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

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 third vector that is orthogonal to both. In the context of intersecting surfaces, the tangent vector to the intersection curve at a point can be found by taking the cross product of the gradients (normal vectors) of the surfaces at that point, ensuring the tangent direction lies within both tangent planes.

Gradient and Normal Vectors

The gradient of a scalar function defines a vector that is perpendicular (normal) to its level surface. For surfaces defined implicitly, the gradients at a given point can be used to determine the direction vectors for tangent lines along the curves of intersection by taking into account how the surfaces change locally.

Curve of Intersection of Surfaces

The curve of intersection is the set of points that simultaneously satisfy the equations of two surfaces. When two surfaces intersect, the resulting curve can be analyzed by considering the geometric properties of each surface, and its tangent direction at any point lies in the intersection of the tangent planes to both surfaces.

Parametric Equations

Parametric equations represent a line (or curve) in space by expressing each coordinate as a function of a single parameter, typically t. This form is useful in vector calculus and geometry because it clearly shows a fixed point on the line and a direction vector that indicates the line’s orientation.

Tangent Line to a Curve

The tangent line to a curve at a point is the line that best approximates the curve near that point. Geometrically, it is the limit position of secant lines through the point, and its direction vector is given by the derivative (or velocity) of the curve at that point.

Transcript

00:03 For this problem we have two functions representing surfaces in three dimensions.

00:09 We have x plus y squared plus 2 z equals 4.

00:16 I'm going to refer to that as function f.

00:20 And then we have the second function which i'm going to refer to as function g, which is x equals 1.

00:27 And these two surfaces meet along some sort of a curve, and one of the points on that curve is the point one, one, one.

00:38 And we want to find the equation for the tangent line to that curve at our point 111.

00:46 So we need to find the gradient vector for each of those functions at that point.

00:52 So we will find the gradient of f, and then evaluate that at our point.

00:57 Point 111.

01:01 And we find the gradient by doing the partial derivative with respect to each variable and taking that times the vector component for that variable.

01:11 So the partial derivative, we are keeping the other variables as constants.

01:17 When we do our partial derivative with respect to x, y and z will be treated as constants.

01:23 So that partial derivative is just 1 times vector i.

01:29 And then we have partial derivative with respect to y, so x and z are treated as constants, and that partial derivative is 2y times vector component j, and then the partial derivative with respect to z gives us 2 times vector component k, and now we have to evaluate that at our point one -one -one -one, so plugging one in for y, we end up with the gradient of vector fxer.

02:00 Is going to be vector i plus two vector j plus two vector k now we have to do the same thing for g which will be really simple because it's a very simple function we've got our gradient of g also evaluated at our point one one one and the partial derivative with respect to x is just one times our vector component i no y no z no z so those partial derivatives are zero, so there's no vector component there.

02:38 And there's no place to plug in our coordinates.

02:41 So the gradient for g is just going to be vector i.

02:49 The vector representing our tangent line is going to be the cross product of those two vectors.

02:59 It has to be perpendicular to both of those at the same time.

03:03 So v is going to be the gradient of f cross product with the gradient of g...

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