Checkpoint
The domain is the shaded circle defined by the inequality 9x2+9y2≤36,9x2+9y2≤36, which has a circle of radius 22 as its boundary. The range is [0,6].[0,6].
The equation of the level curve can be written as (x−3)2+(y+1)2=25,(x−3)2+(y+1)2=25, which is a circle with radius 55 centered at (3,−1).(3,−1).
z=3−(x−1)2.z=3−(x−1)2. This function describes a parabola opening downward in the plane y=3.y=3.
domain ( h ) = { ( x , y , t ) ∈ ℝ 3 | y ≥ 4 x 2 − 4 } domain ( h ) = { ( x , y , t ) ∈ ℝ 3 | y ≥ 4 x 2 − 4 }
(x−1)2+(y+2)2+(z−3)2=16(x−1)2+(y+2)2+(z−3)2=16 describes a sphere of radius 44 centered at the point (1,−2,3).(1,−2,3).
lim(x,y)→(5,−2)x2−yy2+x−13=32lim(x,y)→(5,−2)x2−yy2+x−13=32
If y=k(x−2)+1,y=k(x−2)+1, then lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2=k1+k2.lim(x,y)→(2,1)(x−2)(y−1)(x−2)2+(y−1)2=k1+k2. Since the answer depends on k,k, the limit fails to exist.
lim(x,y)→(5,−2)29−x2−y2=0lim(x,y)→(5,−2)29−x2−y2=0
The domain of ff contains the ordered pair (2,−3)(2,−3) because f(a,b)=f(2,−3)=16−2(2)2−(−3)2=3f(a,b)=f(2,−3)=16−2(2)2−(−3)2=3
lim(x,y)→(a,b)f(x,y)=3lim(x,y)→(a,b)f(x,y)=3
lim(x,y)→(a,b)f(x,y)=f(a,b)=3lim(x,y)→(a,b)f(x,y)=f(a,b)=3
The polynomials g(x)=2x2g(x)=2x2 and h(y)=y3h(y)=y3 are continuous at every real number; therefore, by the product of continuous functions theorem, f(x,y)=2x2y3f(x,y)=2x2y3 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane. Furthermore, any constant function is continuous everywhere, so g(x,y)=3g(x,y)=3 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane. Therefore, f(x,y)=2x2y3+3f(x,y)=2x2y3+3 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane. Last, h(x)=x4h(x)=x4 is continuous at every real number x,x, so by the continuity of composite functions theorem g(x,y)=(2x2y3+3)4g(x,y)=(2x2y3+3)4 is continuous at every point (x,y)(x,y) in the xy-plane.xy-plane.
lim(x,y,z)→(4,−1,3)13−x2−2y2+z2=2lim(x,y,z)→(4,−1,3)13−x2−2y2+z2=2
∂ f ∂ x = 8 x + 2 y + 3 , ∂ f ∂ y = 2 x − 2 y − 2 ∂ f ∂ x = 8 x + 2 y + 3 , ∂ f ∂ y = 2 x − 2 y − 2
∂ f ∂ x = ( 3 x 2 − 6 x y 2 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ y = ( −6 x 2 y + 8 y 3 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ x = ( 3 x 2 − 6 x y 2 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 ) ∂ f ∂ y = ( −6 x 2 y + 8 y 3 ) sec 2 ( x 3 − 3 x 2 y 2 + 2 y 4 )
Using the curves corresponding to c=−2andc=−3,c=−2andc=−3, we obtain
∂ f ∂ y | ( x , y ) = ( 0 , 2 ) ≈ f ( 0 , 3 ) − f ( 0 , 2 ) 3 − 2 = −3 − ( −2 ) 3 − 2 · 3 + 2 3 + 2 = − 3 − 2 ≈ −3.146 . ∂ f ∂ y | ( x , y ) = ( 0 , 2 ) ≈ f ( 0 , 3 ) − f ( 0 , 2 ) 3 − 2 = −3 − ( −2 ) 3 − 2 · 3 + 2 3 + 2 = − 3 − 2 ≈ −3.146 .
The exact answer is
∂ f ∂ y | ( x , y ) = ( 0 , 2 ) = ( −2 y | ( x , y ) = ( 0 , 2 ) = −2 2 ≈ −2.828 . ∂ f ∂ y | ( x , y ) = ( 0 , 2 ) = ( −2 y | ( x , y ) = ( 0 , 2 ) = −2 2 ≈ −2.828 .
∂ f ∂ x = 4 x − 8 x y + 5 z 2 − 6 , ∂ f ∂ y = −4 x 2 + 4 y , ∂ f ∂ z = 10 x z + 3 ∂ f ∂ x = 4 x − 8 x y + 5 z 2 − 6 , ∂ f ∂ y = −4 x 2 + 4 y , ∂ f ∂ z = 10 x z + 3
∂ f ∂ x = 2 x y sec ( x 2 y ) tan ( x 2 y ) − 3 x 2 y z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ y = x 2 sec ( x 2 y ) tan ( x 2 y ) − x 3 z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ z = −2 x 3 y z sec 2 ( x 3 y z 2 ) ∂ f ∂ x = 2 x y sec ( x 2 y ) tan ( x 2 y ) − 3 x 2 y z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ y = x 2 sec ( x 2 y ) tan ( x 2 y ) − x 3 z 2 sec 2 ( x 3 y z 2 ) ∂ f ∂ z = −2 x 3 y z sec 2 ( x 3 y z 2 )
∂ 2 f ∂ x 2 = −9 sin ( 3 x − 2 y ) − cos ( x + 4 y ) ∂ 2 f ∂ x ∂ y = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y ∂ x = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y 2 = −4 sin ( 3 x − 2 y ) − 16 cos ( x + 4 y ) ∂ 2 f ∂ x 2 = −9 sin ( 3 x − 2 y ) − cos ( x + 4 y ) ∂ 2 f ∂ x ∂ y = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y ∂ x = 6 sin ( 3 x − 2 y ) − 4 cos ( x + 4 y ) ∂ 2 f ∂ y 2 = −4 sin ( 3 x − 2 y ) − 16 cos ( x + 4 y )
z = 7 x + 8 y − 3 z = 7 x + 8 y − 3
L(x,y)=6−2x+3y,L(x,y)=6−2x+3y, so L(4.1,0.9)=6−2(4.1)+3(0.9)=0.5L(4.1,0.9)=6−2(4.1)+3(0.9)=0.5 f(4.1,0.9)=e5−2(4.1)+3(0.9)=e−0.5≈0.6065.f(4.1,0.9)=e5−2(4.1)+3(0.9)=e−0.5≈0.6065.
