Consider the parametric curve: x = 16 - t2 y = t3 - 16t At which t value(s) is t

Question

Consider the parametric curve: x = 16 - t2 y = t3 - 16t At which t value(s) is the tangent to this curve vertical?( Note: If there is more than one write them as a comma separated list) Write the point(s) at which the tangent is vertical.( Note: If there is more than one write them as a comma separated list., e.g. (1, 2), (3, 4)) At which t value(s) is the tangent to this curve horizontal?( Note: If there is more than one write them as a comma separated list). Write the point(s) at which the tangent is horizontal.( Note: If there is more than one write them as a comma separated list; e.g. (1, 2), (3, 4))

Explanation / Answer

dy/dx = infinity
dx/dy = 0
dx/dy = dx/dt * dt/dy
dx/dt = -2t; dy/dt = 3t^2 -16 ; dt/dy = 1/ (3t^2 -16)
so, dx/dy = -2t/(3t^2 -16) = 0
so, t = 0
point is ( 16, 0)
FOr tangent to be horizontal,
dy/dx = 0
dy/dx = (3t^2 -16)/(-2t)
so, (3t^2 -16) = 0
t = + sqrt (16/3) , -sqrt (16/3)
t = 2.3094, -2.3094.
points are (10.66666, -24.63361) ; (10.66666,24.633611)

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