Section 14.4 Tangent Planes and Linear Approximation
9. Find an equation of the tangent plane to the given surface at the specified point. z = x sin(x + y), (−1, 1, 0) Solution:
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 1387
2. = ( ) = 2sin ⇒ ( ) = 2cos , ( ) = 2 sin , so (2 −2) = 0 and (2 −2) = −4. By Equation 2, an equation of the tangent plane is − 4 = (2 −2)( − 2) + (2 −2)( − (−2)) ⇒
− 4 = −4( + 2), or = −4 − 4.
3. = ( ) = 22+ 2− 5 ⇒ ( ) = 4, ( ) = 2 − 5, so (1 2) = 4and (1 2) = −1.
By Equation 2, an equation of the tangent plane is − (−4) = (1 2)( − 1) + (1 2)( − 2) ⇒
+ 4 = 4( − 1) + (−1)( − 2), or = 4 − − 6.
4. = ( ) = ( + 2)2− 2( − 1)2− 5 ⇒ ( ) = 2( + 2), ( ) = −4( − 1), so (2 3) = 8and
(2 3) = −8. By Equation 2, an equation of the tangent plane is − 3 = (2 3)( − 2) + (2 3)( − 3) ⇒
− 3 = 8( − 2) + (−8)( − 3), or = 8 − 8 + 11.
5. = ( ) = − ⇒ ( ) = −(1) = −, ( ) = −(−1) = −−, so (2 2) = 1and
(2 2) = −1. Thus, an equation of the tangent plane is − 1 = (2 2)( − 2) + (2 2)( − 2) ⇒
− 1 = 1( − 2) + (−1)( − 2), or = − + 1.
6. = ( ) = 2 ⇒ ( ) = 2, ( ) = 2, so (0 3) = 9and (0 3) = 6. Thus, an equation of the tangent plane is − 9 = (0 3)( − 0) + (0 3)( − 3) ⇒ − 9 = 9 + 6( − 3), or = 9 + 6 − 9.
7. = ( ) = 2
⇒ ( ) = −2
2 , ( ) = 1
, so (−1 1) = −2 and (−1 1) = −1. Thus, an equation of the tangent plane is − (−2) = (−1 1)( − (−1)) + (−1 1)( − 1) ⇒
+ 2 = −2( + 1) − 1( − 1), or = −2 − − 3.
8. = ( ) = 2= −2 ⇒ ( ) = 12, ( ) = −2−3= −23, so (−4 2) = 14 and
(−4 2) = 1. Thus, an equation of the tangent plane is − (−1) = (−4 2) [ − (−4)] + (−4 2)( − 2) ⇒
+ 1 = 14( + 4) + 1( − 2), or = 14 + − 2.
9. = ( ) = sin( + ) ⇒ ( ) = · cos( + ) · 1 + sin( + ) · 1 = cos( + ) + sin( + ) and
( ) = cos( + ) · 1, so (−1 1) = (−1) cos 0 + sin 0 = −1, (−1 1) = (−1) cos 0 = −1. Thus, an equation of the tangent plane is − 0 = (−1 1)( − (−1)) + (−1 1)( − 1) ⇒ = (−1)( + 1) + (−1)( − 1), or
+ + = 0.
10. = ( ) = ln( − 2) ⇒ ( ) = 1( − 2), ( ) = −2( − 2), so (3 1) = 1and (3 1) = −2.
Thus, an equation of the tangent plane is − 0 = (3 1)( − 3) + (3 1)( − 1) ⇒ = 1( − 3) + (−2)( − 1), or = − 2 − 1.
11. = ( ) = 2+ + 32, so ( ) = 2 + ⇒ (1 1) = 3, ( ) = + 6 ⇒ (1 1) = 7and an equation of the tangent plane is − 5 = 3( − 1) + 7( − 1), or = 3 + 7 − 5. After zooming in, the surface and the
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20. Explain why the function is differentiable at the given point. Then find the linearization L(x, y), yd of the function at that point.
f (x, y) = 1 + y
1 + x, (1, 3) Solution:
424 ¤ CHAPTER 14 PARTIAL DERIVATIVES
The surface and tangent plane are shown in the first graph below. After zooming in, the surface and the tangent plane become almost indistinguishable, as shown in the second graph. (Here, the tangent plane is above the surface.) If we zoom in farther, the surface and the tangent plane will appear to coincide.
