Question

The function x=3t^(2) y=t^(3)-t is concave upward on the interval (a) (-oo 0) (b) (0 oo) (c) (-oo oo) (d) None of the above​

Answer 1

The function x = 3t² , y = t³ - t is concavely upward on the interval...

(a) (-∞, 0) (b) (0, ∞) (c) (-∞, ∞) (d) none of the above.

solution : here x = 3t²

differentiating x with respect to t,

dx/dt = 6t .....(1)

again, y = t³ - t

differentiating y with respect to t,

dy/dt = 3t² - 1

now dy/dx = (dy/dt)/(dx/dt) = (3t² - 1)/6t = (t/2 - 1/6t)

differentiating both sides we get,

d²y/dx² = d[dy/dx]/dt/(dx/dt)

⇒d²y/dx² = d(t/2 - 1/6t)/dt/6t

⇒d²y/dx² = d(1/2 + 1/6t²)/6t = (3t² + 1)/36t³

curve is concavely upward when d²y/dx² > 0

so, (3t² + 1)/36t³ > 0 ⇒t > 0

therefore the correct option is (b) (0, ∞)