Parametric Equations
x = 9s^2 - 1
y = 1 - 8s^3
find dy/dx in terms of s
find the coordinates of stationary points
To find dy/dx in terms of s, we will differentiate both equations with respect to s and solve for dy/dx.
Given:
x = 9s^2 - 1
y = 1 - 8s^3
Differentiating x with respect to s:
dx/ds = 18s
Differentiating y with respect to s:
dy/ds = -24s^2
Now, we can find dy/dx by dividing dy/ds by dx/ds:
(dy/dx) = (dy/ds) / (dx/ds)
Substituting the values, we get:
(dy/dx) = (-24s^2) / (18s)
(dy/dx) = -4s/3
Therefore, dy/dx in terms of s is -4s/3.
To find the coordinates of the stationary points, we need to find the values of s for which dy/dx is equal to zero.
Setting dy/dx = 0:
-4s/3 = 0
This implies that s = 0.
Now, substitute s = 0 into the parametric equations to find the corresponding coordinates.
When s = 0,
x = 9(0)^2 - 1 = -1
y = 1 - 8(0)^3 = 1
Therefore, the coordinates of the stationary point are (-1, 1).