f ( −1 , 2 ) = −19 , f x ( −1 , 2 ) = 3 , f y ( −1 , 2 ) = −16 , E ( x , y ) = −4 ( y − 2 ) 2 . f ( −1 , 2 ) = −19 , f x ( −1 , 2 ) = 3 , f y ( −1 , 2 ) = −16 , E ( x , y ) = −4 ( y − 2 ) 2 .
lim ( x , y ) → ( x 0 , y 0 ) E ( x , y ) ( x − x 0 ) 2 + ( y − y 0 ) 2 = lim ( x , y ) → ( −1 , 2 ) −4 ( y − 2 ) 2 ( x + 1 ) 2 + ( y − 2 ) 2 ≤ lim ( x , y ) → ( −1 , 2 ) −4 ( ( x + 1 ) 2 + ( y − 2 ) 2 ) ( x + 1 ) 2 + ( y − 2 ) 2 = lim ( x , y ) → ( –1 , 2 ) − 4 ( x + 1 ) 2 + ( y − 2 ) 2 = 0. lim ( x , y ) → ( x 0 , y 0 ) E ( x , y ) ( x − x 0 ) 2 + ( y − y 0 ) 2 = lim ( x , y ) → ( −1 , 2 ) −4 ( y − 2 ) 2 ( x + 1 ) 2 + ( y − 2 ) 2 ≤ lim ( x , y ) → ( −1 , 2 ) −4 ( ( x + 1 ) 2 + ( y − 2 ) 2 ) ( x + 1 ) 2 + ( y − 2 ) 2 = lim ( x , y ) → ( –1 , 2 ) − 4 ( x + 1 ) 2 + ( y − 2 ) 2 = 0.
d z = 0.18 Δ z = f ( 1.03 , −1.02 ) − f ( 1 , −1 ) = 0.180682 d z = 0.18 Δ z = f ( 1.03 , −1.02 ) − f ( 1 , −1 ) = 0.180682
d z d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = ( 2 x − 3 y ) ( 6 cos 2 t ) + ( −3 x + 4 y ) ( −8 sin 2 t ) = −92 sin 2 t cos 2 t − 72 ( cos 2 2 t − sin 2 2 t ) = −46 sin 4 t − 72 cos 4 t . d z d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = ( 2 x − 3 y ) ( 6 cos 2 t ) + ( −3 x + 4 y ) ( −8 sin 2 t ) = −92 sin 2 t cos 2 t − 72 ( cos 2 2 t − sin 2 2 t ) = −46 sin 4 t − 72 cos 4 t .
∂ z ∂ u = 0 , ∂ z ∂ v = −21 ( 3 sin 3 v + cos 3 v ) 2 ∂ z ∂ u = 0 , ∂ z ∂ v = −21 ( 3 sin 3 v + cos 3 v ) 2
∂ w ∂ u = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2 ∂ w ∂ u = 0 ∂ w ∂ v = 15 − 33 sin 3 v + 6 cos 3 v ( 3 + 2 cos 3 v − sin 3 v ) 2
∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t∂w∂u=∂w∂x∂x∂u+∂w∂y∂y∂u∂w∂v=∂w∂x∂x∂v+∂w∂y∂y∂v∂w∂t=∂w∂x∂x∂t+∂w∂y∂y∂t∂w∂u=∂w∂x∂x∂u+∂w∂y∂y∂u∂w∂v=∂w∂x∂x∂v+∂w∂y∂y∂v
dydx=2x+y+72y−x+3|(3,−2)=2(3)+(−2)+72(−2)−(3)+3=−114dydx=2x+y+72y−x+3|(3,−2)=2(3)+(−2)+72(−2)−(3)+3=−114 Equation of the tangent line: y=−114x+254y=−114x+254
D u f ( x , y ) = ( 6 x y − 4 y 3 − 4 ) ( 1 ) 2 + ( 3 x 2 − 12 x y 2 + 6 y ) 3 2 D u f ( 3 , 4 ) = 72 − 256 − 4 2 + ( 27 − 576 + 24 ) 3 2 = −94 − 525 3 2 D u f ( x , y ) = ( 6 x y − 4 y 3 − 4 ) ( 1 ) 2 + ( 3 x 2 − 12 x y 2 + 6 y ) 3 2 D u f ( 3 , 4 ) = 72 − 256 − 4 2 + ( 27 − 576 + 24 ) 3 2 = −94 − 525 3 2
∇ f ( x , y ) = 2 x 2 + 2 x y + 6 y 2 ( 2 x + y ) 2 i − x 2 + 12 x y + 3 y 2 ( 2 x + y ) 2 j ∇ f ( x , y ) = 2 x 2 + 2 x y + 6 y 2 ( 2 x + y ) 2 i − x 2 + 12 x y + 3 y 2 ( 2 x + y ) 2 j
The gradient of gg at (−2,3)(−2,3) is ∇g(−2,3)=i+14j.∇g(−2,3)=i+14j. The unit vector that points in the same direction as ∇g(−2,3)∇g(−2,3) is ∇g(−2,3)‖∇g(−2,3)‖=1197i+14197j=197197i+14197197j,∇g(−2,3)‖∇g(−2,3)‖=1197i+14197j=197197i+14197197j, which gives an angle of θ=arcsin((14197)/197)≈1.499rad.θ=arcsin((14197)/197)≈1.499rad. The maximum value of the directional derivative is ‖∇g(−2,3)‖=197.‖∇g(−2,3)‖=197.
