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 ▸If 1/(x(a-x)) is equivalent to B(1/x + 1/(a-x)), Express 'B' in terms of 'a'
The point A lies on the curve with equation y=x^0.5. The tangent to this curve at A is parallel to the line 3y-2x=1 . Find an equation of this tangent at A. [5 marks]
How do i differentiate the equation y = x^2 + 6x + 2 with respect to x.
Differentiate 3x^2 with respect to x