A curve has parametric equations: x = 3t +8, y = t^3 - 5t^2 + 7t. Find the co-ordinates of the stationary points.
First differentiate: dx/dt = 3, dy/dt = 3t2 - 10t + 7
Using the chain rule: dy/dx = dy/dt * dt/dx = (3t2 - 10t + 7)/3
At stationary points, the gradient is equal to zero: 3t2 - 10t + 7 = 0
Solve for t using the quadratic formula and substituting these values: a = 3, b = -10, c = 7
Solutions: t = 7/3 and 1
For the co-ordinates, substitute the values of t into the parametric equations: (15, 49/27) and (11, 3)
RB
Related Maths A Level answers
All answers ▸Find the surface area of a hand held fan (modeled with negligible depth) with radius 8 cm and a 60 degree angle at the centre
What is the gradient of the curve y = 2x^3 at the point (2,2)?
How do I integrate by parts?
Find the equation of the tangent to the curve y = (2x -3)^3 at the point (1, - 1), giving your answer in the form y = mx + c.