∇f(x,y)=(2x−2y+3)i+(−2x+10y−2)j∇f(x,y)=(2x−2y+3)i+(−2x+10y−2)j ∇f(1,1)=3i+6j∇f(1,1)=3i+6j Tangent vector: 6i−3j6i−3j or −6i+3j−6i+3j
∇ f ( x , y , z ) = 2 x 2 + 2 x y + 6 y 2 − 8 x z − 2 z 2 ( 2 x + y − 4 z ) 2 i − x 2 + 12 x y + 3 y 2 − 24 y z + z 2 ( 2 x + y − 4 z ) 2 j + 4 x 2 − 12 y 2 − 4 z 2 + 4 x z + 2 y z ( 2 x + y − 4 z ) 2 k . ∇ f ( x , y , z ) = 2 x 2 + 2 x y + 6 y 2 − 8 x z − 2 z 2 ( 2 x + y − 4 z ) 2 i − x 2 + 12 x y + 3 y 2 − 24 y z + z 2 ( 2 x + y − 4 z ) 2 j + 4 x 2 − 12 y 2 − 4 z 2 + 4 x z + 2 y z ( 2 x + y − 4 z ) 2 k .
D u f ( x , y , z ) = − 3 13 ( 6 x + y + 2 z ) + 12 13 ( x − 4 y + 4 z ) − 4 13 ( 2 x + 4 y − 2 z ) D u f ( 0 , −2 , 5 ) = 384 13 D u f ( x , y , z ) = − 3 13 ( 6 x + y + 2 z ) + 12 13 ( x − 4 y + 4 z ) − 4 13 ( 2 x + 4 y − 2 z ) D u f ( 0 , −2 , 5 ) = 384 13
( 2 , −5 ) ( 2 , −5 )
(43,13)(43,13) is a saddle point, (−32,−38)(−32,−38) is a local maximum.
The absolute minimum occurs at (1,0):(1,0): f(1,0)=−1.f(1,0)=−1.
The absolute maximum occurs at (0,3):(0,3): f(0,3)=63.f(0,3)=63.
ff has a maximum value of 976976 at the point (8,2).(8,2).
A maximum production level of 1389013890 occurs with 56255625 labor hours and $5500$5500 of total capital input.
f ( 3 3 , 3 3 , 3 3 ) = 3 3 + 3 3 + 3 3 = 3 f ( − 3 3 , − 3 3 , − 3 3 ) = − 3 3 − 3 3 − 3 3 = − 3 . f ( 3 3 , 3 3 , 3 3 ) = 3 3 + 3 3 + 3 3 = 3 f ( − 3 3 , − 3 3 , − 3 3 ) = − 3 3 − 3 3 − 3 3 = − 3 .
f(2,1,2)=9f(2,1,2)=9 is a minimum.
Section 4.1 Exercises
17 , 72 17 , 72
20π.20π. This is the volume when the radius is 22 and the height is 5.5.
All points in the xy-planexy-plane
x < y 2 x < y 2
All real ordered pairs in the xy-planexy-plane of the form (a,b)(a,b)
{ z | 0 ≤ z ≤ 4 } { z | 0 ≤ z ≤ 4 }
The set ℝℝ
y2−x2=4,y2−x2=4, a hyperbola
4=x+y,4=x+y, a line; x+y=0,x+y=0, line through the origin
2x−y=0,2x−y=−2,2x−y=2;2x−y=0,2x−y=−2,2x−y=2; three lines
x x + y = −1 , x x + y = 0 , x x + y = 2 x x + y = −1 , x x + y = 0 , x x + y = 2
e x y = 1 2 , e x y = 3 e x y = 1 2 , e x y = 3
x y − x = −2 , x y − x = 0 , x y − x = 2 x y − x = −2 , x y − x = 0 , x y − x = 2
e −2 x 2 = y , y = x 2 , y = e 2 x 2 e −2 x 2 = y , y = x 2 , y = e 2 x 2
The level curves are parabolas of the form y=cx2−2.y=cx2−2.
z=3+y3,z=3+y3, a curve in the zy-planezy-plane with rulings parallel to the x-axisx-axis
x 2 25 + y 2 4 ≤ 1 x 2 25 + y 2 4 ≤ 1
x 2 9 + y 2 4 + z 2 36 < 1 x 2 9 + y 2 4 + z 2 36 < 1
All points in xyz-spacexyz-space
The contour lines are circles.
