Asked by Anonymous
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.
I get a slope of 7.5.
Answers
Answered by
Steve
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.
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.
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