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 the concavity of the curve parametrically defined by $x = 6t^2$ and $y = t^3 - 3t$, we need to find the second derivative of $y$ with respect to $x$ ($d^2y/dx^2$) and analyze its sign. First, find $dy/dx$ by dividing $dy/dt$ by $dx/dt$. $dy/dt = 3t^2 - 3$ and $dx/dt = 12t$. Thus, $dy/dx = (3t^2 - 3)/(12t) = (t^2 - 1)/(4t)$. Then, find $d^2y/dx^2$ by differentiating $dy/dx$ with respect to $t$ and dividing by $dx/dt$. Simplifying this gives a function where the sign depends on the value of $t$. The curve is concave upward where $d^2y/dx^2 > 0$ and concave downward where $d^2y/dx^2 < 0$. The analysis shows that the curve is concave downward for $t < 0$ and concave upward for $t > 0$, matching option A.

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 $x = 6 t ^ { 2 } , y = t ^ { 3 } - 3 t \text { is concave }$