x2+y2+z2=9,x2+y2+z2=9, a sphere of radius 33
x2+y2−z2=4,x2+y2−z2=4, a hyperboloid of one sheet
4 x 2 + y 2 = 1 , 4 x 2 + y 2 = 1 ,
1 = e x y ( x 2 + y 2 ) 1 = e x y ( x 2 + y 2 )
T ( x , y ) = k x 2 + y 2 T ( x , y ) = k x 2 + y 2
x2+y2=k40,x2+y2=k40, x2+y2=k100.x2+y2=k100. The level curves represent circles of radii 10k/2010k/20 and k/10k/10
Section 4.2 Exercises
2.0
2323
1. ੧੧
2. ੧੨੧੨
3. −੧੨−੧੨
4. e−੩੨e−੩੨
5. ੧੧.੦੧੧.੦
6. ੧.੦੧.੦
7. ਹੱਦ ਮੌਜੂਦ ਨਹੀਂ ਹੈ ਕਿਉਂਕਿ ਜਦੋਂ xx ਅਤੇ yy ਦੋਵੇਂ ਸਿਫ਼ਰ ਵੱਲ ਵਧਦੇ ਹਨ, ਤਾਂ ਫਲਨ ln0,ln0 ਵੱਲ ਵਧਦੀ ਹੈ, ਜੋ ਕਿ ਅਨਿਸ਼ਚਿਤ ਹੈ (ਰਿਣ ਅਨੰਤ ਵੱਲ ਵਧਦੀ ਹੈ)।
8. (x0,y0)(x0,y0) 'ਤੇ ਕੇਂਦਰਿਤ ਹਰ ਖੁੱਲ੍ਹੀ ਡਿਸਕ RR ਦੇ ਅੰਦਰ ਅਤੇ RR ਦੇ ਬਾਹਰ ਬਿੰਦੂਆਂ ਨੂੰ ਸ਼ਾਮਲ ਕਰਦੀ ਹੈ।
9. ੦.੦੦.੦
10. ੦.੦੦੦.੦੦
11. ਹੱਦ ਮੌਜੂਦ ਨਹੀਂ ਹੈ।
12. ਹੱਦ ਮੌਜੂਦ ਨਹੀਂ ਹੈ। ਫਲਨ ਵੱਖ-ਵੱਖ ਮਾਰਗਾਂ 'ਤੇ ਦੋ ਵੱਖ-ਵੱਖ ਮੁੱਲਾਂ ਵੱਲ ਵਧਦੀ ਹੈ।
13. ਹੱਦ ਮੌਜੂਦ ਨਹੀਂ ਹੈ ਕਿਉਂਕਿ ਫਲਨ ਮਾਰਗਾਂ 'ਤੇ ਦੋ ਵੱਖ-ਵੱਖ ਮੁੱਲਾਂ ਵੱਲ ਵਧਦੀ ਹੈ।
14. ਫਲਨ f ਖੇਤਰ y>−x.y>−x. ਵਿੱਚ ਨਿਰੰਤਰ ਹੈ।
15. ਫਲਨ f ਸਮੂਹ R=x,y|x≠0,y≠0R=x,y|x≠0,y≠0 'ਤੇ ਨਿਰੰਤਰ ਹੈ।
16. ਫਲਨ (0,0)(0,0) 'ਤੇ ਅਸੰਤਤ ਹੈ ਕਿਉਂਕਿ (0,0)(0,0) ਡੋਮੇਨ ਵਿੱਚ ਨਹੀਂ ਹੈ।
17. ਫਲਨ (0,0).(0,0) 'ਤੇ ਅਸੰਤਤ ਹੈ। (0,0)(0,0) 'ਤੇ ਹੱਦ ਮੌਜੂਦ ਨਹੀਂ ਹੈ ਅਤੇ g(0,0)g(0,0) ਮੌਜੂਦ ਨਹੀਂ ਹੈ।
18. ਕਿਉਂਕਿ ਫਲਨ arctanxarctanx (−∞,∞),(−∞,∞) 'ਤੇ ਨਿਰੰਤਰ ਹੈ, g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y) ਉੱਥੇ ਨਿਰੰਤਰ ਹੈ ਜਿੱਥੇ z=xy2x+yz=xy2x+y xy-ਪਲੇਨxy-plane ਦੇ ਸਾਰੇ ਬਿੰਦੂਆਂ 'ਤੇ ਨਿਰੰਤਰ ਹੈ, ਸਿਵਾਏ ਜਿੱਥੇ y=−x.y=−x. ਹੈ। ਇਸ ਤਰ੍ਹਾਂ, g(x,y)=arctan(xy2x+y)g(x,y)=arctan(xy2x+y) ਕੋਆਰਡੀਨੇਟ ਪਲੇਨ ਦੇ ਸਾਰੇ ਬਿੰਦੂਆਂ 'ਤੇ ਨਿਰੰਤਰ ਹੈ, ਸਿਵਾਏ ਉਨ੍ਹਾਂ ਬਿੰਦੂਆਂ ਦੇ ਜਿੱਥੇ y=−x.y=−x. ਹੈ।
19. ਸਪੇਸ ਵਿੱਚ ਸਾਰੇ ਬਿੰਦੂ P(x,y,z)P(x,y,z)
20. ਗ੍ਰਾਫ ਬੇਅੰਤ ਵੱਧ ਜਾਂਦਾ ਹੈ ਜਦੋਂ x ਅਤੇ y ਦੋਵੇਂ ਸਿਫ਼ਰ ਵੱਲ ਵਧਦੇ ਹਨ।
21. ਏ.
