at any point on the curve the vector
v = <x,y> = <t^3+1,t^2>
for values of t.
Consider the plane curve y^2=x^3+1. Represent the curve as a vector-valued function.
No idea how to even begin for this one. never had to change anything with a cube in it to be a vector valued function or set of parametric equations before.
3 answers
Afraid that doesn't work Steve, since, solving x=t^3+1 for t gives (x-1)^(1/3) = t so for y=t^2 that's sqrt(y)=(x-1)^(1/3)... or how i actually saw it was wrong. graphed both equations. graphs didn't match up.
Wow, just answered my own question. For anyone else that might have trouble with this. let y = t than solve the equation t^2=x^3+1 for x to get x=(t^2-1)^(1/3) and so the vector valued function is V = (t^2-1)^(1/3)i + tj