Lesson 20 (Section 15.4)Tangent Planes

Math 20

November 5, 2007

Announcements

I Problem Set 7 on the website. Due November 7.

I No class November 12. Yes class November 21.

I OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)

I Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)

Outline

Tangent Lines in one variable

Tangent lines in two or more variables

Tangent planes in two variables

One more example

Tangent Lines in one variable

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Summary

FactThe tangent line to y = f (x) through the point (x0, y0) hasequation

y = f (x0) + f ′(x0)(x − x0)

I The expression f ′(x0) is a number, not a function!

I This is the best linear approximation to f near x0.

I This is the first-degree Taylor polynomial for f .

Summary

FactThe tangent line to y = f (x) through the point (x0, y0) hasequation

y = f (x0) + f ′(x0)(x − x0)

I The expression f ′(x0) is a number, not a function!

I This is the best linear approximation to f near x0.

I This is the first-degree Taylor polynomial for f .

Example

Example

Find the equation for the tangent line to y =√

x at x = 4.

SolutionWe have

dy

dx

∣∣∣∣x=4

=1

2√

x

∣∣∣∣x=4

=1

2√

4=

1

4

So the tangent line has equation

y = 2 + 14(x − 4) = 1

4x + 1

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Example

Example

Find the equation for the tangent line to y =√

x at x = 4.

SolutionWe have

dy

dx

∣∣∣∣x=4

=1

2√

x

∣∣∣∣x=4

=1

2√

4=

1

4

So the tangent line has equation

y = 2 + 14(x − 4) = 1

4x + 1

Outline

Tangent Lines in one variable

Tangent lines in two or more variables

Tangent planes in two variables

One more example

Recall

Any line in Rn can be described by a point a and a direction v andgiven parametrically by the equation

x = a + tv

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Last time we differentiated the curve

x 7→ (x , y0, f (x , y0))

at x = x0 and said that was a line tangent to the graph at (x0, y0).

That vector is (1, 0, f ′1(x0, y0)).So a line tangent to the graph is given by

(x , y , z) = (x0, y0, z0) + t(1, 0, f ′1(x0, y0))

Another line is

(x , y , z) = (x0, y0, z0) + t(0, 1, f ′2(x0, y0))

ExampleLet f = f (x , y) = 4− x2 − 2y4. Look at the point P = (1, 1, 1)on the graph of f .

-2

-1

0

1

2

-2

-1

0

1

2

-30

-20

-10

0

-2

-1

0

1

2

z

xy

There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

Last time we differentiated the curve

x 7→ (x , y0, f (x , y0))

at x = x0 and said that was a line tangent to the graph at (x0, y0).That vector is (1, 0, f ′1(x0, y0)).

So a line tangent to the graph is given by

(x , y , z) = (x0, y0, z0) + t(1, 0, f ′1(x0, y0))

Another line is

(x , y , z) = (x0, y0, z0) + t(0, 1, f ′2(x0, y0))

Last time we differentiated the curve

x 7→ (x , y0, f (x , y0))

at x = x0 and said that was a line tangent to the graph at (x0, y0).That vector is (1, 0, f ′1(x0, y0)).So a line tangent to the graph is given by

(x , y , z) = (x0, y0, z0) + t(1, 0, f ′1(x0, y0))

Another line is

(x , y , z) = (x0, y0, z0) + t(0, 1, f ′2(x0, y0))

There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

Last time we differentiated the curve

x 7→ (x , y0, f (x , y0))

at x = x0 and said that was a line tangent to the graph at (x0, y0).That vector is (1, 0, f ′1(x0, y0)).So a line tangent to the graph is given by

(x , y , z) = (x0, y0, z0) + t(1, 0, f ′1(x0, y0))

Another line is

(x , y , z) = (x0, y0, z0) + t(0, 1, f ′2(x0, y0))

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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There are two interesting curves going through the point P:

x 7→ (x , 1, f (x , 1)) y 7→ (1, y , f (1, y))

Each of these is a one-variable function, so makes a curve, and hasa slope!

d

dxf (x , 1)

∣∣∣∣x=1

=d

dx

(2− x2

)∣∣∣∣x−1

= −2x |x=1 = −2.

d

dyf (1, y)

∣∣∣∣y=1

=d

dx

(3− 2y4

)∣∣∣∣y=1

= −8y3∣∣y=1

= −8.

We see that the tangent plane is spanned by these twovectors/slopes.

Summary

Let f a function of n variables differentiable at (a1, a2, . . . , an).Then the line given by

(x1, x2, . . . , xn) = f (a1, a2, . . . , an)+t(0, . . . , 1︸︷︷︸i

, . . . , 0, f ′i (a1, a2, . . . , an))

is tangent to the graph of f at (a1, a2, . . . , an).

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Example

Find the equations of two lines tangent to z = xy2 at the point(2, 1, 2).

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Outline

Tangent Lines in one variable

Tangent lines in two or more variables

Tangent planes in two variables

One more example

Recall

DefinitionA plane (in three-dimensional space) through a that is orthogonalto a vector p 6= 0 is the set of all points x satisfying

p · (x− a) = 0.

QuestionGiven a function and a point on the graph of the function, how dowe find the equation of the tangent plane?

Recall

DefinitionA plane (in three-dimensional space) through a that is orthogonalto a vector p 6= 0 is the set of all points x satisfying

p · (x− a) = 0.

