Help with this question please:
Find the values of t when the function defined by x = 5t^3 + 2 and y = 3t^2 -5t is concave down.
Refer this site: Sign of 2nd derivative, Maths First, Institute of Fundamental Sciences, Massey University
Concave down = Slopes decreasing
$\displaystyle \frac{dx}{dt}=15t^2$
$\displaystyle \frac{d^2x}{dt^2}=30t$
$\displaystyle \frac{dy}{dt}=6t-5$
$\displaystyle \frac{d^2y}{dt^2}=6$
$\displaystyle \frac{d^2y}{dx^2}=\frac{6}{30t}$
Since the slopes are decreasing,
$\displaystyle \frac{d^2y}{dx^2}<0$
$\displaystyle \frac{6}{30t}<0$
$\displaystyle t<0$