Chapter 9 Differential Vector Calculus

8/12/2019 Chapter 9 Differential Vector Calculus

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In the gure in Example 9.4, observe that at each point, the gradient vector is orthog-onal to the level curve through the point. This illustrates a general phenomenon which isdescribed more carefully in Propositions 9.1 and 9.2.

Proposition 9.1. If f : D 2 → R is differentiable at (a 1 , a 2 ) and f (a 1 , a 2 ) = 0, thenf (a 1 , a 2 ) is orthogonal to the tangent line to the level set of f which passes through

(a 1 , a 2 ).

Proof. Let C denote the level set f − 1 (f (a 1 , a 2 )) . Suppose that g : I → R 2 is a parame-terized curve whose image is C and which satises g (t 0 ) = ( a 1 , a 2 ) for t0 I . Then thetangent line to C at (a 1 , a 2 ) has direction vector g ′ (t 0 ).

f gC (a , )1

t0

a2

f (C)

g’ (t )0

The composition f ◦ g is a constant function, so its derivative is zero at all values of t:

0 = ddt

(f ◦ g )( t) = D (f ◦ g )( t).

On the other hand, we can also compute this derivative using the Chain Rule:

D (f ◦ g )( t 0 ) = Df (g (t 0 ))D g (t 0 )= Df (a 1 , a 2 )g ′ (t 0 )= f (a 1 , a 2 ) · g ′ (t 0 ).

Comparing the two equations shows that

0 = f (a 1 , a 2 ) · g ′ (t 0 ),

so the tangent line to C at t 0 is orthogonal to f (a 1 , a 2 ).

In fact, a similar proof shows the following more general result:

Proposition 9.2. Suppose that f : D n → R is differentiable at a D n and that f (a ) =0 . Denote by S the level set f − 1 (f (a )) . Then if C is any curve in S which passes througha , the tangent line to C at a is orthogonal to f (a ).

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Example 9.6. The sphere dened by x 2 + y2 + z 2 = 26 intersects the plane x + 7y + 7 z = 50 ina curve C . Find the tangent line to C at the point (1, 3, 4).

We can use Proposition 9.2 to solve this problem. The sphere is the level set at height26 of the function G(x,y,z ) = x2 + y2 + z 2 , so the tangent line to C is orthogonal to

G (1, 3, 4). On the other hand, the plane is the level set at height 50 of the functionH (x,y,z ) = x + 7y + 7 z , so the tangent line to C is also orthogonal to H (1, 3, 4).

The cross product of G and H is a direction vector for the tangent line to C :

G(1, 3, 4) × H (1, 3, 4) =268

×177

=− 14− 68

Therefore, we can parameterize the tangent line to C at (1, 3, 4) by

L (t) =13

4

+ t− 7− 3

4

.

9.5 Exercises

In Exercises 1 through 4, use Denition 9.1 to compute the directional derivative D v f (a )for the given function f , direction v , and point a .

1. f (x, y ) = 2 xy , v = 3− 4 , a = (7 , − 2).

2. f (x, y ) = 1

xy, v = 3

2, a = (1 , 2)

3. f (x,y,z ) = 4 xz − 1

y, v =

1− 22

, a = (0 , 1, 1)

4. f (x,y,z ) = x 2 yz − xz 2 + 3 y2 , v =50

− 12, a = ( − 2, − 1, 1)

In Exercises 5 through 18, use Theorem 17 to compute the directional derivative D v f (a )for the given function f , direction v , and point a .

5. f (x, y ) = 2 xy , v = 3− 4 , a = (7 , − 2).

6. f (x, y ) = 1

xy, v = 3

2 , a = (1 , 2)

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7. f (x,y,z ) = 4 xz −1

y, v =

1

−22

, a = (0 , 1, 1)

8. f (x,y,z ) = x2

yz − xz 2

+ 3 y2

, v

=

5

0−12 ,

a

= ( −2, −1, 1)

9. f (x, y) = 2 x7

− 7x2 y + 4 xy2 + y6 , v = −43

, a = ( −1, −1)

10. f (x, y) = e− x2 − y

2

, v = −11

, a = (0 , 1)

11. f (x, y) = xexy , v = 1

−2, a = ( −2, 0)

12. f (x, y) = sin( x2

− 5xy − 2y2 + 3 x + 3y), v = −35

, a = (1 , −2)

13. f (x,y,z ) = 1

√ x 2 + y2 + z

2, v =

111

, a = (2 , 1, −2)

14. f (x,y,z ) = exyz sin(xyz ), v =1

−25

, a = (0 , 3, 10)

15. f (x,y,z ) = x cos(yz ) + y cos(xz ) + z cos(yx), v =1

−2

−2, a = (0 , 2, π/ 4)

16. f (x,y,z ) = ln( x2 + y2 + z 2 ), v = −222

, a = (3 , 4, −3)

17. f (w,x ,y,z ) = 6 w2 z −

xy3

−3wy + 7 x2 z 2 , v =

1

−1

−11

, a = (1 , 0, 2,−

5)

18. f (x1 ,...,x n ) = 1

x21 + ... + x

2n

, v =

11...1

, a = (1 , 1,..., 1)

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25. Let C be the curve in R 2 dened by the equation sin x + cos(2y) = 1 . Find an equa-tion for the line perpendicular to the tangent line to C at the point (π/ 6, π/ 6).

26. Let C be the curve in R 2 dened by 3x3 − x2 y2 + y4 = 21 . Find an equation for the

tangent line to C at the point (2, − 1).

27. Let C be the curve given by 4x2 − xy + y2 = 4 . At what points is the tangent line to

C parallel to the vector 11 ?

28. Let S be the hyperboloid of two sheets given by 3x2 − 2y2 + z 2 = − 5.

(a) Find two parametrized curves C 1 , C 2 on S through the point (2, 3, − 1) whosetangent lines L1 , L 2 are distinct.

(b) Let P be the plane containing the lines L1 , L 2 . Find an equation for P .(c) Describe the relationship between the normal vector to P and f (2, 3, − 1).(d) On one set of axes, sketch S and P .

29. Let S be the hyperbolic paraboloid given by 7x2 − y2 − 2z = 1 . Let g 1 and g2 beparametrized curves whose images lie in S and which pass through a = ( − 1, 2, 1).Recall that for i = 1, 2, the tangent line to g i at the point a is parallel to g ′

i (a ). Find avector parallel to g ′

1 (a ) × g ′

2 (a ).

30. Let C be the curve which is the intersection of the ellipsoid 2x2 +3 y2 + z 2 = 9 and thehyperboloid − 3x2 + 6 y2 + z 2 = 7 . Find a parametric representation for the tangentline to C at the point (1, − 1, 2).

31. Let f (x ,y,z ) = x2 + 3 y2 + 4 z 2 , and let E be the ellipsoid given by f (x,y,z ) = 8 . LetP be the plane given by P = {(0, 0, 2) + su + tv |s, t R } where

u =11

− 1 and v =

2− 21

.

Notice that (1, 1, 1) lies in the intersection E ∩ P .

(a) What geometric relationships can you nd between u , v , and f (1, 1, 1)?(Hint: Take dot products of pairs of these vectors.)

(b) Use your answer from the rst part to describe or sketch the intersection of E and P .

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Chapter 9 Differential Vector Calculus