Trajectory and parametric equation

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SUMMARY

The discussion focuses on calculating the velocity and acceleration components of a particle defined by the parametric equations x(t) = 2t², y(t) = t² - 4t, and z(t) = 3t - 5 at time t = 1. The velocity vector is derived as r'(t) = 4t i + (2t - 4) j + 3 k, and the acceleration vector is r''(t) = 4 i + 2 j + 0 k. To find the components of these vectors in the direction of the vector A = i - 3j + 2k, the dot product is utilized, followed by normalization using the magnitude of vector A.

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Homework Statement


a particle is on a trajectory defined by the parametric equations x(t)= 2t^2,y(t)= t^2-4t,
z(t)= 3t-5, where t is the time.Find the components of its velocity and acceleration at time =1, in the direcion i-3j+2k.


Homework Equations


what i thought is r(t)=x(t) i +y(t) j +z(t) k, then take first derivative and get velocity vector. take the second derivative and get the acceleration vector.
However, the question provides a direction vector and i have no idea how to deal with the direction vector.

The Attempt at a Solution


r(t)'=4t i + (2t-4) j+(3) k.
r(t)''= (4) i + (2)j +(0) k
 
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Find the direction cosines between velocity and acceleration vectors and given direction vector.
 
You are on the correct path to solving the problem. Once you find r'(1) and r''(1) you will project the components of these two vectors on to the vector given by i-3j+2k. The dot product gives the component of one vector onto another. Since this component is multiplied by the magnitude of other vector, you must divide by the magnitude of the vector i.e.

Let A = i-3j+2k

then

[tex]\mbox{component}=\frac{r'(1)\bullet\mbox{A}}{\mid\mbox{A}\mid}[/tex]

Use the same approach for r''(1).
 

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