Find the interval for which x=t^2/2-3t+5 and y=t^2+4t-1 is concave up and concave down
So, concave up (-â,3) and concave down (3,â)???
To find the intervals for which the given functions are concave up and concave down, we need to calculate the second derivative of each function and then analyze the sign of the second derivative. Let's begin by finding the second derivative of x and y with respect to t.
Given:
x = t^2/2 - 3t + 5
y = t^2 + 4t - 1
To find the second derivative of x, we need to differentiate it twice with respect to t.
First derivative of x:
x' = d(x)/dt = d(t^2/2)/dt - d(3t)/dt + d(5)/dt
= t - 3
Second derivative of x:
x'' = d^2(x)/dt^2 = d(t - 3)/dt
= 1
As we can see, the second derivative of x is a constant, 1. Therefore, x is neither concave up nor concave down. It is a linear function.
Now, let's find the second derivative of y.
First derivative of y:
y' = d(y)/dt = d(t^2)/dt + d(4t)/dt - d(1)/dt
= 2t + 4
Second derivative of y:
y'' = d^2(y)/dt^2 = d(2t + 4)/dt
= 2
The second derivative of y is a constant, 2. Thus, y is both concave up and concave down over its entire domain.
In summary:
- The function x = t^2/2 - 3t + 5 is a linear function, so it is neither concave up nor concave down.
- The function y = t^2 + 4t - 1 is both concave up and concave down over its entire domain.
you need to check y" for positive or negative
y' = (dy/dt) / (dx/dt) = (2t+4)/(t-3)
y" = dy'/dx = d/dt (2t+4)/(t-3) / (dx/dt) = -10/(t-3)^3
So it looks like it is concave
up for t<3
down for t>3