(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
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)=
1 answer