A tool is programmed to move, such that x=2cos3t and y=cos2t. Find the velocity of the tool when t=4.1s.
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The velocity of the machine can be divided into two components, x and y.
It moves at x = 2 cos 3t in the x-direction.
Therefore the x component of the velocity is dx/dt = -2*3 sin 3t.
At t = 4.1, this is equal to 1.5793.
In the y- direction it moves at y = cos 2t.
dy/dt = -2* sin 2t
At t = 4.1, this is equal to -1.8814
The direction of the velocity is at angle of -0.8724 radians to the x-axis.
Therefore the magnitude of the velocity at t = 4.1 is sqrt ( 1.5793^2 + 1.8814^2) = 2.4563.
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We assume X and Yare perpendicular direction representing x and y axis.
The component of the velocity of the machine in x direction is at time t isgiven by: dx/dt = d/dt(2cos2t) = 2(-sin3t)3= -6sin2t.
The component of velocity of the velocity of the machine in y direction at time t is given by: dy/dt = d/dt(cos2t) = (-sin2t)(2) = -2sin2t.
Therefore the v velocity of the machine at time t is given by : v(t) = {(x component of the velocity)^2+(y component of the velocity)^2}^(1/2) = (-6sin3t)^2 +(-2sin2t)^2}^(1/2) = {36(csin3t)^2+4sin2t)^2 }^(1/2)
v(t) = 2{9(sin3t)^2 + (sin2t)^2}^(1/2)
v(4.1) = 2{9sin12.3))^2 + (sin8.2)^2}^(1/2)
v(4.1) = 2{0.62362+0.88497}^(1/2)
v(t) = 2.4565 length units/sec is the velocity of the machine tool.
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