# Thread: If x(t)=(t, sin 4t, cost 4t), 0<= t<= Pi, then how do you find the following?

1. ## If x(t)=(t, sin 4t, cost 4t), 0<= t<= Pi, then how do you find the following?

a) the velocity, speed, acceleration at any time t
b) the length of arc of the curve
c) a parameterization of the curve by arc length
d) T, N, and K at t=pi/2
e) the tangential and normal components of acceleration

2. Originally Posted by keysar7
a) the velocity, speed, acceleration at any time t
b) the length of arc of the curve
c) a parameterization of the curve by arc length
d) T, N, and K at t=pi/2
e) the tangential and normal components of acceleration
What have you tried?

For the first one, just like in 1d calculus the velocity is the first derivative with respect to time and the acceleration is the 2nd derivative. Speed is the magnitude of velocity
$\text{speed }=||\vec{v}||$

The arc length formula is
$s=\int_{0}^{t}||\vec{v}||dt$

for c solve part b) for t and substitute into the original equation.

This should get you started.

3. ## There you go... But double check!

This question is relevant to the stuff I'm trying to learn too. So I took a stab at it.
But could somebody check my answers below? Hope this helps Keysar too.

Part a)
v(t) = (1, 4 cos 4t, -4 sin 4t)
v(t)| = √(1 + 16 cos^2 4t + 16 sin^2 4t) = √(1 + 16) = √17
a(t) = (0, -16 sin 4t, -16 cos 4t)

Part b)
Length of arc = √(17) pi

Part c)
x(t) = (s/√17, sin (4s)/√17, cos (4s)/√17)

Part d)

T(pi/2) =1/√17 (1,4,0)
N(pi/2) =(0,0,-1)
k=16/17

Part e)
Tangential component of acceleration (everywhere) = 0
Normal component of acceleration (at pi/2) = (0,0,-16)