for the position of the particle as it moves along the tangent line is:
L(t) = cos(πt)i + (sin(πt) - πt)j + (2t - tπ)k
To answer these questions, we'll first need to find the value of t when the particle has a height of 6. Then, we'll calculate the velocity of the particle when its height is 6. Finally, we'll use this velocity to find a vector parametric equation for the position of the particle as it moves along the tangent line.
(a) To find t when the particle has a height of 6, we set the z-component of r(t) equal to 6 and solve for t:
z = 2t
6 = 2t
t = 3
So, when the particle has a height of 6, t = 3.
(b) To find the velocity of the particle when its height is 6, we'll calculate the derivative of r(t) with respect to t. The velocity vector v(t) is given by:
v(t) = dr(t)/dt
Taking the derivative of each component of r(t):
v(t) = -sin(πt)i + cos(πt)j + 2k
Now, to find the velocity when the height is 6, we substitute t = 3 into the velocity equation:
v(3) = -sin(3π)i + cos(3π)j + 2k
Since sin(3π) = 0 and cos(3π) = -1, the velocity becomes:
v(3) = 0i - j + 2k
v(3) = -j + 2k
So, the velocity of the particle when its height is 6 is -j + 2k.
(c) To find a vector parametric equation for the position of the particle as it moves along the tangent line with the velocity found in part (b), we'll use the point-slope form of a line equation. The equation is given by:
L(t) = r(t) + t * v(3)
L(t) = r(t) + t * (-j + 2k)
Substituting the expression for r(t):
L(t) = (cos(πt)i + sin(πt)j + 2tk) + t * (-j + 2k)
Expanding this equation gives us the vector parametric equation for the position of the particle as it moves along the tangent line:
L(t) = cos(πt)i + (sin(πt) - tj + 2tk)