On what intervals of t is the curve described by the given parametric equation concave up? Concave down?
x=t^2; y=t^(3)+3t
To determine when the curve described by the given parametric equation is concave up or concave down, we need to analyze the second derivative of the equation with respect to t.
First, let's find the first derivative of x and y with respect to t:
x' = 2t
y' = 3t^2 + 3
Next, let's find the second derivative of x and y with respect to t:
x'' = 2
y'' = 6t
Since we are interested in the intervals of t where the curve is concave up or down, we need to examine the sign of y''.
When y'' > 0, the curve is concave up.
When y'' < 0, the curve is concave down.
To find the intervals where the curve is concave up or down, we set y'' equal to zero and solve for t:
6t = 0
t = 0
Now we can create a number line to understand the concavity:
(-∞) ---(concave down)--- (0) ---(concave up)--- (+∞)
From the number line, we can see that the curve is concave down for t < 0 and concave up for t > 0.
Therefore, the curve described by the given parametric equation is concave down for t < 0 and concave up for t > 0.