11. ( ) = 1 + ln( − 5). The partial derivatives are ( ) = · 1
− 5() + ln( − 5) · 1 =
− 5 + ln( − 5) and ( ) = · 1
− 5() = 2
− 5, so (2 3) = 6and (2 3) = 4. Both and are continuous functions for
5, so by Theorem 8, is differentiable at (2 3). By Equation 3, the linearization of at (2 3) is given by
( ) = (2 3) + (2 3)( − 2) + (2 3)( − 3) = 1 + 6( − 2) + 4( − 3) = 6 + 4 − 23.
12. ( ) = √ = ()12. The partial derivatives are ( ) = 12()−12() =
2√ and
( ) = 12()−12() =
2√, so (1 4) = 4 2√
4
= 1and (1 4) = 1 2√
4
= 14. Both and are continuous functions for 0, so is differentiable at (1 4) by Theorem 8. The linearization of at (1 4) is
( ) = (1 4) + (1 4)( − 1) + (1 4)( − 4) = 2 + 1( − 1) +14( − 4) = +14.
13. ( ) = 2. The partial derivatives are ( ) = 2and ( ) = 2, so (1 0) = 2and (1 0) = 1. Both
and are continuous functions, so by Theorem 8, is differentiable at (1 0). By Equation 3, the linearization of at (1 0)is given by ( ) = (1 0) + (1 0)( − 1) + (1 0)( − 0) = 1 + 2( − 1) + 1( − 0) = 2 + − 1.
14. ( ) = 1 +
1 + = (1 + )(1 + )−1. The partial derivatives are ( ) = (1 + )(−1)(1 + )−2= − 1 + (1 + )2 and
( ) = (1)(1 + )−1= 1
1 + , so (1 3) = −1 and (1 3) = 12. Both and are continuous functions for
6= −1, so is differentiable at (1 3) by Theorem 8. The linearization of at (1 3) is
( ) = (1 3) + (1 3)( − 1) + (1 3)( − 3) = 2 + (−1)( − 1) +12( − 3) = − + 12 +32.
15. ( ) = 4 arctan(). The partial derivatives are ( ) = 4 · 1
1 + ()2() = 4
1 + 22, and
( ) = 4
1 + 22, so (1 1) = 2and (1 1) = 2. Both and are continuous functions, so is differentiable at (1 1) by Theorem 8. The linearization of at (1 1) is
( ) = (1 1) + (1 1)( − 1) + (1 1)( − 1) = 4(4) + 2( − 1) + 2( − 1) = 2 + 2 + − 4.
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24. Verify the linear approximation at (0, 0). y−1x+1 ≈ x + y − 1 Solution:
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 425 16. ( ) =√
+ 4 = ( + 4)12. The partial derivatives are ( ) = 12( + 4)−12and
( ) = 12( + 4)−12(44) = 24( + 4)−12, so (3 0) = 12(3 + 0)−12= 14and
(3 0) = 20(3 + 0)−12 = 1. Both and are continuous functions near (3 0), so is differentiable at (3 0) by Theorem 8. The linearization of at (3 0) is
( ) = (3 0) + (3 0)( − 3) + (3 0)( − 0) = 2 + 14( − 3) + 1( − 0) = 14 + +54.
17. Let ( ) = cos(). Then ( ) = [− sin()]() + cos() = [cos() − sin()] and
( ) = [− sin()]() = −sin(). Both and are continuous functions, so by Theorem 8, is differentiable at (0 0). We have (0 0) = 0(cos 0 − 0) = 1, (0 0) = 0and the linear approximation of at (0 0) is
( ) ≈ (0 0) + (0 0)( − 0) + (0 0)( − 0) = 1 + 1 + 0 = + 1.
