# Tangent Plane to the Surface

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• 0:04 Describing a Support Surface
• 0:47 Finding a Tangent Line
• 3:26 Tangent Plane to a Surface
• 4:51 Lesson Summary

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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

In this lesson, we extend the idea of finding the equation of a tangent line to a very nice procedure for finding the equation of a plane tangent to a surface. Applications include the design of support surfaces for three-dimensional objects.

## Describing a Support Surface

If you were allowed only one orientation for a sled sliding down a snowy hill, you might want to consider a plane tangent to the surface of this hill. This approach requires some knowledge of the three-dimensional equation describing the hill's surface or at least an approximation to this equation. But the idea is cool all by itself. Imagine sliding down the plane tangent to a surface!

In this lesson, we start by reviewing how to find the tangent line to a curve. Then, we cast this approach in terms of a derivative operator called the gradient. The gradient involves partial derivatives and unit vectors. From the two-dimensional scenario of tangent line to a curve, we will naturally stretch our analytic powers to the three-dimensional problem of finding the equation of the plane tangent to a surface.

## Finding a Tangent Line

We start with a nice curve called a parabola. This parabola looks somewhat like a hill:

And yes, we know the equation for this ''hill'':

Your sled is the red dot located on the curve:

From calculus, we know the derivative, y', is the slope of the tangent line at a point. The derivative of the equation y = - x2/2 is -2x/2 or just -x:

And at the point (1, 0.5), the value of x is 1. Substituting this value for x gives:

We will use the point-slope form for the equation of the tangent line:

Okay, m is the slope of the tangent line at the point (1, 0.5); m = -1. The other two constants in the equation are xo and yo. These constants are just the coordinates of the point: xo = 1 and yo = 0.5 = 1/2. Substituting for m, xo and yo:

We can move everything over to the left-hand side:

The ''1''s in front of (x - 1) and (y - 1/2) will be useful later.

Using some algebra on this equation we get:

This equation is the tangent line to the curve! Leaving the variables on the left-hand side is for a reason.`

Now, for a powerful way to do this type of problem. We start with the same y = -x2/2 + 1 equation but write all the variables on the left-hand side. The -x2/2 on the right-hand side becomes x2/2 on the left-hand side:

We let f be a function equal to the left-hand side:

The gradient of f is the partial derivative of f' with respect to x in the i direction plus the partial derivative of f with respect to y in the j direction:

The upside-down triangle next to f is shorthand for ''gradient of f.''

The partial derivatives evaluate to 2x/2 and 1:

But 2x/2 is just x:

Next, we evaluate the gradient of f at the point (1, 0.5). The gradient only has x as the variable:

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