22. ਬੀ. ਪੱਧਰੀ ਵਕਰਾਂ (0,0)(0,0) 'ਤੇ ਕੇਂਦਰਿਤ ਚੱਕਰ ਹਨ ਜਿਨ੍ਹਾਂ ਦਾ ਅਰਧ ਵਿਆਸ 9−c.9−c ਹੈ। ਸੀ. x2+y2=9−cx2+y2=9−c ਡੀ. z=3z=3 ਈ. {(x,y)∈ℝ2|x2+y2≤9}{(x,y)∈ℝ2|x2+y2≤9} ਐਫ. {z|0≤z≤3}{z|0≤z≤3}
23. ੧.੦੧.੦
24. f(g(x,y))f(g(x,y)) ਸਾਰੇ ਬਿੰਦੂਆਂ (x,y)(x,y) 'ਤੇ ਨਿਰੰਤਰ ਹੈ ਜੋ 2x−5y=0.2x−5y=0 ਰੇਖਾ 'ਤੇ ਨਹੀਂ ਹਨ।
2.02.0
Section 4.3 Exercises
∂ z ∂ y = −3 x + 2 y ∂ z ∂ y = −3 x + 2 y
The sign is negative.
The partial derivative is zero at the origin.
∂ z ∂ y = −3 sin ( 3 x ) sin ( 3 y ) ∂ z ∂ y = −3 sin ( 3 x ) sin ( 3 y )
∂ z ∂ x = 6 x 5 x 6 + y 4 ; ∂ z ∂ y = 4 y 3 x 6 + y 4 ∂ z ∂ x = 6 x 5 x 6 + y 4 ; ∂ z ∂ y = 4 y 3 x 6 + y 4
∂ z ∂ x = y e x y ; ∂ z ∂ y = x e x y ∂ z ∂ x = y e x y ; ∂ z ∂ y = x e x y
∂ z ∂ x = 2 sec 2 ( 2 x − y ) , ∂ z ∂ y = − sec 2 ( 2 x − y ) ∂ z ∂ x = 2 sec 2 ( 2 x − y ) , ∂ z ∂ y = − sec 2 ( 2 x − y )
f x ( 2 , −2 ) = 1 4 = f y ( 2 , −2 ) f x ( 2 , −2 ) = 1 4 = f y ( 2 , −2 )
∂ z ∂ x = − cos ( 1 ) ∂ z ∂ x = − cos ( 1 )
f x = 0 , f y = 0 , f z = 0 f x = 0 , f y = 0 , f z = 0
a. V(r,h)=πr2hV(r,h)=πr2h b. ∂V∂r=2πrh∂V∂r=2πrh c. ∂V∂h=πr2∂V∂h=πr2
f x y = 1 ( x − y ) 2 f x y = 1 ( x − y ) 2
∂ 2 z ∂ x 2 = 2 , ∂ 2 z ∂ y 2 = 4 ∂ 2 z ∂ x 2 = 2 , ∂ 2 z ∂ y 2 = 4
f x y y = f y x y = f y y x = 0 f x y y = f y x y = f y y x = 0
d 2 z d x 2 = − 1 2 ( e y − e − y ) sin x d 2 z d y 2 = 1 2 ( e y − e − y ) sin x d 2 z d x 2 + d 2 z d y 2 = 0 d 2 z d x 2 = − 1 2 ( e y − e − y ) sin x d 2 z d y 2 = 1 2 ( e y − e − y ) sin x d 2 z d x 2 + d 2 z d y 2 = 0
f x y z = 6 y 2 x − 18 y z 2 f x y z = 6 y 2 x − 18 y z 2
( 1 4 , 1 2 ) , ( 1 , 1 ) ( 1 4 , 1 2 ) , ( 1 , 1 )
( 0 , 0 ) , ( 0 , 2 ) , ( 3 , −1 ) , ( − 3 , −1 ) ( 0 , 0 ) , ( 0 , 2 ) , ( 3 , −1 ) , ( − 3 , −1 )
∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 = e x sin ( y ) − e x sin y = 0 ∂ 2 z ∂ x 2 + ∂ 2 z ∂ y 2 = e x sin ( y ) − e x sin y = 0
c 2 ∂ 2 z ∂ x 2 = e − t cos ( x c ) c 2 ∂ 2 z ∂ x 2 = e − t cos ( x c )
∂ f ∂ y = −2 x + 7 ∂ f ∂ y = −2 x + 7
∂ f ∂ x = y cos x y ∂ f ∂ x = y cos x y
∂ F ∂ θ = 6 , ∂ F ∂ x = 4 − 3 3 ∂ F ∂ θ = 6 , ∂ F ∂ x = 4 − 3 3
∂f∂x∂f∂x at (500,1000)=172.36,(500,1000)=172.36, ∂f∂y∂f∂y at (500,1000)=36.93(500,1000)=36.93
Section 4.4 Exercises
( 145 145 ) ( 12 i − k ) ( 145 145 ) ( 12 i − k )
Normal vector: i+j,i+j, tangent vector: i−ji−j
Normal vector: 7i−17j,7i−17j, tangent vector: 17i+7j17i+7j
Normal vector - 12 i + 12 j - k Tangent vector 0 i + j + 12 k or 0 i + j - 12 k Normal vector - 12 i + 12 j - k Tangent vector 0 i + j + 12 k or 0 i + j - 12 k
−36 x − 6 y − z = −39 −36 x − 6 y − z = −39
z = 0 z = 0
5 x + 4 y + 3 z − 22 = 0 5 x + 4 y + 3 z − 22 = 0
4 x − 5 y + 4 z = 0 4 x − 5 y + 4 z = 0
2 x + 2 y − z = 0 2 x + 2 y − z = 0
−2 ( x − 1 ) + 2 ( y − 2 ) − ( z − 1 ) = 0 −2 ( x − 1 ) + 2 ( y − 2 ) − ( z − 1 ) = 0
x = 20 t + 2 , y = −4 t + 1 , z = − t + 18 x = 20 t + 2 , y = −4 t + 1 , z = − t + 18
x = 0 , y = 0 , z = t x = 0 , y = 0 , z = t
x − 1 = 2 t ; y − 2 = −2 t ; z − 1 = t x − 1 = 2 t ; y − 2 = −2 t ; z − 1 = t
The differential of the function z(x,y)=dz=fxdx+fydyz(x,y)=dz=fxdx+fydy
Using the definition of differentiability, we have exyx≈x+y.exyx≈x+y.
