1) Let r(t) = 5cos(t)i+5sin(t)j+12tk a) Graph the curve determined by the vector

Question

1) Let r(t) = 5cos(t)i+5sin(t)j+12tk

a) Graph the curve determined by the vector equation.

b) Find the velocity, acceleration, and speed at time t.

c) Find the length of the curve over the interval 0 %u2264 t %u2264 2pi

2) A particle has position function r(t) = (2t, t^2, 1/3 (t^3) Find the tangential and normal components of acceleration when t = 3.

3) Find the position function of a particle that hasthe given acceleration and the given

initial velocity and position:

a(t) = 6 i + 12 t^2j - 6t k

v(0) = 2i + 3k

r(0) = j

Explanation / Answer

Hope you have given the normal vector.

r (t) = cos(4t) i + sin(4t) j - 8 k

v= r'(t) = [-4 sin 4 t] i + [4 cos 4t] j

a =r"(t) = [-16 cos 4t] i + [-16 sin 4t] j = (-16) [cos(4t) i + sin(4t) j]

Now, |v| = %u221A ( [-4 sin 4 t]^2 + [4 cos 4t]^2) =4

Also, v.a = [64 sin 4t cos 4t] i - [64 sin 4t cos 4t] j = [64 sin 4t cos 4t] [i - j]

v x a = 64 k

Now, Tangential Component aT = [v.a]/|v| = [4 sin 4t cos 4t] [i-j]

Normal Component aN = [v x a]/|v| = 16 k

It looks right to me except you took off the vector units. You probably should have left them on.
v(t) = 2ti +(1/t)j +2k
and acceleration should be
a(t) = 2i -(1/t^2)j
speed should be the magnitude of the vector of velocity. You can find it by thepythagorean theorem
s(t) = sqrt((2t)^2+(1/t)^2 +2^2)

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