Asked by Anonymous
If x = t^2 + 1 and y = t^3, then d^2y/dx^2 =
I know I can solve for t in terms of x and substitute that into y = t^3 and find the double derivative.
I also know that I can take the derivative of x and y then divide dy/dt by dx/dt. Then take the derivative to get the answer.
However, my question is why can't you find d^2x/dt^2 and d^2y/dt^2 and divide to find d^2y/dx^2?
Like
x = t^2 + 1
dx/dt = 2t(dt/dt) + 0(dt/dt)
d^2x/dt^2 = 2(dt/dt) = 2
y = t^3
dy/dt = 3t^2(dt/dt)
d^2y/dt^2 = 6t(dt/dt)
d^2y/dx^2 = (d^2y/dt^2)/(d^2x/dt^2)
d^2y/dx^2 = 6t / 2 = 3t
The correct answer, however, is 3/4t. Why? Am I doing something wrong -.-?
I know I can solve for t in terms of x and substitute that into y = t^3 and find the double derivative.
I also know that I can take the derivative of x and y then divide dy/dt by dx/dt. Then take the derivative to get the answer.
However, my question is why can't you find d^2x/dt^2 and d^2y/dt^2 and divide to find d^2y/dx^2?
Like
x = t^2 + 1
dx/dt = 2t(dt/dt) + 0(dt/dt)
d^2x/dt^2 = 2(dt/dt) = 2
y = t^3
dy/dt = 3t^2(dt/dt)
d^2y/dt^2 = 6t(dt/dt)
d^2y/dx^2 = (d^2y/dt^2)/(d^2x/dt^2)
d^2y/dx^2 = 6t / 2 = 3t
The correct answer, however, is 3/4t. Why? Am I doing something wrong -.-?
Answers
Answered by
bobpursley
See this
http://en.wikipedia.org/wiki/Parametric_derivative
http://en.wikipedia.org/wiki/Parametric_derivative
Answered by
Anonymous
Ah, I see. Just for confirmation, d^2x != dx^2, correct?
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