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1 0 D f x, y dx dy Sketch the region iterated integral with reversed order of integration. and express the double integral as an D 63–67 Use geometry or symmetry, or both, to evaluate the double integral. CAS yy x 2 dA D x, y 0 y s9 x 2 , 63. D 15.4 Double Integrals in Polar Coordinates Suppose that we want to evaluat... |
enclosed by both of the cardioids r 1 cos 1 x and r 1 cos 17. The region inside the circle x 2 y 2 1 circle x 12 y 2 1 and outside the 18. The region inside the cardioid r 3 cos circle r 1 cos and outside the 19–27 Use polar coordinates to find the volume of the given solid. x 2 y 2 4 19. Under the cone and above the di... |
a occupying the y (0, 2) y=2-2x ” , ’ 11 16 3 8 D 0 (1, 0) x FIGURE 5 x My m 1 m yy D x x, y dA y Mx m 1 m yy D y x, y dA where the mass m is given by m yy x, y dA D v EXAMPLE 2 0, 0 1, 0 , , and Find the mass and center of mass of a triangular lamina with vertices 0, 2 x, y 1 3x y if the density function is . SOLUTION... |
l learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_15_ch15_p1026-1035.qk_97817_15_ch15_p1026-1035 11/8/10 3:35 PM Page 1034 1034 CHAPTER 15 MULTIPLE INTEGRALS Suppose X is a random variable w... |
2 ) 2, Y 1 Y and (ii) . 29. Suppose function X and Y are random variables with joint density f x, y 0.1e0.5x0.2y 0 if x 0, y 0 otherwise (a) Verify that (b) Find the following probabilities. is indeed a joint density function. f (i) PY 1 (ii) (c) Find the expected values of PX 2, Y 4 and X Y . 30. (a) A lamp has two b... |
linder x 2 y 2 z2 a 2 xy and above the -plane that lies within the 12. The part of the sphere z x 2 y 2 paraboloid x 2 y 2 z2 4z that lies inside the 13–14 Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral.... |
1044 1044 CHAPTER 15 MULTIPLE INTEGRALS z 0 E D y x=u¡(y, z) x x=u™(y, z) FIGURE 7 A type 2 region where, this time, face is D x u1y, z E yz is the projection of onto the x u2y, z , the front surface is , and we have -plane (see Figure 7). The back sur- 10 yyy f x, y, z dV yy E D Finally, a type 3 region is of the for... |
ibe as a z x E y E x, y, z 1 y 1, y2 x 1, 0 z x Then, if the density is x, y, z , the mass is m yyy E dV y1 1 y1 y 2 y x 0 dz dx dy y1 1 y1 y 2 x dx dy y1 1 x 2 2 x1 xy 2 dy y1 2 1 1 y 4 dy y1 0 1 y 4 dy y y 5 5 1 0 4 5 E Because of the symmetry of y 0 Mxz 0 . The other moments are and about the and therefore xz -plane... |
volume enclosed by the hypersphere formulas for x sinnx dx in x cosnx dx or .) . (Use only trigonometric substitution and the reduction 4. Use an -tuple integral to find the volume enclosed by a hypersphere of radius in . [Hint: The formulas are different for even and odd.] n n n n r n -dimensional space 15.8 Triple Int... |
uate xxx z x 2 y 2 , where E z dV E and the plane is enclosed by the paraboloid z 4 . 19. Evaluate , where octant that lies under the paraboloid x y z dV E is the solid in the first E z 4 x 2 y 2 . xxx 20. Evaluate and x 2 y 2 9 xxx E x dV z x y 5 . E , where is enclosed by the planes x 2 y 2 4 and by the cylinders z 0 ... |
ting x sin cos y sin sin z cos using the appropriate limits of integration, and replacing illustrated in Figure 8. dV by 2 sin d d d . This is z 0 x ∏ sin ˙ d¨ d∏ ∏ ˙ ∏ d˙ d¨ y FIGURE 8 Volume element in spherical coordinates: dV=∏@ sin ˙ d∏ d¨ d˙ This formula can be extended to include more general spherical regions s... |
Find the great-circle distance 118.25 360 1 1 N, long. N, long. 34.06 and the is P models for tumors. The “bumpy sphere” with n 5 volume it encloses. is shown. Use a computer algebra system to find the have been used as and m 6 CAS 45. The surfaces APPLIED PROJECT ROLLER DERBY 1063 sx 2 y 2 z2 ex 2y 2z 2 dx dy dz 2 46. ... |
