For the curve x=t^2, y=t^3-3t, -? < t

Summary### Determine where the curve is concave up and concave down? in

Full Text / Question### Determine where the curve is concave up and concave down? in **Science **

*Determine where the curve is concave up and concave down?*:For the curve x=t^2, y=t^3-3t, -? < t …

__Determine where the curve is concave up and concave down?__
nautz_mst says

Hi

dx = 2 t dt ; dy = 3 t^2 – 3 = 3 (t^2 -1)

dy/dx = 3 (t^2 -1) / (2 t) = (3/2) (t^2 – 1) / t = (3/2) ( t – (1/t))

d(dy/dx) =(3/2) d (( t – (1/t))) = (3/2) (dt + (1/t^2) dt)

d(dy/dx) =(3/2) ((t^2 + 1)/t^2) dt

d^2y/dx^2 = d(dy/dx) / dx = (3/2) ((t^2 + 1)/t^2) dt / (2 t dt)

d^2y/dx^2 = (3/4) ( t^2 + 1) / t^3

For t = 0 ;

t-> 0- d^2y/dx^2–> -inf

t-> 0+ d^2y/dx^2–> +inf

Vertical tangent

For t < 0 ---> d^2y/dx^2 < 0 Concave down For t > 0 —> d^2y/dx^2 > 0 Concave up

Graphmatica verified.

regards.