1. Lesson 20 (Section 15.4)
Tangent Planes
Math 20
November 5, 2007
Announcements
Problem Set 7 on the website. Due November 7.
No class November 12. Yes class November 21.
OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
Prob. Sess.: Sundays 6–7 (SC B-10), Tuesdays 1–2 (SC 116)
2. Outline
Tangent Lines in one variable
Tangent lines in two or more variables
Tangent planes in two variables
One more example
6. Summary
Fact
The tangent line to y = f (x) through the point (x0 , y0 ) has
equation
y = f (x0 ) + f (x0 )(x − x0 )
7. Summary
Fact
The tangent line to y = f (x) through the point (x0 , y0 ) has
equation
y = f (x0 ) + f (x0 )(x − x0 )
The expression f (x0 ) is a number, not a function!
This is the best linear approximation to f near x0 .
This is the first-degree Taylor polynomial for f .
8. Example
Example
√
Find the equation for the tangent line to y = x at x = 4.
9.
10. Example
Example
√
Find the equation for the tangent line to y = x at x = 4.
Solution
We have
dy 1 1 1
=√ =√=
dx 4
2x 24
x=4 x=4
So the tangent line has equation
y = 2 + 4 (x − 4) = 1 x + 1
1
4
11. Outline
Tangent Lines in one variable
Tangent lines in two or more variables
Tangent planes in two variables
One more example
12. Recall
Any line in Rn can be described by a point a and a direction v and
given parametrically by the equation
x = a + tv
13.
14. Last time we differentiated the curve
x → (x, y0 , f (x, y0 ))
at x = x0 and said that was a line tangent to the graph at (x0 , y0 ).
15. Example
Let f = f (x, y ) = 4 − x 2 − 2y 4 . Look at the point P = (1, 1, 1)
on the graph of f .
0
-10
z
-20
-30
-2 -2
-1 -1
0 0
y x
1 1
16. There are two interesting curves going through the point P:
x → (x, 1, f (x, 1)) y → (1, y , f (1, y ))
Each of these is a one-variable function, so makes a curve, and has
a slope!
17. There are two interesting curves going through the point P:
x → (x, 1, f (x, 1)) y → (1, y , f (1, y ))
Each of these is a one-variable function, so makes a curve, and has
a slope!
d d
2 − x2 = −2x|x=1 = −2.
f (x, 1) =
dx dx
x=1 x−1
18. Last time we differentiated the curve
x → (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, f1 (x0 , y0 )).
19. Last time we differentiated the curve
x → (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, f1 (x0 , y0 )).
So a line tangent to the graph is given by
(x, y , z) = (x0 , y0 , z0 ) + t(1, 0, f1 (x0 , y0 ))
20. There are two interesting curves going through the point P:
x → (x, 1, f (x, 1)) y → (1, y , f (1, y ))
Each of these is a one-variable function, so makes a curve, and has
a slope!
d d
2 − x2 = −2x|x=1 = −2.
f (x, 1) =
dx dx
x=1 x−1
d d
3 − 2y 4 = −8y 3 = −8.
f (1, y ) = y =1
dy dx
y =1 y =1
21. Last time we differentiated the curve
x → (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, f1 (x0 , y0 )).
So a line tangent to the graph is given by
(x, y , z) = (x0 , y0 , z0 ) + t(1, 0, f1 (x0 , y0 ))
Another line is
(x, y , z) = (x0 , y0 , z0 ) + t(0, 1, f2 (x0 , y0 ))
22.
23. There are two interesting curves going through the point P:
x → (x, 1, f (x, 1)) y → (1, y , f (1, y ))
Each of these is a one-variable function, so makes a curve, and has
a slope!
d d
2 − x2 = −2x|x=1 = −2.
f (x, 1) =
dx dx
x=1 x−1
d d
3 − 2y 4 = −8y 3 = −8.
f (1, y ) = y =1
dy dx
y =1 y =1
We see that the tangent plane is spanned by these two
vectors/slopes.
24. 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 , . . . , 0, fi (a1 , a2 , . . . , an ))
i
is tangent to the graph of f at (a1 , a2 , . . . , an ).
28. Outline
Tangent Lines in one variable
Tangent lines in two or more variables
Tangent planes in two variables
One more example
29. Recall
Definition
A plane (in three-dimensional space) through a that is orthogonal
to a vector p = 0 is the set of all points x satisfying
p · (x − a) = 0.
30. Recall
Definition
A plane (in three-dimensional space) through a that is orthogonal
to a vector p = 0 is the set of all points x satisfying
p · (x − a) = 0.
Question
Given a function and a point on the graph of the function, how do
we find the equation of the tangent plane?
31.
32.
33. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )).
34. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )). Then p must
satisfy
p · (1, 0, f1 (x0 , y0 )) = 0
p · (0, 1, f2 (x0 , y0 )) = 0
35.
36. Let p = (p1 , p2 , p3 ), a = (x0 , y0 , z0 = f (x0 , y0 )). Then p must
satisfy
p · (1, 0, f1 (x0 , y0 )) = 0
p · (0, 1, f2 (x0 , y0 )) = 0
A solution is to let p1 = f1 (x0 , y0 ), p2 = f2 (x0 , y0 ), p3 = −1.
37. 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 (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation
f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0
38. 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 (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation
f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0
or
z = f (x0 , y0 ) + f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 )
39.
40. 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 (f1 (x0 , y0 ), f2 (x0 , y0 ), −1) and equation
f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 ) − (z − z0 ) = 0
or
z = f (x0 , y0 ) + f1 (x0 , y0 )(x − x0 ) + f2 (x0 , y0 )(y − y0 )
This is the best linear approximation to f near (x0 , y0 ).
is is the first-degree Taylor polynomial (in two variables) for f .
41. Outline
Tangent Lines in one variable
Tangent lines in two or more variables
Tangent planes in two variables
One more example
42. Example
The number of units of output per day at a factory is
−1/2
1 −2 9
x + y −2
P(x, y ) = 150 ,
10 10
where x denotes capital investment (in units of $1000), and y
denotes the total number of hours (in units of 10) the work force is
employed 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.
43. Example
The number of units of output per day at a factory is
−1/2
1 −2 9
x + y −2
P(x, y ) = 150 ,
10 10
where x denotes capital investment (in units of $1000), and y
denotes the total number of hours (in units of 10) the work force is
employed 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.
44. Solution
−3/2
−2
∂P 1 1 −2 9
x + y −2 x −3
(x, y ) = 150 −
∂x 2 10 10 10
−3/2
1 −2 9
x + y −2 x −3
= 15
10 10
∂P
(50, 50) = 50
∂x
45. Solution
−3/2
−2
∂P 1 1 −2 9
x + y −2 x −3
(x, y ) = 150 −
∂x 2 10 10 10
−3/2
1 −2 9
x + y −2 x −3
= 15
10 10
∂P
(50, 50) = 50
∂x
−3/2
∂P 1 1 −2 9 9
x + y −2 (−2)y −3
(x, y ) = 150 −
∂y 2 10 10 10
−3/2
1 −2 9
x + y −2 x −3
= 15
10 10
∂P
(50, 50) = 135
∂y
So the linear approximation is
L = 7500 + 15(x − 50) + 135(y − 50)
46. 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
47. 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 is
P(55, 49) ≈ 7427
13
≈ 1.75%
So we are off by 7427
