$\text { Determine the } t \text { intervals on which the curve } x = 7 t ^ { 2 } , y = \ln t \text { is concave downward }$ or concave upward.
A) concave downward: $- \infty

Question

$\text { Determine the } t \text { intervals on which the curve } x = 7 t ^ { 2 } , y = \ln t \text { is concave downward }$ 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: $- \infty < t < \infty$
D) concave upward: $0 < t < \infty$
E) concave downward: $0 < t < \infty$

Quizplus · Accepted Answer

To determine the concavity of the curve, we need to find the second derivative of y with respect to x. Note that $y = \ln t \implies \dfrac{dy}{dt} = \dfrac{1}{t} \implies \dfrac{d^2y}{dt^2} = -\dfrac{1}{t^2}$ Also, $x = 7t^2 \implies \dfrac{dx}{dt} = 14t \implies \dfrac{d^2x}{dt^2} = 14$ Using the second derivative test for concavity, we can express the concavity of the curve in terms of the sign of $\dfrac{d^2y}{dt^2}\left(\dfrac{d{x}}{dt}\right)^{-2}$ as follows: $\dfrac{d^2y}{dt^2}\left(\dfrac{d{x}}{dt}\right)^{-2} > 0 \implies$ the curve is concave upward. $\dfrac{d^2y}{dt^2}\left(\dfrac{d{x}}{dt}\right)^{-2} < 0 \implies$ the curve is concave downward. Now, since $\dfrac{d{x}}{dt} = 14t$, we have $\left(\dfrac{d{x}}{dt}\right)^{-2} > 0$. Therefore, we only need to consider the sign of $\dfrac{d^2y}{dt^2}$. We have $\dfrac{d^2y}{dt^2} < 0$ for $0 < t < \infty$. Hence, the curve is concave downward for $0 < t < \infty$. Answer choice E is the correct choice.

Determine the t intervals on which the curve x=7t2,y=ln⁡t is concave downward \text { Determine the } t \text { intervals on which the curve } x = 7 t ^ { 2 } , y = \ln t \text { is concave downward } Determine the t intervals on which the curve x=7t2,y=lnt is concave downward

$\text { Determine the } t \text { intervals on which the curve } x = 7 t ^ { 2 } , y = \ln t \text { is concave downward }$