A curve is defined by the parametric equations x = 2t and y = 4t^2 + t. Find the gradient of the curve when t = 4
the gradient of the curve = dy/dx
and dy/dx = (dy/dt)(dt/dx)
dy/dt = 8t + 1
dx/dt = 2 therefore dt/dx = 1/2
dy/dx as above = (8t + 1) * 1/2 = (8t + 1)/2
where t = 4, dy/dx = (8*4 + 1)/2 = (32 + 1)/2 = 33/2
AB
Related Maths A Level answers
All answers ▸The line L has equation y=5-2x. Find an equation of the line perpendicular to L, which passes through the point P (3,-1).
What is a stationary point and how do I find where they occur and distinguish between them?
Find the area between the curve y = 8 + 2x - x^2 and the line y = 8 - 2x.
Find (dy/dx) of x^3 - x + y^3 = 6 + 2y^2 in terms of x and y