Tangent Planes

Math 55 - Elementray Analysis III

Institute of MathematicsUniversity of the Philippines

Diliman

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Recall

For f(x, y), the gradient of f is

f(x, y) = fx(x, y), fy(x, y)

and the directional derivative of f at (x0, y0) in the directionof the unit vector u = a, b is

Duf(x0, y0) = limh0

f(x0 + ha, yo + hb) f(x0, y0)h

= f(x0, y0) u

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Recall (from Math 54)

R For a function w = F (x, y, z), the surface F (x, y, z) = k, forsome constant k, is called a level surface of the givenfunction.

R The equation of a plane with normal vector N = a, b, c,containing the point P (x0, y0, z0) is (Point-Normal Form):

a(x x0) + b(y y0) + c(z z0) = 0R A set of equations for the line parallel to the vector

V = a, b, c, passing through P (x0, y0, z0) isSymmetric:

x x0a

=y y0b

=z z0c

Parametric: x = x0 + at, y = y0 + bt, z = z0 + ct

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Tangent Planes to Level Surfaces

Theorem

Let S be a level surface of F (x, y, z) containing the pointP (x0, y0, z0). The gradient vector F (x0, y0, z0) is orthogonalto S at P .

Proof. Suppose S has the equation F (x, y, z) = k and C is a curveon S passing through P . Let C be given by the vector functionR(t) = x(t), y(t), z(t), such that R(t0) = x(t0), y(t0), z(t0). ThenF (x(t), y(t), z(t)) = k. By Chain Rule,

F

x

dx

dt+F

y

dy

dt+F

z

dz

dt= 0

Fx, Fy, Fz x(t), y(t), z(t) = 0F (x, y, z) R(t) = 0

F (x0, y0, z0) R(t0) = 0Since R(t) is tangent to C at P , F (x0, y0, z0) is orthogonal to C.Because C is arbitrary, F (x0, y0, z0) is orthogonal to S at P .

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Tangent Planes to Level Surfaces

If F (x0, y0, z0) 6= 0, then it isorthogonal to the tangentplane to S at P ,

i.e., F (x0, y0, z0) = Fx(x0, y0, z0), Fy(x0, y0, z0), Fz(x0, y0, z0)is the normal vector to the tangent plane to F (x, y, z) = k atthe point P (x0, y0, z0).

By the Point-Normal Form, the equation of the tangent planeto the surface F (x, y, z) = k at P (x0, y0, z0) is

Fx(x0, y0, z0)(xx0)+Fy(x0, y0, z0)(yy0)+Fz(x0, y0, z0)(zz0) = 0

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Example

Example

Find the equation of the tangent plane to the ellipsoid

x2 + 2y2 + 3z2 = 6

at the point (1,1, 1).

Solution. Let F (x, y, z) = x2 + 2y2 + 3z2 (k = 6) andtherefore we have

Fx(x, y, z) = 2x Fx(1,1, 1) = 2Fy(x, y, z) = 4y Fy(1,1, 1) = 4Fz(x, y, z) = 6z Fz(1,1, 1) = 6

So the equation of the tangent plane is

2(x 1) 4(y + 1) + 6(z 1) = 02x 4y + 6z 12 = 0

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Tangent Planes

If the equation of the surface can be written in the formz = f(x, y), then f(x, y) z = 0.

Let F (x, y, z) = f(x, y) z. Therefore,

Fx(x, y, z) = fx(x, y), Fy(x, y, z) = fy(x, y), Fz(x, y, z) = 1

Hence the equation of the tangent plane at (x0, y0, z0) is

fx(x0, y0)(x x0) + fy(x0, y0)(y y0) (z z0) = 0fx(x0, y0)(x x0) + fy(x0, y0)(y y0) + z0 = z

fx(x0, y0)(x x0) + fy(x0, y0)(y y0) + f(x0, y0) = z

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Normal Lines

Definition

The normal line to a surface S at a point P is the line passingthrough P perpendicular to the tangent plane to S at P .

The direction of the normal line is therefore given by a normalvector of the tangent plane i.e., the gradient vectorF (x0, y0, z0).

So a set of symmetric equations for the normal line is

x x0Fx(x0, y0, z0)

=y y0

Fy(x0, y0, z0)=

z z0Fz(x0, y0, z0)

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Example

Example

Find symmetric equations for the normal line to the surface

z = x2 + 2y2

at the point (2, 1, 6).

Solution. Let F (x, y, z) = x2 + 2y2 z (k = 0).Fx(x, y, z) = 2x Fx(2, 1, 6) = 4Fy(x, y, z) = 4y Fy(2, 1, 6) = 4

Fz(x, y, z) = 1 Fx(2, 1, 6) = 1

Hence the symmeteric equations for the normal line is

x + 2

4 =y 1

4=z 61

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Gradient vector, revisited

What we know about the gradient vector so far:

R F gives the direction of the fastest increase of FR F (x0, y0, z0) is orthogonal to a level surface of F through

P (x0, y0, z0)

On a level surface S, moving away from P in the perpendiculardirection, we get the maximum change in F .

If f is a function of x and y and P (x0, y0) is in the domain of f ,it can be shown that f(x0, y0) is perpendicular to the levelcurve f(x, y) = k containing P .

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Exercises

1 Find the equation of the tangent plane to the following surfacesat the given point.

a. f(x, y) = ln(x 2y); (3, 1)b. x2 2y2 + z = 2xy; (1, 2, 3)c. z + 1 = xey cos z; (1, 0, 0)

2 Find symmetric and parametric equations for the normal line tothe following surfaces at the given point.

a. z + 1 = xey cos z; (1, 0, 0) b. yz = ln(x + z); (0, 0, 1)

3 At what point on the graph of y = x2 + z2 is the tangent planeparallel to x + 2y + 3z = 1?

4 Find parametric equations for the tangent line to the curve ofintersection of the paraboloid z = x2 + y2 and the ellipsoid4x2 + y2 + z2 = 9 at (1, 1, 2)

5 Show that the equation of the tangent plane to the ellipsoidx2

a2+y2

b2+z2

c2= 1 at (x0, y0, z0) can be written as

xx0a2

+yy0b2

+zz0c2

= 1.

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References

1 Stewart, J., Calculus, Early Transcendentals, 6 ed., ThomsonBrooks/Cole, 2008

2 Dawkins, P., Calculus 3, online notes available athttp://tutorial.math.lamar.edu/

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