Find dy/dx and d^2y/dx^2 and then determine for which t values the curve is concave up:

x=t^3 - 18t y=t^2 - 1

1 answer

dy/dt = 2t

dx/dt = 3t^2 - 18

dy/dx = (3t^2 - 18)/(2t)

d^2y/d^2x = (2t(6t) - 2(3t^2-18))/(4t^2) by quotient rule
> 0 for concave up
the denominator will always be positive for x≠0
so
12t^2 - 6t^2 + 36 >
6t^2 + 6 > 0
t^2 + 6 > 0

which is true for all values of t
Similar Questions
    1. answers icon 3 answers
  1. For f(x)=2(x+5)^3 +7Find and classify the extreme values, determine where the function is increasing and decreasing, where it is
    1. answers icon 3 answers
  2. Consider the function of defined by f(x)=x^3/3-4/xa. Find the X values for the points of inflection. b. Determine the intervals
    1. answers icon 1 answer
    1. answers icon 2 answers
more similar questions