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.

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- Jun 15th 2009, 04:06 AMguest12345Differentiation
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. - Jun 15th 2009, 04:17 AMalexmahone
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,

$\displaystyle \frac{d^2y}{dx^2}<0$

$\displaystyle \frac{6}{30t}<0$

$\displaystyle t<0$ - Jun 15th 2009, 04:26 AMguest12345
Thanks, I kinda understand what the questions asking but I don't think I'd be able to answer it in a test lol, didn't realise it was an excellence question, nvm. :]