Partial derivatives and the calculus of surfaces
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The tangent plane
To solve this problem we need to realise that for the surfaces we are dealing with here - surfaces without surprising "wrinkles", or other odd features, like cuts or tears, the set of tangents to a given point form a tangent plane - that is, they all lie in a plane that just touches the surface at the given point. [Diagram goes here - download the original to see it.] So, any vector that is tangent to the surface will be a linear combination of vectors which are tangent to the surface along the direction of each of the coordinates. Here, the vector r is a linear combination of the vectors v and w which are vectors parallel to the x-axis and the y-axis respectively. [Equation goes here - download the original to see it.] The next question to ask, is how do we obtain representations of the vectors at a given point that are parallel to the coordinate axes? We illustrate the solution to this problem by considering our example once again. The function we were considering was [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] This gives the gradient of a vector parallel to the x-axis. In other words, this vector will have no y-component. If its x-component is 1 unit, then its z-component will be 4. So this vector is given by [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] This gives the gradient of a vector parallel to the y-axis, without x-component. If its y-component is 1 unit, then its z-component will be 3. So this vector is given by [Equation goes here - download the original to see it.] Any vector, r, lying in the tangent plane to the surface at the point will be a linear combination of these two vectors: [Equation goes here - download the original to see it.] From our study of vector planes, we will be aware that a plane can be defined by a vector equation. The general equation of a plane in vector form is [Equation goes here - download the original to see it.] Where is the position vector of any point lying in the plane and is any vector that is perpendicular to the plane; k is a real number. [Diagram goes here - download the original to see it.] A vector perpendicular to the surface at a given point, will be perpendicular to any two vectors lying in the tangent plane. [Diagram goes here - download the original to see it.] This perpendicular vector can be found by finding the cross product of any two independent vectors lying in the tangent plane. Of course, we are usually starting with the vectors that are parallel to the x and y-axes respectively. In our example, the normal to the tangent plane at the point [Equation goes here - download the original to see it.] is given by [Equation goes here - download the original to see it.] [Diagram goes here - download the original to see it.] Of course, if u is a vector normal (perpendicular) to a surface, then -u is also normal to that surface, and so is any scalar multiple of u. [Diagram goes here - download the original to see it.] In our example the vector equation of the normal at the point is given by [Equation goes here - download the original to see it.] The tangent plane at this point is (we calculate the scalar-product of these vectors:) [Equation goes here - download the original to see it.]
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Contents of
Partial derivatives and the calculus of surfaces
1 Multiple variable calculus. Functions of more than one variable
2 Two-dimensional surfaces embedded in three-dimensional space
3 The tangent plane
4 Stationary points
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