18. Let ( ) = − 1
+ 1. Then ( ) = ( − 1)(−1)( + 1)−2= 1 −
( + 1)2 and ( ) = 1
+ 1. Both and are continuous functions for 6= −1, so by Theorem 8, is differentiable at (0 0). We have (0 0) = 1, (0 0) = 1and the linear approximation of at (0 0) is ( ) ≈ (0 0) + (0 0)( − 0) + (0 0)( − 0) = −1 + 1 + 1 = + − 1.
19. We can estimate (22 49) using a linear approximation of at (2 5), given by
( ) ≈ (2 5) + (2 5)( − 2) + (2 5)( − 5) = 6 + 1( − 2) + (−1)( − 5) = − + 9. Thus
(22 49) ≈ 22 − 49 + 9 = 63.
20. ( ) = 1 − cos ⇒ ( ) = − cos and
( ) = −[(− sin ) + (cos )(1)] = sin − cos , so (1 1) = 1, (1 1) = 1. Then the linear approximation of at (1 1) is given by
( ) ≈ (1 1) + (1 1)( − 1) + (1 1)( − 1)
= 2 + (1)( − 1) + (1)( − 1) = + Thus (102 097) ≈ 102 + 097 = 199. We graph and its
tangent plane near the point (1 1 2) below. Notice near = 1 the surfaces are almost identical.
21. ( ) =
2+ 2+ 2 ⇒ ( ) =
2+ 2+ 2, ( ) =
2+ 2+ 2, and
( ) =
2+ 2+ 2, so (3 2 6) = 37, (3 2 6) = 27, (3 2 6) = 67. Then the linear approximation of
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52. Suppose you need to know an equation of the tangent plane to a surface S at the point P (2, 1, 3). You don’t have an equation for S but you know that the curves
r1(t) =< 2 + 3t, 1 − t2, 3 − 4t + t2>
r2(u) =< 1 + u2, 2u3− 1, 2u + 1 >
both lie on S. Find an equation of the tangent plane at P . Solution:
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS ¤ 429 38. Here = ∆ = 03, = ∆ = −5, = 831
, so
=
831
−831 ·
2 = 831
− 5 12− 310
144· 3 10
≈ −883. Thus the pressure will drop by about 883 kPa.
39. First we find
1
implicitly by taking partial derivatives of both sides with respect to 1:
1
1
= [(11) + (12) + (13)]
1 ⇒ −−2
1 = −−21 ⇒
1
= 2
21. Then by symmetry,
2
= 2
22,
3
= 2
23. When 1= 25, 2= 40and 3= 50, 1
= 17
200 ⇔ = 20017 Ω. Since the possible error for each is 05%, the maximum error of is attained by setting ∆ = 0005. So
∆ ≈ =
1
∆1+
2
∆2+
3
∆3= (0005)2
1
1
+ 1
2
+ 1
3
= (0005) = 171 ≈ 0059 Ω.
40. The errors in measurement are at most 2%, so
∆
≤ 002 and
∆
≤ 002. The relative error in the calculated surface area is
∆
≈
= +
= 7209(04250425−1)0725 + 72090425(07250725−1)
720904250725
= 0425
+ 0725
To estimate the maximum relative error, we use
=
∆
= 002 and
=
∆
= 002 ⇒
= 0425(002) + 0725(002) = 0023.
Thus the maximum percentage error is approximately 23%.
41. (a) ( ) = 2 ⇒ ( ) = 12 and ( ) = −23. Since
0, both and are continuous functions, so is differentiable at (23 110). We have (23 110) = 23(110)2 ≈ 1901, (23 110) = 1(110)2 ≈ 08264, and
(23 110) = −2(23)(110)3≈ −3456, so the linear approximation of at (23 110) is
( ) ≈ (23 110)+(23 110)(−23)+(23 110)(−110) ≈ 1901+08264(−23)−3456(−110) or ( ) ≈ 08264 − 3456 + 3802.
