Tangent Planes and Normal Lines

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Tangent Plane and Normal Line to a Surface

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Calculus I

If we zoom in on a point on a smooth curve (one described by a differentiable function), the curve looks more and more like a tangent line at that point. Once we have the tangent line at a point, it can be used to approximate function values and to estimate changes in the dependent variable.

Now we see that differentiability at a point implies the existence of a tangent plane at that point.

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Calculus I

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Calculus III

Consider a smooth surface described by a differentiable function f and focus on a single point on the surface. As we zoom in on that point, the surface appears more and more like a plane.

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Calculus III

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Tangent Planes

Recall that a surface in R3 may be defined in at least two different ways:

1.Explicitly in the form z = f(x, y) or

2.Implicitly in the form F(x, y, z) = 0

Tangent Planes for F(x, y, z) = 0

To find an equation of the tangent plane, consider a smooth curve C: r(t) = <x(t), y(t), z(t)> that lies on the surface F(x, y, z) = 0.

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Since the points of C lie on the surface, we have F(x(t), y(t), z(t)) = 0. Differentiating both sides of this equation with respect to t, a useful relationship emerges. The derivative of the right side is 0. The Chain Rule applied to the left side yields……

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Therefore, and at any point on the curve, the tangent vector r ‘(t) is orthogonal to the gradient.

Now fix a point P0(a, b, c) on the surface, assume that

and let C be any smooth curve on the surface passing through P0. We have shown that the vector tangent to C is orthogonal to at P0. Since this argument applies to all smooth curves on the surface passing through P0, the tangent vectors for all these curves (with their tails at P0) are orthogonal to , and thus they all lie in the same plane.

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This plane is called the tangent plane to P0. We can easily find an equation of the tangent plane since we know both a point on the plane P0(a, b, c) and a normal vector

an equation is simply……

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Example 1

Consider the ellipsoid

Find the equation of the plane tangent to the ellipsoid at (0, 4, 3/5).

Notice that the equation of the ellipsoid is written in implicit form, F(x, y, z) = 0. The gradient of F is……

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Evaluated at (0, 4, 3/5), we have……

An equation of the tangent plane at this point is……

or 4y + 15z = 25. The equation does not involve x, so the tangent plane is parallel to the x-axis.

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At what points on the ellipsoid is the tangent plane horizontal?

A horizontal plane has a normal vector of the form <0, 0, c>, where c ≠ 0. A plane tangent to the ellipsoid has a normal vector……

Therefore, the ellipsoid has a horizontal tangent plane when Fx = (2x)/9 = 0 and Fy = (2y)/25 = 0, or when x = 0 and y = 0. Substituting these values into the original equation for the ellipsoid, we find that horizontal planes occur at (0, 0, 1) and (0, 0, -1).

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Observation

The surface F(x, y, z) = 0 is a level surface of the function w = F(x, y, z) (corresponding to w = 0). At any point on that surface, the tangent plane has a normal vector

Therefore, the gradient is orthogonal to the level surface F(x, y, z) = 0 at all points of the domain at which F is differentiable.

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Tangent Planes for z = f(x, y)

Surfaces in R3 are often defined explicitly in the form z = f(x, y). In this situation, the equation of the tangent plane is a special case of the general equation just derived. The equation z = f(x, y) is written F(x, y, z) = z – f(x, y) = 0, and the gradient of F at the point (a, b, f(a, b)) is……

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Proceeding as before, an equation of the plane tangent to the surface z = f(x, y) at the point (a, b, f(a, b)) is……

After some rearranging, we obtain an equation of the tangent plane.

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Example 2

Find an equation of the plane tangent to the paraboloid z = f(x, y) = 32 – 3x2 – 4y2 at (2, 1, 16).

The partial derivatives are fx = -6x and fy = -8y. Evaluating the partial derivatives at (2, 1), we have……

fx(2, 1) = -12 and fy(2, 1) = -8

Therefore, an equation of the tangent plane is……

z = fx(a, b)(x – a) + fy(a, b)(y – b) + f(a, b)

= -12(x – 2) – 8(y – 1) + 16

= -12x – 8y + 48

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Example 3

Find a set of symmetric equations for the normal line at the point (-2, 1, -3) to the ellipsoid

The ellipsoid is the level surface (with k = 3) of the function

The partial derivatives are……

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Evaluated at (-2, 1, -3)……

The symmetric equations are……

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The Angle of Inclination of a Plane

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The Angle of Inclination of a Plane

The gradient F(x, y, z) is used to determine the angle of inclination of the tangent plane to a surface.

The angle of inclination of a plane is defined as the angle (0 2) between the given plane and the xy-plane, as shown below. (The angle of inclination of a horizontal plane is defined as zero.)

The angle of inclination

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Because the vector k is normal to the xy-plane, you can use the formula for the cosine of the angle between two planes to conclude that the angle of inclination of a plane with normal vector n is given by

The Angle of Inclination of a Plane

Angle of inclination of a plane

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Example 4 – Finding the Angle of Inclination of a Tangent Plane

Find the angle of inclination of the tangent plane to the ellipsoid given by

at the point (2, 2, 1).

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Because F(2, 2, 1) is normal to the tangent plane and k is

normal to the xy-plane, it follows that the angle of

inclination of the tangent plane is given by

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which implies that

as shown.

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A Comparison of the Gradients f (x, y) and F(x, y, z)

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A Comparison of the Gradients f (x, y) and F (x, y, z)

This section concludes with a comparison of the gradients f (x, y) and F(x, y, z).

You know that the gradient of a function f of two variables is normal to the level curves of f.

Specifically, if f is differentiable at (x0, y0) and f(x0, y0) ≠ 0,

then f(x0, y0) is normal to the level curve through (x0, y0).

Having developed normal lines to surfaces, you can now extend this result to a function of three variables.

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A Comparison of the Gradients f (x, y) and F (x, y, z)