Trajectory and parametric equation

AI Thread Summary
To find the velocity and acceleration components of a particle defined by the parametric equations x(t)= 2t^2, y(t)= t^2-4t, and z(t)= 3t-5 at time t=1, the first derivative of the position vector r(t) yields the velocity vector, while the second derivative provides the acceleration vector. The direction vector given is i-3j+2k, which requires calculating the dot product of the velocity and acceleration vectors with this direction vector to find their components. The magnitude of the direction vector is also essential for normalization. The final components can be determined using the formula for projection onto the direction vector.
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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

\mbox{component}=\frac{r'(1)\bullet\mbox{A}}{\mid\mbox{A}\mid}

Use the same approach for r''(1).
 
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