TANGENT PLANES AND NORMAL LINES PDF

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MATH2225

MODULE 4

TANGENT PLANES AND NORMAL LINES Overview The tangent plane and normal line to a surface, as well as the angle of inclination of a plane, will be introduced in this module. Time allotment: 1 week Objectives: Upon completion of this module, you are expected to: 1. find equations of tangent planes and normal lines to surfaces; 2. find the angle of inclination of a plane in space.

PRE-ASSESSMENT 1. Is the process in finding equations of tangent planes to surfaces of equations of two variables the same with the process for single variable?

1. Tangent Plane and Normal Line to a Surface In the case of function of a single variable, the derivative helped us to formalize the concept of tangent and normal to the graph of the function. Similar results hold for functions of several variables, the role of the derivative being played by the gradient. You can represent the surfaces in space primarily by equations of the form 𝒛 = 𝒇(𝒙, 𝒚) Equation of a surface S In the development to follow, however, it is convenient to use the more general representation 𝐹(𝑥, 𝑦, 𝑧 ) = 0.

For a surface 𝑆 given by 𝑧 = 𝑓(𝑥, 𝑦), you can convert to the general form by defining 𝐹 as 𝐹(𝑥, 𝑦, 𝑧 ) = 𝑓(𝑥, 𝑦) − 𝑧. Because 𝑓(𝑥, 𝑦 ) − 𝑧 = 0, you can consider 𝑆 to be the level surface of 𝐹 given by 𝐹(𝑥, 𝑦, 𝑧 ) = 0 alternative equation of surface S

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Example 1. For the function given by

𝐹(𝑥, 𝑦, 𝑧 ) = 𝑥 2 + 𝑦 2 + 𝑧 2 − 4 describe the level surface given 𝐹(𝑥, 𝑦, 𝑧 ) = 0. SOLUTION: The level surface given by 𝐹(𝑥, 𝑦, 𝑧 ) = 0 can be written as 𝑥 2 + 𝑦 2 + 𝑧 2 = 4 which is a sphere of radius 2 whose center is at the origin.

In the process of finding a normal line to a surface, you are also able to solve the problem of finding a tangent plane to the surface. Let 𝑆 be a surface given by

and let 𝑃(𝑥0 , 𝑦0 , 𝑧0 ) be a point on 𝑆.

𝐹(𝑥, 𝑦, 𝑧 ) = 0

Let 𝐶 be a curve on 𝑆 through 𝑃 that is defined by the vector-valued function 𝑟(𝑡) = 𝑥 (𝑡)𝒊 + 𝑦 (𝑡)𝒋 + 𝑧(𝑡)𝒌. Then, for all 𝑡, 𝐹(𝑥 (𝑡)), 𝑦 (𝑡), 𝑧 (𝑡) = 0.

If 𝐹 is differentiable and 𝑥 ′ (𝑡), 𝑦 ′ (𝑡), and 𝑧′(𝑡)all exist, it follows from the chain rule that 0 = 𝐹 ′ (𝑡) = 𝐹𝑥 (𝑥, 𝑦, 𝑧 )𝑥 ′ (𝑡) + 𝐹𝑦 (𝑥, 𝑦, 𝑧 )𝑦 ′ (𝑡) + 𝐹𝑧 (𝑥, 𝑦, 𝑧)𝑧′(𝑡) At (𝑥0 , 𝑦0 , 𝑧0 ), the equivalent vector form is 0 = ∇𝐹 (𝑥0 , 𝑦0 , 𝑧0 ) ⋅ 𝑟 ′ (𝑡0 )

Definitions of Tangent Plane and Normal Line Let 𝐹 be differentiable at the point 𝑃(𝑥0 , 𝑦0 , 𝑧0 ) on the surface 𝑆 given by 𝐹(𝑥, 𝑦, 𝑧 ) = 0 such that ∇𝐹 (𝑥0 , 𝑦0 , 𝑧0 ) ≠ 0. 1. The plane through 𝑃 that is normal to ∇𝐹 (𝑥0 , 𝑦0 , 𝑧0 ) is called the tangent plane to 𝑺 at 𝑷. 2. The line through 𝑃 having the direction of ∇𝐹 (𝑥0 , 𝑦0 , 𝑧0 ) is called the normal line to 𝑺 and 𝑷. THEOREM. EQUATION OF TANGENT PLANE If 𝑓 is differentiable at (𝑥0 , 𝑦0 , 𝑧0 , ), then an equation of the tangent plane to the surface given by 𝐹(𝑥, 𝑦, 𝑧 ) = 0 at (𝑥0 , 𝑦0 , 𝑧0 ) is 𝐹𝑥 (𝑥0 , 𝑦0 , 𝑧0 )(𝑥 − 𝑥0 ) + 𝐹𝑦 (𝑥0 , 𝑦0 , 𝑧0 )(𝑦 − 𝑦0 ) + 𝐹𝑧 (𝑥0 , 𝑦0 , 𝑧0 )(𝑧 − 𝑧0 ) = 0 2

MATH2225

Example 2. Find an equation of the tangent plane to the hyperboloid given by 𝑧 2 − 2𝑥 2 − 2𝑦 2 = 12 at the point (1, −1,4). SOLUTION: Begin by writing the equation of the surface as 𝑧 2 − 2𝑥 2 − 2𝑦 2 − 12 = 0 Then, by considering 𝐹(𝑥, 𝑦, 𝑧 ) = 𝑧 2 − 2𝑥 2 − 2𝑦 2 − 12,

We can get the partial derivative with respect to 𝑥, 𝑦 and 𝑧. 𝐹𝑥 (𝑥, 𝑦, 𝑧) = −4𝑥, 𝐹𝑦 (𝑥, 𝑦, 𝑧) = −4𝑦 and

𝐹𝑧 (𝑥, 𝑦, 𝑧) = 2𝑧.

