Find parametric equations for the line tangent to the curve of intersection of the surfaces at t

step 1 For this problem, we're given two surfaces. The first surface I'm going to refer to as F, and th

Find parametric equations for the line tangent to the curve...

Key Concepts

Cross Product

The cross product of two vectors in three-dimensional space yields a vector that is orthogonal to both. When applied to the gradient vectors of the two surfaces, the resulting cross product gives a direction vector that lies tangent to the curve defined by their intersection. This fundamental operation is instrumental in deriving the parametric equations of the tangent line.

Parametric Equations of a Line

Parametric equations express a line in terms of a parameter, typically denoted as t, by specifying its coordinates as functions of this parameter. In the context of finding the tangent line, once the direction vector (obtained from the cross product of the gradients) and a point on the curve are known, the line can be expressed in a clear and concise form. These equations are essential for representing the line in a way that can be easily manipulated and analyzed in further applications.

Curve of Intersection

A curve of intersection is formed when two surfaces intersect in three-dimensional space. The set of points that lie on both surfaces defines a curve, and understanding this intersection is fundamental in multivariable calculus and differential geometry. This concept allows one to study the properties of curves that are defined implicitly by the intersection of two or more surfaces.

Gradient Vector

The gradient vector of a scalar field represents the direction of the steepest ascent and is perpendicular (normal) to level surfaces. In the context of finding a tangent to the curve of intersection, the gradients of the defining surfaces are used to determine the normals to the surfaces at a given point. This is crucial because the tangent vector to the curve will be orthogonal to both of these normal vectors.

Transcript

00:03 For this problem, we're given two surfaces.

00:06 The first surface i'm going to refer to as f, and that is x cubed plus 3x squared, y squared, plus y cubed, plus 4xy, minus z squared, equals zero.

00:31 The second surface is x squared plus y squared plus z squared equals 11, and i'll be referring to that as function g.

00:49 And those two surfaces meet somewhere and form a curve.

00:53 And we want to find the equation for the tangent line at the point 1 .1.

01:03 So in order to do that, we have to find the gradient vector for each.

01:07 Of those functions and then take the cross product of that to find the direction that the tangent line is going to be going.

01:16 When we do the gradient of a function, we're doing the partial derivative of the function with respect to each variable times the vector component going along that variable's axis.

01:30 And we have to evaluate those at our point.

01:34 And so we've got the gradient of f being evaluated at the point 1113.

01:42 So partial derivative with respect to x is going to be kind of long because everything that has an x has to be derived and any other variables are treated as constant.

01:54 So we've got three terms with x's enough.

01:56 So deriving the x cubed, that's 3x squared from the power rule, plus deriving the next term, the y squared is treated as a constant just like the three, so that'll be two times the three, then to the one last power, so that'll be six x, y squared.

02:18 The y -quib will be a constant, so that's zero.

02:21 And then the plus four -x -y has a derivative with respect to x of plus four -y, and the z -squared treated as a constant, that derivative is zero.

02:31 So all that is just the i part of this.

02:36 So there's my partial derivative with respect times vector i.

02:40 Now we have to do it with respect to y.

02:44 And again, three terms that have ys in them, so this is going to be a little longer.

02:49 The x cubed derivative is zero.

02:52 The three x squared y squared is 6x squared y.

02:59 The y cubed, derivative of that is going to be 3y squared.

03:05 And then the derivative of the 4xy will be plus 4x...