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.
at any point on the curve the vector
v = <x,y> = <t^3+1,t^2>
for values of t.
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
To represent the plane curve y^2 = x^3 + 1 as a vector-valued function, we can convert it into a set of parametric equations.
1. Start by choosing a parameter, typically denoted as t, to help parameterize the curve.
2. Let's assume that t represents the x-coordinate of a point on the curve. We can write x = t.
3. Substitute the value of x into the equation y^2 = x^3 + 1:
y^2 = (t)^3 + 1
4. Take the square root of both sides to solve for y:
y = ±√(t^3 + 1)
5. Since y can be positive or negative, we can write it as two separate equations:
y = √(t^3 + 1)
y = -√(t^3 + 1)
6. Combine the x and y equations to form a vector-valued function:
r(t) = (t, √(t^3 + 1))
r(t) = (t, -√(t^3 + 1))
Now, for any value of t, the vector r(t) will give you a point on the curve y^2 = x^3 + 1. You can choose different values of t to find corresponding points on the curve.