Determine the t intervals on which the curve $x = 6 t ^ { 2 } , y = t ^ { 3 } - 3 t \text { is concave }$ downward or concave upward.
A) concave downward: $- \infty

Question

Determine the t intervals on which the curve $x = 6 t ^ { 2 } , y = t ^ { 3 } - 3 t \text { is concave }$ downward or concave upward. 
A) concave downward: $- \infty < t < 0$; concave upward: $0 < t < \infty$
B) concave downward: $0 < t < \infty$; concave upward: $- \infty < t < 0$
C) concave downward: $- 6 < t < 3$; concave upward: $0 < t < 3$
D) concave downward: $- 6 < t < 0$; concave upward: $0 < t < \infty$
E) concave downward: $- \infty < t < 0$; concave upward: $0 < t < 3$

Quizplus · Accepted Answer

To determine concavity, we need to find the second derivative of $y$ with respect to $x$, denoted as $\frac{d^2y}{dx^2}$. First, find $\frac{dy}{dt}$ and $\frac{dx}{dt}$, then use the chain rule to find $\frac{d^2y}{dx^2}$. Given $x = 6t^2$ and $y = t^3 - 3t$, we have $\frac{dx}{dt} = 12t$ and $\frac{dy}{dt} = 3t^2 - 3$. To find $\frac{d^2y}{dx^2}$, we also need $\frac{d^2y}{dt^2} = 6t$ and $\frac{d^2x}{dt^2} = 12$. Then, $\frac{d^2y}{dx^2} = \frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}} = \frac{6t}{12} = \frac{t}{2}$. The curve is concave upward when $\frac{d^2y}{dx^2} > 0$ (i.e., $t > 0$) and concave downward when $\frac{d^2y}{dx^2} < 0$ (i.e., $t < 0$).

Determine the T Intervals on Which the Curve x=6t2,y=t3−3t is concave x = 6 t ^ { 2 } , y = t ^ { 3 } - 3 t \text { is concave }x=6t2,y=t3−3t is concave

Determine the T Intervals on Which the Curve $x = 6 t ^ { 2 } , y = t ^ { 3 } - 3 t \text { is concave }$