(b) From part (a), for values near = 23 and = 110, ( ) ≈ 08264 − 3456 + 3802. If increases by 1 kg to 24 kg and increases by 003 m to 113 m, we estimate the BMI to be
(24 113) ≈ 08264(24) − 3456(113) + 3802 ≈ 18801. This is very close to the actual computed BMI:
(24 113) = 24(113)2≈ 18796.
42. r1() =
2 + 3 1 − 2 3 − 4 + 2
⇒ r01() = h3 −2 −4 + 2i, r2() =
1 + 2 23− 1 2 + 1
⇒ r02() =
2 62 2
. Both curves pass through since r1(0) = r2(1) = h2 1 3i, so the tangent vectors r01(0) = h3 0 −4i and r02(1) = h2 6 2i are both parallel to the tangent plane to at . A normal vector for the tangent plane is
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430 ¤ CHAPTER 14 PARTIAL DERIVATIVES
r01(0) × r02(1) = h3 0 −4i × h2 6 2i = h24 −14 18i, so an equation of the tangent plane is 24( − 2) − 14( − 1) + 18( − 3) = 0 or 12 − 7 + 9 = 44.
43. ∆ = ( + ∆ + ∆) − ( ) = ( + ∆)2+ ( + ∆)2− (2+ 2)
= 2+ 2 ∆ + (∆)2+ 2+ 2 ∆ + (∆)2− 2− 2= 2 ∆ + (∆)2+ 2 ∆ + (∆)2
But ( ) = 2and ( ) = 2and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ + ∆ ∆, which is Definition 7 with 1= ∆and 2= ∆. Hence is differentiable.
44. ∆ = ( + ∆ + ∆) − ( ) = ( + ∆)( + ∆) − 5( + ∆)2− ( − 52)
= + ∆ + ∆ + ∆ ∆ − 52− 10 ∆ − 5(∆)2− + 52
= ( − 10) ∆ + ∆ + ∆ ∆ − 5 ∆ ∆,
but ( ) = and ( ) = − 10 and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ − 5∆ ∆, which is Definition 7 with 1= ∆and 2= −5 ∆. Hence is differentiable.
45. To show that is continuous at ( ) we need to show that lim
()→() ( ) = ( )or equivalently lim
(∆∆)→(00) ( + ∆ + ∆) = ( ). Since is differentiable at ( ),
( + ∆ + ∆) − ( ) = ∆ = ( ) ∆ + ( ) ∆ + 1∆ + 2∆, where 1and 2→ 0 as
(∆ ∆) → (0 0). Thus ( + ∆ + ∆) = ( ) + ( ) ∆ + ( ) ∆ + 1∆ + 2∆. Taking the limit of both sides as (∆ ∆) → (0 0) gives lim
(∆∆)→(00) ( + ∆ + ∆) = ( ). Thus is continuous at ( ).
46. (a) lim
→0
( 0) − (0 0)
= lim
→0
0 − 0
= 0and lim
→0
(0 ) − (0 0)
= lim
→0
0 − 0
= 0. Thus (0 0) = (0 0) = 0.
To show that isn’t differentiable at (0 0) we need only show that is not continuous at (0 0) and apply Exercise 45. As ( ) → (0 0) along the -axis ( ) = 02= 0for 6= 0 so ( ) → 0 as ( ) → (0 0) along the -axis. But as ( ) → (0 0) along the line = , ( ) = 2
22
= 12for 6= 0 so ( ) → 12as ( ) → (0 0) along this line. Thus lim
()→(00) ( )doesn’t exist, so is discontinuous at (0 0) and thus not differentiable there.
(b) For ( ) 6= (0 0), ( ) = (2+ 2) − (2)
(2+ 2)2 = (2− 2)
(2+ 2)2. If we approach (0 0) along the -axis, then
( ) = (0 ) =3
4 = 1
, so ( ) → ±∞ as ( ) → (0 0). Thus lim
()→(00)( )does not exist and
( )is not continuous at (0 0) Similarly, ( ) = (2+ 2) − (2)
(2+ 2)2 = (2− 2)
(2+ 2)2 for ( ) 6= (0 0), and if we approach (0 0) along the -axis, then ( ) = ( 0) = 3
4 = 1
. Thus lim
()→(00)( )does not exist and
( )is not continuous at (0 0)
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1
54. (a) The function
f (x, y) =
xy
x2+y2 if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0)
was graphed in Figure 4. Show that fx(0, 0) and fy(0, 0) both exist but f is not differentiable at (0, 0). [Hint: Use the result of Exercise 53.]
