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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
Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007
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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).
Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007
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Example
Find the equations of two lines tangent to z = xy2 at the point(2, 1, 2).
Math 20 - November 05, 2007.GWBMonday, Nov 5, 2007
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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)
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