Question

Parametric equation stuff: When finding a tangent of two equations x(t),y(t), how do you find d^2y/dx^2?

x = 1+ln(t); y = t^2+2

Answer 1

dy/dx = 2t^2

Answer 2

Ok, if anyone was wondering it is asking for the derivative of dy/dx = (d/dt)(dy/dx). And all of that over (dx/dt).

In the case of my example the answer would have been 4t/(1/t)= 4t^2/.

Answer 3

\[x=1+\ln{t}\quad\Rightarrow\quad t=e^{x-1}\]\[y=t^2+2=e^{2(x-1)}+2\]\[y'=2e^{2(x-1)}=2t^2\]\[y''=4e^{2(x-1)}=4t^2\]