A particle moves along the parabolic path given by y^2=4x in such a way that when it is at the point (1,-2), its horizontal velocity (in the direction of the x-axis) is 3ft/s. What is its vertical velocity (in the direction of the y-axis) at this instant? PLEASE SHOW WORK!
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it is given that y^2=4x
differentiating both sides with time
2y(dy/dt)=4(dx/dt)
as dy/dt is velocity in y direction
and dx/dt is velocity in x direction
we put values
y=-2
dx/dt=3
so the value of velocity in y direction is -3ft/s
pleas vote brainliest
differentiating both sides with time
2y(dy/dt)=4(dx/dt)
as dy/dt is velocity in y direction
and dx/dt is velocity in x direction
we put values
y=-2
dx/dt=3
so the value of velocity in y direction is -3ft/s
pleas vote brainliest
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