Parametric Equations
x = 9s^2 - 1
y = 1 - 8s^3
find dy/dx in terms of s
find the coordinates of stationary points
dy/dx = (dy/ds)/(dx/ds)
= (-24s^2)/(18s)
= -4s/3
naturally, dy/dx=0 at s = 0
see the graph
http://www.wolframalpha.com/input/?i=plot+x+%3D+9s^2+-+1,++y+%3D+1+-+8s^3
To find dy/dx in terms of s, we need to take the derivative of y with respect to x and express it in terms of s.
To do this, we'll use the chain rule. The derivative of y with respect to x can be written as dy/dx. Using the chain rule, we have:
dy/dx = (dy/ds) / (dx/ds)
First, let's find dx/ds:
dx/ds = d/ds (9s^2 - 1)
= 18s
Now, let's find dy/ds:
dy/ds = d/ds (1 - 8s^3)
= -24s^2
Finally, we can find dy/dx:
dy/dx = (dy/ds) / (dx/ds)
= (-24s^2) / (18s)
= -4s/3
So, dy/dx in terms of s is -4s/3.
Now, let's find the coordinates of the stationary points. A stationary point occurs when dy/dx = 0. Therefore, we need to solve the equation -4s/3 = 0.
Setting -4s/3 = 0, we find that s = 0.
To find the corresponding coordinates, substitute the value of s into the parametric equations:
x = 9s^2 - 1
= 9(0)^2 - 1
= -1
y = 1 - 8s^3
= 1 - 8(0)^3
= 1
Therefore, the coordinates of the stationary point are (-1, 1).