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Find the interval for which x=t^2/2-3t+5 and y=t^2+4t-1 is concave up and concave down
4 years ago

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oobleck
you need to check y" for positive or negative
y' = (dy/dt) / (dx/dt) = (2t+4)/(t-3)
y" = dy'/dx = d/dt (2t+4)/(t-3) / (dx/dt) = -10/(t-3)^3
So it looks like it is concave
up for t<3
down for t>3
4 years ago
Will
So, concave up (-∞,3) and concave down (3,∞)???
4 years ago

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