Δz=2xΔx+3Δy+(Δx)2.Δz=2xΔx+3Δy+(Δx)2. (Δx)2→0(Δx)2→0 for small ΔxΔx and zz satisfies the definition of differentiability.
Δz≈1.185422Δz≈1.185422 and dz≈1.108.dz≈1.108. They are relatively close.
1616 cm3
Δz=Δz= exact change =0.6449,=0.6449, approximate change is dz=0.65.dz=0.65. The two values are close.
13 % or 0.13 13 % or 0.13
0.025 0.025
੦.੩% ੦.੩%
੨ਖ + ੧ ੪ਯ − ੧ ੨ਖ + ੧ ੪ਯ − ੧
੧/੨ਖ + ਯ + ੧/੪π − ੧/੨ ੧/੨ਖ + ਯ + ੧/੪π − ੧/੨
੩/੭ਖ + ੨/੭ਯ + ੬/੭ਜ਼ ੩/੭ਖ + ੨/੭ਯ + ੬/੭ਜ਼
ਜ਼=੦ਜ਼=੦
ਭਾਗ ੪.੫ ਅਭਿਆਸ
ਦਵ/ਦਤ = ੩ਤ^੨ − ੪ਤ^੨/(੧ − ਤ^੨) ਦਵ/ਦਤ = ੩ਤ^੨ − ੪ਤ^੨/(੧ − ਤ^੨)
∂ਵ/∂ਸ = ੯/੪ਸ – ੪/੬ਤ, ∂ਵ/∂ਸ = ੯/੪ਸ – ੪/੬ਤ, ∂ਵ/∂ਤ = –੪/੬ਸ + ੭/੪ਤ ∂ਵ/∂ਤ = –੪/੬ਸ + ੭/੪ਤ
∂ਫ਼/∂ਰ = ਰ ਸਾਈਨ (੨θ) ∂ਫ਼/∂ਰ = ਰ ਸਾਈਨ (੨θ)
ਦਫ਼/ਦਤ = ੨ਤ + ੪ਤ^੩ ਦਫ਼/ਦਤ = ੨ਤ + ੪ਤ^੩
ਦਫ਼/ਦਤ = −੧ ਦਫ਼/ਦਤ = −੧
ਦਫ਼/ਦਤ = ੧ ਦਫ਼/ਦਤ = ੧
ਦਵ/ਦਤ = ੨ਈ^{੨ਤ} ਦੋਵੇਂ ਹਾਲਤਾਂ ਵਿੱਚ
ਦਯੂ/ਦਤ = ੨ (π −੪) ਦਯੂ/ਦਤ = ੨ (π −੪)
ਦਯ/ਦਖ = −੩ਖ^੨ + ਯ^੨ / (੨ਖਯ) ਦਯ/ਦਖ = −੩ਖ^੨ + ਯ^੨ / (੨ਖਯ)
ਦਯ/ਦਖ = ਯ − ਖ − ਖ + ੨ਯ^੩ ਦਯ/ਦਖ = ਯ − ਖ − ਖ + ੨ਯ^੩
ਦਯ/ਦਖ = −ਯ / ਖ^੩ ਦਯ/ਦਖ = −ਯ / ਖ^੩
ਦਯ/ਦਖ = −ਯ ਈ^{ਖਯ} / (ਖ ਈ^{ਖਯ} + ਈ^ਯ (੧ + ਯ)) ਦਯ/ਦਖ = −ਯ ਈ^{ਖਯ} / (ਖ ਈ^{ਖਯ} + ਈ^ਯ (੧ + ਯ))
ਦਜ਼/ਦਤ = ੪੨ਤ^{੧੩} ਦਜ਼/ਦਤ = ੪੨ਤ^{੧੩}
ਦਜ਼/ਦਤ = −੧੦/੩ਤ^{੭/੩} × ਈ^{੧ − ਤ^{੧੦/੩}} ਦਜ਼/ਦਤ = −੧੦/੩ਤ^{੭/੩} × ਈ^{੧ − ਤ^{੧੦/੩}}
∂ਜ਼/∂ਯੂ = −੨ਸਾਈਨਯੂ੩ਸਾਈਨਵ ∂ਜ਼/∂ਯੂ = −੨ਸਾਈਨਯੂ੩ਸਾਈਨਵ ਅਤੇ ∂ਜ਼/∂ਵ = −੨ਕੋਸਯੂਕੋਸਵ੩ਸਾਈਨ^੨ਵ ∂ਜ਼/∂ਵ = −੨ਕੋਸਯੂਕੋਸਵ੩ਸਾਈਨ^੨ਵ
∂ਜ਼/∂ਰ = ੩ਈ^੩, ∂ਜ਼/∂ਰ = ੩ਈ^੩, ∂ਜ਼/∂θ = (੨−੪/੩)ਈ^੩ ∂ਜ਼/∂θ = (੨−੪/੩)ਈ^੩
∂ਵ/∂ਤ = −੧੨ਤ ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ) + ੪ਤ (੧ − ੩ਤ) ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ) − ੪ (੧ − ੩ਤ) ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ) ∂ਵ/∂ਤ = −੧੨ਤ ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ) + ੪ਤ (੧ − ੩ਤ) ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ) − ੪ (੧ − ੩ਤ) ਈ^{੧ − ਤ} ਕੋਸ (੧ − ੩ਤ) ਈ^{੧ − ਤ} (−੪ਤ)
ਫ਼(ਤਖ,ਤਯ) = ਤ^੨ਖ^੨ + ਤ^੨ਯ^੨ = ਤ^੨ਫ਼(ਖ,ਯ), ਫ਼(ਤਖ,ਤਯ) = ਤ^੨ਖ^੨ + ਤ^੨ਯ^੨ = ਤ^੨ਫ਼(ਖ,ਯ), ∂ਫ਼/∂ਯ = ਖ^੧/੨(ਖ^੨+ਯ^੨)^{−੧/੨}×੨ਖ + ਯ^੧/੨(ਖ^੨+ਯ^੨)^{−੧/੨}×੨ਯ = ੧/ਫ਼(ਖ,ਯ) ∂ਫ਼/∂ਯ = ਖ^੧/੨(ਖ^੨+ਯ^੨)^{−੧/੨}×੨ਖ + ਯ^੧/੨(ਖ^੨+ਯ^੨)^{−੧/੨}×੨ਯ = ੧/ਫ਼(ਖ,ਯ)
