Determine the concavity of the curve with parametric equations x = t2, y = t3 - 3t at t = 2. justify your answer.

Question

asked Jun 23, 2023 175k views

1 Answer

Jamie Marshall · Answer 1 · 2023-06-28T23:46:39+0000

Use the chain rule to find the first and second derivatives of
with respect to
.

$(dy)/(dx) = (dy)/(dt)\cdot(dt)/(dx) = (dy/dt)/(dx/dt)$

Differentiate the parametric equations with respect to
.

$x(t) = t^2 \implies (dx)/(dt) = 2t$

$y(t) = t^3 - 3t \implies (dy)/(dt) = 3t^2 - 3$

$\implies (dy)/(dx) = (3t^2 - 3)/(2t)$

Let
f(t) = (dy)/(dx) . By the chain rule,

$(d^2y)/(dx^2) = (df)/(dx) = (df)/(dt) \cdot (dt)/(dx) = \frac[df/dt}{dx/dt}$

Differentiate
.

$f(t) = (3t^2 - 3)/(2t) = \frac32t - \frac3{2t} \implies (df)/(dt) = (d^2y)/(dx^2) = \frac32 + \frac3{2t^2}$

At
t=2 , the second derivative has a positive sign, so the curve is concave upward at this point.