QuestionGiven a function and a point on the graph of the function, how dowe find the equation of the tangent plane?

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Let p = (p1, p2, p3), a = (x0, y0, z0 = f (x0, y0)).

Then p mustsatisfy

p · (1, 0, f ′1(x0, y0)) = 0

p · (0, 1, f ′2(x0, y0)) = 0

A solution is to let p1 = f ′1(x0, y0), p2 = f ′2(x0, y0), p3 = −1.

Let p = (p1, p2, p3), a = (x0, y0, z0 = f (x0, y0)). Then p mustsatisfy

p · (1, 0, f ′1(x0, y0)) = 0

p · (0, 1, f ′2(x0, y0)) = 0

A solution is to let p1 = f ′1(x0, y0), p2 = f ′2(x0, y0), p3 = −1.

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Let p = (p1, p2, p3), a = (x0, y0, z0 = f (x0, y0)). Then p mustsatisfy

p · (1, 0, f ′1(x0, y0)) = 0

p · (0, 1, f ′2(x0, y0)) = 0

A solution is to let p1 = f ′1(x0, y0), p2 = f ′2(x0, y0), p3 = −1.

Summary

Fact (tangent planes in two variables)

The tangent plane to z = f (x , y) through (x0, y0, z0 = f (x0, y0))has normal vector (f ′1(x0, y0), f ′2(x0, y0),−1) and equation

f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)− (z − z0) = 0

orz = f (x0, y0) + f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)

I This is the best linear approximation to f near (x0, y0).

I is is the first-degree Taylor polynomial (in two variables) for f .

Summary

Fact (tangent planes in two variables)

The tangent plane to z = f (x , y) through (x0, y0, z0 = f (x0, y0))has normal vector (f ′1(x0, y0), f ′2(x0, y0),−1) and equation

f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)− (z − z0) = 0

orz = f (x0, y0) + f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)

I This is the best linear approximation to f near (x0, y0).

I is is the first-degree Taylor polynomial (in two variables) for f .

Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007

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Summary

Fact (tangent planes in two variables)

The tangent plane to z = f (x , y) through (x0, y0, z0 = f (x0, y0))has normal vector (f ′1(x0, y0), f ′2(x0, y0),−1) and equation

f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)− (z − z0) = 0

orz = f (x0, y0) + f ′1(x0, y0)(x − x0) + f ′2(x0, y0)(y − y0)

I This is the best linear approximation to f near (x0, y0).

I is is the first-degree Taylor polynomial (in two variables) for f .

Outline

Tangent Lines in one variable

Tangent lines in two or more variables

Tangent planes in two variables

One more example

Example

The number of units of output per day at a factory is

P(x , y) = 150

[1

10x−2 +

9

10y−2

]−1/2

,

where x denotes capital investment (in units of $1000), and ydenotes the total number of hours (in units of 10) the work force isemployed per day. Suppose that currently, capital investment is$50,000 and the total number of working hours per day is 500.Estimate the change in output if capital investment is increased by$5000 and the number of working hours is decreased by 10 per day.

Example

The number of units of output per day at a factory is

P(x , y) = 150

[1

10x−2 +

9

10y−2

]−1/2

,

where x denotes capital investment (in units of $1000), and ydenotes the total number of hours (in units of 10) the work force isemployed per day. Suppose that currently, capital investment is$50,000 and the total number of working hours per day is 500.Estimate the change in output if capital investment is increased by$5000 and the number of working hours is decreased by 10 per day.

Solution

∂P

∂x(x , y) = 150

(−1

2

)[1

10x−2 +

9

10y−2

]−3/2(−2

10

)x−3

= 15

[1

10x−2 +

9

10y−2

]−3/2

x−3

∂P

∂x(50, 50) = 50

∂P

∂y(x , y) = 150

(−1

2

)[1

10x−2 +

9

10y−2

]−3/2( 9

10

)(−2)y−3

= 15

[1

10x−2 +

9

10y−2

]−3/2

x−3

∂P

∂y(50, 50) = 135

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

Solution

∂P

∂x(x , y) = 150

(−1

2

)[1

10x−2 +

9

10y−2

]−3/2(−2

10

)x−3

= 15

[1

10x−2 +

9

10y−2

]−3/2

x−3

∂P

∂x(50, 50) = 50

∂P

∂y(x , y) = 150

(−1

2

)[1

10x−2 +

9

10y−2

]−3/2( 9

10

)(−2)y−3

= 15

[1

10x−2 +

9

10y−2

]−3/2

x−3

∂P

∂y(50, 50) = 135

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

Solution, continued

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

If ∆x = 5 and ∆y = −1, then

L = 7500 + 15 · 5 + 135 · (−1) = 7440

The actual value isP(55, 49) ≈ 7427

So we are off by 137427 ≈ 1.75%

Solution, continued

So the linear approximation is

L = 7500 + 15(x − 50) + 135(y − 50)

If ∆x = 5 and ∆y = −1, then

L = 7500 + 15 · 5 + 135 · (−1) = 7440

The actual value isP(55, 49) ≈ 7427

So we are off by 137427 ≈ 1.75%

Contour plot of P

0 20 40 60 80 1000

20

40

60

80

100

Contour plot of L

0 20 40 60 80 1000

20

40

60

80

100

Contour plots, superimposed

0 20 40 60 80 1000

20

40

60

80

100

Animation of P and its linear approximation at (50, 50)