(b) Explain why fx and fy are not continuous at (0, 0).
Solution:
430 ¤ CHAPTER 14 PARTIAL DERIVATIVES
r01(0) × r02(1) = h3 0 −4i × h2 6 2i = h24 −14 18i, so an equation of the tangent plane is 24( − 2) − 14( − 1) + 18( − 3) = 0 or 12 − 7 + 9 = 44.
43. ∆ = ( + ∆ + ∆) − ( ) = ( + ∆)2+ ( + ∆)2− (2+ 2)
= 2+ 2 ∆ + (∆)2+ 2+ 2 ∆ + (∆)2− 2− 2= 2 ∆ + (∆)2+ 2 ∆ + (∆)2
But ( ) = 2and ( ) = 2and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ + ∆ ∆, which is Definition 7 with 1= ∆and 2= ∆. Hence is differentiable.
44. ∆ = ( + ∆ + ∆) − ( ) = ( + ∆)( + ∆) − 5( + ∆)2− ( − 52)
= + ∆ + ∆ + ∆ ∆ − 52− 10 ∆ − 5(∆)2− + 52
= ( − 10) ∆ + ∆ + ∆ ∆ − 5 ∆ ∆,
but ( ) = and ( ) = − 10 and so ∆ = ( ) ∆ + ( ) ∆ + ∆ ∆ − 5∆ ∆, which is Definition 7 with 1= ∆and 2= −5 ∆. Hence is differentiable.
45. To show that is continuous at ( ) we need to show that lim
()→() ( ) = ( )or equivalently lim
(∆∆)→(00) ( + ∆ + ∆) = ( ). Since is differentiable at ( ),
( + ∆ + ∆) − ( ) = ∆ = ( ) ∆ + ( ) ∆ + 1∆ + 2∆, where 1and 2→ 0 as
(∆ ∆) → (0 0). Thus ( + ∆ + ∆) = ( ) + ( ) ∆ + ( ) ∆ + 1∆ + 2∆. Taking the limit of both sides as (∆ ∆) → (0 0) gives lim
(∆∆)→(00) ( + ∆ + ∆) = ( ). Thus is continuous at ( ).
46. (a) lim
→0
( 0) − (0 0)
= lim
→0
0 − 0
= 0and lim
→0
(0 ) − (0 0)
= lim
→0
0 − 0
= 0. Thus (0 0) = (0 0) = 0.
To show that isn’t differentiable at (0 0) we need only show that is not continuous at (0 0) and apply Exercise 45. As ( ) → (0 0) along the -axis ( ) = 02= 0for 6= 0 so ( ) → 0 as ( ) → (0 0) along the -axis. But as ( ) → (0 0) along the line = , ( ) = 2
22
= 12 for 6= 0 so ( ) → 12 as ( ) → (0 0) along this line. Thus lim
()→(00) ( )doesn’t exist, so is discontinuous at (0 0) and thus not differentiable there.
(b) For ( ) 6= (0 0), ( ) = (2+ 2) − (2)
(2+ 2)2 = (2− 2)
(2+ 2)2. If we approach (0 0) along the -axis, then
( ) = (0 ) = 3
4 = 1
, so ( ) → ±∞ as ( ) → (0 0). Thus lim
()→(00)( )does not exist and
( )is not continuous at (0 0) Similarly, ( ) = (2+ 2) − (2)
(2+ 2)2 = (2− 2)
(2+ 2)2 for ( ) 6= (0 0), and if we approach (0 0) along the -axis, then ( ) = ( 0) = 3
4 = 1
. Thus lim
()→(00)( )does not exist and
( )is not continuous at (0 0)
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