At the point (1, −1,4) [substitute values of 𝑥 = 1, 𝑦 = −1 and 𝑧 = 1 to 𝐹𝑥 , 𝐹𝑦 and 𝐹𝑧 ], the partial derivatives are 𝐹𝑥 (1, −1,4) = −4, 𝐹𝑦 (1, −1,4) = 4 and 𝐹𝑧 (1, −1,4) = 8.

So, an equation of the tangent line at (1, −1,4) is 𝐹𝑥 (𝑥0 , 𝑦0 , 𝑧0 )(𝑥 − 𝑥0 ) + 𝐹𝑦 (𝑥0 , 𝑦0 , 𝑧0 )(𝑦 − 𝑦0 ) + 𝐹𝑧 (𝑥0 , 𝑦0 , 𝑧0 )(𝑧 − 𝑧0 ) = 0 −4(𝑥 − 1) + 4(𝑦 + 1) + 8(𝑧 − 4) = 0 −4𝑥 + 4 + 4𝑦 + 4 + 8𝑧 − 32 = 0 −4𝑥 + 4𝑦 + 8𝑧 − 24 = 0 −𝑥 + 𝑦 + 2𝑧 − 6 = 0 𝑥 − 𝑦 − 2𝑧 + 6 = 0

To find the equation of the tangent plane at a point on a surface given by 𝑧 = 𝑓 (𝑥, 𝑦 ), you can define the function 𝐹 by 𝑓(𝑥, 𝑦, 𝑧 ) = 𝑓 (𝑥, 𝑦 ) − 𝑧. Then 𝑆 is given by the level surface 𝐹(𝑥, 𝑦, 𝑧 ) = 0, and by Theorem above an equation of the tangent plane to 𝑆 at the point (𝑥0 , 𝑦0 , 𝑧0 ). 𝐹𝑥 (𝑥0 , 𝑦0 , 𝑧0 )(𝑥 − 𝑥0 ) + 𝐹𝑦 (𝑥0 , 𝑦0 , 𝑧0 )(𝑦 − 𝑦0 ) + 𝐹𝑧 (𝑥0 , 𝑦0 , 𝑧0 )(𝑧 − 𝑧0 ) = 0

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MATH2225 2. 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 𝜃 (0 ≤ 𝜃 ≤ ) between the 𝜋

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given plane and the 𝑥𝑦-plane. (The angle of inclination of a horizontal plane is defined as zero.) Because the vector 𝒌 is normal to 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 cos 𝜃 = ‖𝑛‖‖𝑘‖ = |𝑛⋅𝑘|

|𝑛⋅𝑘| ‖𝑛‖

angle of inclination of a plane

Example 2. Find the angle of inclination of the tangent plane to the ellipsoid given by 𝑥2

𝑦2

+ 12 + 12

𝑧2 3

SOLUTION: We let

= 1 at the point (2,2,1).

𝐹(𝑥, 𝑦, 𝑧) =

𝑥2

12

+

𝑦2

12

+

𝑧2 3

−1

And solving for the partial derivatives with respect to 𝑥, 𝑦 and 𝑧 we have

𝐹𝑥 (𝑥, 𝑦, 𝑧) =

𝑥

6

𝐹𝑦 (𝑥, 𝑦, 𝑧) = 6

𝑦

𝐹𝑧 (𝑥, 𝑦, 𝑧) =

hence, the gradient of 𝐹 at the point (2,2,1) is given by 𝑥 𝑦 2𝑧 ∇𝐹 (𝑥, 𝑦, 𝑧 ) = 𝒊 + 𝒋 + 𝒌 3 6 6 2 2 2(1) 𝒌 ∇𝐹 (2,2,1) = 𝒊 + 𝒋 + 3 6 6 1 1 2 ∇𝐹 (2,2,1) = 𝒊 + 𝒋 + 𝒌 3 3 3

2𝑧 3

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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cos 𝜃 = which implies that

|∇𝐹(2,2,1) ⋅ 𝒌|

2/3 = √( 1 2 2 1 2 ‖∇𝐹(2,2,1)‖ ) + ( 3) + ( 3 𝜃 = arccos √ = 35.3°. 2

3

3

)

2

=√

3 2

ASSESSMENT: A. Find a unit normal vector to the surface at the given point. [Hint: Normalize the gradient vector ∇𝐹(𝑥, 𝑦, 𝑧)]. 1. 2. 3. 4.

3𝑥 + 4𝑦 + 12𝑧 = 0, 𝑥 + 𝑦 + 𝑧 = 4, 𝑥 2 + 𝑦 2 + 𝑧 2 = 6, 𝑥2𝑦4 − 𝑧 = 0

5. 𝑧 − 𝑥 sin 𝑦 = 4

(0,0,0) (2,0,2) (1,1,2) (1,2,16)

(6, , 7) 6 𝜋

B. Find an equation of the tangent plane to the surface at the given point. 6. 𝑓(𝑥, 𝑦 ) = 𝑥 2 − 2𝑥𝑦 + 𝑦 2 , 7. 𝑥𝑦 2 + 3𝑥 − 𝑧 2 = 8

(1,2,1) (1, −3,2)

REFERENCES [1] James Stewart, Calculus, Cengage Learning 2011. [2] Ron Larson and Bruce H. Edwards, Multivariable Calculus, ninth edition. Cengage Learning 2010. Note: Most of the examples were taken from Ron Larson and Bruce H. Edwards, Multivariable Calculus, ninth edition. Cengage Learning 2010.

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