Question

(1 point) Given that the acceleration vector is a(t)=(−9cos(3t))i+(−9sin(3t))j+(−5t)k

, the initial velocity is v(0)=i+k
, and the initial position vector is r(0)=i+j+k
, compute:

A. The velocity vector v(t)=

i+

j+

k

B. The position vector r(t)=

i+

j+

k

Note: the coefficients in your answers must be entered in the form of expressions in the variable t
; e.g. "5 cos(2t)"

Answer 1

Therefore, the position vector r(t) is:

r(t) = (cos(3t) + t)i + (sin(3t) + 1)j + ((-5/6)t^3 + t + 1)k

Explanation:

To find the velocity vector v(t) and the position vector r(t), we will integrate the given acceleration vector a(t) with respect to time (t) starting from the initial conditions.

A. To find the velocity vector v(t), we integrate each component of the acceleration vector a(t) with respect to time (t).

∫(-9cos(3t)) dt = -3sin(3t) + C1 (integrating the x-component)

∫(-9sin(3t)) dt = 3cos(3t) + C2 (integrating the y-component)

∫(-5t) dt = (-5/2)t^2 + C3 (integrating the z-component)

Since the initial velocity is v(0) = i + k, we can equate the components:

-3sin(0) + C1 = 1 (from the x-component)

3cos(0) + C2 = 0 (from the y-component)

(-5/2)(0)^2 + C3 = 1 (from the z-component)

Simplifying these equations, we get:

C1 = 1

C2 = 0

C3 = 1

Therefore, the velocity vector v(t) is:

v(t) = (-3sin(3t) + 1)i + (3cos(3t))j + (-5/2)t^2 + 1)k

B. To find the position vector r(t), we integrate each component of the velocity vector v(t) with respect to time (t).

∫(-3sin(3t) + 1) dt = (cos(3t) + t) + C4 (integrating the x-component)

∫(3cos(3t)) dt = sin(3t) + C5 (integrating the y-component)

∫((-5/2)t^2 + 1) dt = (-5/6)t^3 + t + C6 (integrating the z-component)

Since the initial position is r(0) = i + j + k, we can equate the components:

(cos(0) + 0) + C4 = 1 (from the x-component)

(sin(0) + 0) + C5 = 1 (from the y-component)

(-5/6)(0)^3 + 0 + C6 = 1 (from the z-component)

Simplifying these equations, we get:

C4 = 0

C5 = 1

C6 = 1

Therefore, the position vector r(t) is:

r(t) = (cos(3t) + t)i + (sin(3t) + 1)j + ((-5/6)t^3 + t + 1)k