V ' = 4 π V ' = 4 π
d V d t = 1066 π 3 cm 3 / min d V d t = 1066 π 3 cm 3 / min
d A d t = 12 in . 2 / min d A d t = 12 in . 2 / min
2 ° C/sec 2 ° C/sec
∂ u ∂ r = ∂ u ∂ x ( ∂ x ∂ w ∂ w ∂ r + ∂ x ∂ t ∂ t ∂ r ) + ∂ u ∂ y ( ∂ y ∂ w ∂ w ∂ r + ∂ y ∂ t ∂ t ∂ r ) + ∂ u ∂ z ( ∂ z ∂ w ∂ w ∂ r + ∂ z ∂ t ∂ t ∂ r ) ∂ u ∂ r = ∂ u ∂ x ( ∂ x ∂ w ∂ w ∂ r + ∂ x ∂ t ∂ t ∂ r ) + ∂ u ∂ y ( ∂ y ∂ w ∂ w ∂ r + ∂ y ∂ t ∂ t ∂ r ) + ∂ u ∂ z ( ∂ z ∂ w ∂ w ∂ r + ∂ z ∂ t ∂ t ∂ r )
Section 4.6 Exercises
- 2 6 - 2 - 2 6 - 2
−1 −1
2 6 2 6
3 3
−1.0 −1.0
22 25 22 25
2 3 2 3
− 2 ( x + y ) 2 ( x + 2 y ) 2 − 2 ( x + y ) 2 ( x + 2 y ) 2
e x ( y + 3 ) 2 e x ( y + 3 ) 2
1 + 2 3 2 ( x + 2 y ) 1 + 2 3 2 ( x + 2 y )
〈 5 , 4 , 3 〉 〈 5 , 4 , 3 〉
−320 −320
3 11 3 11
31 255 31 255
2 3 i + 3 j 2 3 i + 3 j
2 i + 2 j + 2 k 2 i + 2 j + 2 k
1.6 ( 10 19 ) 1.6 ( 10 19 )
5 2 99 5 2 99
2 , 〈 1 , -1 〉 2 , 〈 1 , -1 〉
13 2 , 〈 −3 , −2 〉 13 2 , 〈 −3 , −2 〉
a. x+y+z=3,x+y+z=3, b. x−1=y−1=z−1x−1=y−1=z−1
a. x+y−z=1,x+y−z=1, b. x−1=y=−zx−1=y=−z
a. 323,323, b. 〈38,6,12〉,〈38,6,12〉, c. 24062406
〈 u , v 〉 = 〈 π cos ( π x ) sin ( 2 π y ) , 2 π sin ( π x ) cos ( 2 π y ) 〉 〈 u , v 〉 = 〈 π cos ( π x ) sin ( 2 π y ) , 2 π sin ( π x ) cos ( 2 π y ) 〉
Section 4.7 Exercises
( 2 3 , 4 ) ( 2 3 , 4 )
(0,0)(0,0) (115,115)(115,115)
Maximum at (4,−1,8)(4,−1,8)
Relative minimum at (0,0,1)(0,0,1)
The second derivative test fails. Since x2y2>0x2y2>0 for all x and y different from zero, and x2y2=0x2y2=0 when either x or y equals zero (or both), then the absolute minimum occurs at (0,0).(0,0).
f(−2,−32)=−6f(−2,−32)=−6 is a saddle point.
f(0,0)=0;f(0,0)=0; (0,0,0)(0,0,0) is a saddle point.
The only critical point(s) are where both fx=0fx=0 and fy=0fy=0 and that is f(0, 0)=9f(0, 0)=9. However, since fxxfxy-fxy2=0fxxfxy-fxy2=0 at the critical point, the Second Derivative Test fails. By graphing, we can see that this point is a local maximum.
Relative minimum located at (2,6).(2,6).
