Question

Find the equation of the line that is tangent to the curve x(t) = t^2+1 and y(t)=t^3-1 at the point (5,7).

I get a slope of 7.5.

Answer 1

dy/dt = 3t^2
dx/dt = 2t
at(5,7), t=2, so dy/dx = (dy/dt)/(dx/dt) = 3t/2 = 6/2 = 3
now use that and the point-slope form to find the tangent line.