How do you find the velocity and position vectors if you are given that the acceleration vector is #a(t)= (-4 cos (-2t))i + (-4sin (-2t))j + (-2t)k# and the initial velocity is #v(0)=i+k# and the initial position vector is #r(0)= i+j+k#?

Answer 1
First, note that #cos(-2t)=cos(2t)# and #sin(-2t)=-sin(2t)# since cosine is an even function and sine is an odd function.
Now integrate to get #v(t)=-2sin(2t)i-2cos(2t)j-t^2k+\vec{c}#, where #\vec{c}# is a constant of integration. Since #v(0)=i+k# and since #sin(0)=0# and #cos(0)=1#, it follows that #\vec{c}=i+2j+k# so that #v(t)=(-2sin(2t)+1)i+(-2cos(2t)+2)j+(1-t^2)k#.
Now integrate again to get #r(t)=(cos(2t)+t)i+(-sin(2t)+2t)j+(t-\frac{1}{3}t^{3})k+\vec{c}#. Since #r(0)=i+j+k#, it follows that #\vec{c}=j+k# so that #r(t)=(cos(2t)+t)i+(-sin(2t)+2t+1)j+(t-\frac{1}{3}t^{3}+1)k#
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Answer 2

To find the velocity and position vectors, integrate the acceleration vector with respect to time to obtain the velocity vector, and then integrate the velocity vector to obtain the position vector.

The velocity vector v(t) is given by: v(t) = ∫a(t) dt

Integrating each component of the acceleration vector separately, we get: v(t) = ∫(-4 cos(-2t)) dt * i + ∫(-4 sin(-2t)) dt * j + ∫(-2t) dt * k

Integrating each component: v(t) = (2 sin(-2t)) i + (-2 cos(-2t)) j + (-t^2) k + C

Where C is the constant of integration.

Given that v(0) = i + k, we can solve for C: v(0) = (2 sin(0)) i + (-2 cos(0)) j + (0) k + C v(0) = 2i - 2j + C

Comparing coefficients, we find C = 1 + 2j.

Thus, the velocity vector is: v(t) = (2 sin(-2t)) i + (-2 cos(-2t)) j + (-t^2) k + (1 + 2j)

To find the position vector r(t), integrate the velocity vector: r(t) = ∫v(t) dt

Integrating each component of the velocity vector separately: r(t) = ∫(2 sin(-2t)) dt * i + ∫(-2 cos(-2t)) dt * j + ∫(-t^2) dt * k + ∫(1 + 2j) dt

Integrating each component: r(t) = (-cos(-2t)) i + (-sin(-2t)) j + (-1/3 * t^3) k + (t + 2tj) + C'

Where C' is the constant of integration.

Given that r(0) = i + j + k, we can solve for C': r(0) = (-cos(0)) i + (-sin(0)) j + (-1/3 * 0^3) k + (0 + 2(0)) + C' r(0) = -i + j + C'

Comparing coefficients, we find C' = 1.

Thus, the position vector is: r(t) = (-cos(-2t)) i + (-sin(-2t)) j + (-1/3 * t^3) k + (t + 2tj + 1)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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