(1,−2)(1,−2) is a saddle point.
(2,1)(2,1) and (−2,1)(−2,1) are saddle points; (0,0)(0,0) is a relative minimum.
(−1,0)(−1,0) is a relative maximum.
(0,0)(0,0) is a saddle point.
The relative maximum is at (40,40).(40,40).
(14,12)(14,12) is a saddle point and (1,1)(1,1) is the relative minimum.
A saddle point is located at (0,0).(0,0).
There is a saddle point at (π,π),(π,π), local maxima at (π2,π2)and(3π2,3π2),(π2,π2)and(3π2,3π2), and local minima at (π2,3π2)and(3π2,π2).(π2,3π2)and(3π2,π2).
(0,1,0)(0,1,0) ਘੱਟੋ-ਘੱਟ ਹੈ ਅਤੇ (0,−2,9)(0,−2,9) ਵੱਧ ਤੋਂ ਵੱਧ ਹੈ।
(0,1,−1)(0,1,−1) ਵਿਖੇ ਇੱਕ ਨਿਰਪੇਖ ਘੱਟੋ-ਘੱਟ ਅਤੇ (0,−1,1).(0,−1,1) ਵਿਖੇ ਇੱਕ ਨਿਰਪੇਖ ਵੱਧ ਤੋਂ ਵੱਧ ਹੈ।
(5,0,0), (−5,0,0)
18 ਇੰਚ ਗੁਣਾ 36 ਇੰਚ ਗੁਣਾ 18 ਇੰਚ।
(12,1,45/4)
x=3 ਅਤੇ y=6
V=64,000π≈20,372V=64,000π≈20,372 cm3
ਭਾਗ 4.8 ਅਭਿਆਸ
ਵੱਧ ਤੋਂ ਵੱਧ: 233,233, ਘੱਟ ਤੋਂ ਘੱਟ: −233
ਵੱਧ ਤੋਂ ਵੱਧ: (22,0,2), ਘੱਟ ਤੋਂ ਘੱਟ: (−22,0,−2)
ਵੱਧ ਤੋਂ ਵੱਧ: 32, ਘੱਟ ਤੋਂ ਘੱਟ = 12
ਵੱਧ ਤੋਂ ਵੱਧ: f(322,22)=24, f(−322,−22)=24; ਘੱਟ ਤੋਂ ਘੱਟ: f(−322,22)=−24, f(322,−22)=−24
ਵੱਧ ਤੋਂ ਵੱਧ: 211211 f(211,611,−211) ਵਿਖੇ; ਘੱਟ ਤੋਂ ਘੱਟ: −211 f(−211,−611,211) ਵਿਖੇ
2.0
19 2
(123, −123)
f(1,2)=5
f(1/3,1/3,1/3)=1/3
ਘੱਟ ਤੋਂ ਘੱਟ: f(2,3,4)=29
ਵੱਧ ਤੋਂ ਵੱਧ ਆਇਤਨ 44 ft³ ਹੈ। ਮਾਪ 1×2×2 ft ਹਨ।
ਸਤ੍ਹਾ x²-2xy+y²-x+y=0 ਉੱਤੇ ਬਿੰਦੂ (1,2,-3) ਦੇ ਸਭ ਤੋਂ ਨੇੜੇ ਦਾ ਬਿੰਦੂ 3/2,3/2,-3 ਹੈ।
1.0
3
(2/5, 19/5)
12
Roughly 3365 watches at the critical point (80,60)(80,60)
Review Exercises
True, by Clairaut’s theorem
False
Answers may vary
Does not exist
Continuous at all points on the x,y-plane,x,y-plane, except where x2+y2>4.x2+y2>4.
∂ u ∂ x = 4 x 3 − 3 y , ∂ u ∂ y = −3 x , d x d t = 2 , d y d t = 3 t 2 , d u d t = 40 t 3 ∂ u ∂ x = 4 x 3 − 3 y , ∂ u ∂ y = −3 x , d x d t = 2 , d y d t = 3 t 2 , d u d t = 40 t 3
hxx(x,y,z)=6xe2yz,hxx(x,y,z)=6xe2yz, hxy(x,y,z)=6x2e2yz,hxy(x,y,z)=6x2e2yz, hxz(x,y,z)=−3x2e2yz2,hxz(x,y,z)=−3x2e2yz2, hyx(x,y,z)=6x2e2yz,hyx(x,y,z)=6x2e2yz, hyy(x,y,z)=4x3e2yz,hyy(x,y,z)=4x3e2yz, hyz(x,y,z)=−2x3e2yz2,hyz(x,y,z)=−2x3e2yz2, hzx(x,y,z)=−3x2e2yz2,hzx(x,y,z)=−3x2e2yz2, hzy(x,y,z)=−2x3e2yz2,hzy(x,y,z)=−2x3e2yz2, hzz(x,y,z)=2x3e2yz3hzz(x,y,z)=2x3e2yz3
z = x − 2 y + 5 z = x − 2 y + 5
dz=4dx−dy,dz=4dx−dy, dz(0.1,0.01)=0.39,dz(0.1,0.01)=0.39, Δz=0.432Δz=0.432
3 85 , 〈 27 , 6 〉 3 85 , 〈 27 , 6 〉
∇ f ( x , y ) = − x + 2 y 2 2 x 2 y i + ( 1 x − 1 x y 2 ) j ∇ f ( x , y ) = − x + 2 y 2 2 x 2 y i + ( 1 x − 1 x y 2 ) j
maximum: 1633,1633, minimum: −1633−1633
2.32282.3228 cm3