1-2 Determine whether the points P and Q lie on the given surface. 𝐫(u, v)=⟨ u+v, u^2-v, u+

1-2 Determine whether the points P and Q lie on the given...

Key Concepts

Substitution Method for Verification

The substitution method is a technique used to determine if a given point lies on a parameterized surface. By substituting the coordinates of the point into the vector function and setting the resulting expressions equal to the point's coordinates, one obtains a system of equations in the parameters.

Systems of Equations in Parameter Space

When verifying whether a point is on a surface, a system of equations is formed by equating each component of the parameterized function with the corresponding coordinate of the point. Solving this system confirms whether there exist parameters that map to the given point, thereby establishing its presence on the surface.

Surface Parameterization

Surface parameterization involves expressing a surface in space as a vector function of two parameters. This function maps each pair of parameters to a unique point on the surface, enabling easier manipulation and analysis of the surface’s properties including curvature, tangent planes, and area.

Transcript

00:01 For this problem, we are asked to determine whether the points p and q, being 3 -1 -5, and negative -1 -3 -4, lie on the surface defined by r of u -v equals u -plus v, u -squared minus v, u -plus v squared.

00:14 So to begin, we want to solve for u and v such that the different components of r equal the components of each point.

00:23 So we have u -plus v must be equal to 3.

00:25 We have that u squared minus v must be equal to negative one and we have that u plus v squared must be equal to five so first we can rearrange that first equation writing u equals 3 minus v then we can substitute that into the second equation so we'd have uh 3 minus v actually let me change how i'm writing this here uh let's see actually what i'll do is plug this into the third equation instead.

01:00 So we'd have, still we have u equals 3 minus v.

01:04 So we'd have 3 minus v plus v squared equals 5.

01:09 So we can write this then as v squared minus v plus 3 minus 5.

01:16 So that would be negative 2 is equal to 0.

01:24 Now we'll have two solutions to this.

01:26 One would be v equals negative 1 and the other is v equals.

01:30 2 from factoring this as v plus 1 times v minus 2.

01:37 So if v equals 2, then we have that u must be equal to 3 minus 2, so u equals 1.

01:43 Now we can substitute that into the equation that we didn't use.

01:46 U squared minus v equals negative 1.

01:49 So we'd have u squared would be 1, minus v would be well then minus 2, and 1 minus 2 does indeed equal negative 1.

01:58 So point p does lie on the curve.