If #x=at^2# and #y=2at#, what is #(d^2y)/(dx^2)# at #t=1/2#?

1 Answer
Steve M
Jul 14, 2017

# [ (d^2y)/(dx^2) ]_(t=1/2) = -4/a#

Explanation:

We have:

# x = at^2 #
# y=2at #

Differentiating wrt #t# we get:

# dx/dt = 2at #

# dy/dt = 2a #

By the chain rule we have:

# dy/dx = (dy//dt) / (dx//dt) #
# " " = (2a) / (2at) #
# " " = 1/t #

Differentiating again wrt #x#, and applying the chain rule we have:

# (d^2y)/(dx^2) = d/dx(dy/dx) #
# " " = d/dt(dy/dx) * dt/dx \ \ \ \ \ # (chain rule)

# " " = (d/dt(1/t)) / (dx/dt) #

# " " = (-1/t^2) / (2at) #

# " " = (-1/t^2) (2at) #

# " " = -1/(2at^3) #

When #t=1/2# we have:

# [ (d^2y)/(dx^2) ]_(t=1/2) = -1/(2a*1/8) = -4/a#

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