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Easy_as_Pi
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Hey all! Long time lurker, first time poster. Already answered questions on this forum have managed to help me out immensely in calc 1 and 2, but I am truly stumped on this problem o.0 Thanks in advance for your time!
The position of a particle in the plane at time t is
r(t) = t/sqrt(1+t^2)i + 1/sqrt(1+t^2)j + tk
What is the particle's highest speed. When and where does it reach this speed?
Speed = magnitude of velocity vector
velocity = derivative of position vector
acceleration = derivative of velocity vector
Well, I took the derivative of r(t), and got velocity equal to the following vector
<1/(t^2+1)^3/2 i, -t/(1+t^2)^3/2 j, 1k>
I then took the derivative of velocity to get acceleration, which I found to be:
a(t) = <-3t/(t^2+1)^5/2 i, (2t^2-1)/(t^2+1)^5/2 j, 0 k>
I know I need the zeroes of this, because they will be the maximum / minimum values for my velocity. So, I set each part (i,j,k) equal to 0 and solved. I don't think this is right, however. My answers were 0 and 1/4 for t. With 0 being my minimum and 1/4 being the max. If I plug 1/4 in for t in the v(t) equation and find the magnitude I get speed being equal to sqrt(1273)/27. I'm not sure about the where and when? Would that just be the velocity vector at time t (vector gives where, t gives when)?
Sorry for the lack of latex, I tried to use it, but couldn't get my work to format right.
Homework Statement
The position of a particle in the plane at time t is
r(t) = t/sqrt(1+t^2)i + 1/sqrt(1+t^2)j + tk
What is the particle's highest speed. When and where does it reach this speed?
Homework Equations
Speed = magnitude of velocity vector
velocity = derivative of position vector
acceleration = derivative of velocity vector
The Attempt at a Solution
Well, I took the derivative of r(t), and got velocity equal to the following vector
<1/(t^2+1)^3/2 i, -t/(1+t^2)^3/2 j, 1k>
I then took the derivative of velocity to get acceleration, which I found to be:
a(t) = <-3t/(t^2+1)^5/2 i, (2t^2-1)/(t^2+1)^5/2 j, 0 k>
I know I need the zeroes of this, because they will be the maximum / minimum values for my velocity. So, I set each part (i,j,k) equal to 0 and solved. I don't think this is right, however. My answers were 0 and 1/4 for t. With 0 being my minimum and 1/4 being the max. If I plug 1/4 in for t in the v(t) equation and find the magnitude I get speed being equal to sqrt(1273)/27. I'm not sure about the where and when? Would that just be the velocity vector at time t (vector gives where, t gives when)?
Sorry for the lack of latex, I tried to use it, but couldn't get my work to format right.