by x tu, v and 7 y hu, v is x u y u x v y v x, y u With this notation we can use Equation 6 to give an approximation to the area A of R : A x, y u v u, v 8 where the Jacobian is evaluated at in the S Next we divide a region . (See Figure 6.) -plane Rij . u0, v0 uv -plane into rectangles Sij and call their images in the... |
om 97817_15_ch15_p1066-1078.qk_97817_15_ch15_p1066-1078 11/26/10 11:12 AM Page 1072 1072 CHAPTER 15 MULTIPLE INTEGRALS 18. 19. ; 20. R x 2 xy y 2 dA xx by the ellipse x s2 u s23 v y s2 u s23 v , R is the region bounded x 2 xy y 2 2 ; , where xx R xy dA by the lines xy 3 ; R , where is the region in the first quadrant bo... |
olume of the given solid. 29. Under the paraboloid R 0, 2 1, 4 z x 2 4y 2 and above the rectangle 30. Under the surface z x 2 y xy -plane with vertices and above the triangle in the 4, 0 1, 0 2, 1 , , and 31. The solid tetrahedron with vertices and 2, 2, 0 0, 0, 0 0, 0, 1 0, 2, 0 , , , 32. Bounded by the cylinder y z 3... |
wish I had paid more attention in calculus class when we were studying parametric surfaces. It would sure have helped me today.” In this chapter we study the calculus of vector fields. (These are functions that assign vectors to points in space.) In particular we define line integrals (which can be used to find the work d... |
rd the origin, and the unit vector in this direction is could be the mass of the earth and the origin x x, y, z . . The gravitational force exerted on this second object acts is the gravitational constant. (This is r 2 x2 r x , so 3 be M M m G x x Therefore the gravitational force acting on the object at x x, y, z is 3... |
.) Now if f , we evaluate at the point f C , and we assume that is a . See Section 13.3.] If we of equal width and we let C subarcs into Pi*xi*, yi* ith is any function of two vari, multiply by n in the xi*, yi* divide n i1 f xi*, yi* si which is similar to a Riemann sum. Then we take the limit of these sums and make t... |
ify, using the denotes the line segment from 5, 3 0, 2 C1 to curve. If parametrization x 5t y 2 5t 0 t 1 that y C1 y 2 dx x dy 5 6 Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed fr... |
dt t 4 4 5t 7 7 1 0 27 28 Finally, we note the connection between line integrals of vector fields and line integrals 3 is given in component form by the equa- F of scalar fields. Suppose the vector field on tion F P i Q j R k . We use Definition 13 to compute its line integral along C : y C F dr yb a Frt rt dt yb a yb a P ... |
as deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/9/10 9:06 AM Pa... |
ion. A simply-connected region in the plane is a connected region such that every simple closed curve in . Notice from Figure 7 that, intuitively speaking, a simply-connected region contains no hole and can’t consist of two separate pieces. encloses only points that are in D D D In terms of simply-connected regions, we... |
ght to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_16_ch16_p1098-1107.qk_97817_16_ch16_p1098-1107 11/9/10 9:06 AM Page 1107 SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS 1107 16. 17. 18. Fx, y, z y2z 2xz2 i 2 xyz j xy 2 2x 2z k , C x st... |
x 4 dx xy dy Evaluate 0, 0 1, 0 , from to , where 1, 0 C to is the triangular curve consisting of the 0, 1 0, 1 , and from 0, 0 to . y (0, 1) y=1-x C D (0, 0) (1, 0) x FIGURE 4 Instead of using polar coordinates, we could simply use the fact that and write is a disk of radius 3 D 4 dA 4 32 36 yy D SOLUTION Although th... |
erall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_16_ch16_p1108-1117.qk_97817_16_ch16_p1108-1117 11/9/10 9:16 AM Page 1114 1114 CHAPTER 16 VECTOR CALCULUS 4. x C x 2y 2 dx xy dy C y x 2 ... |
y 2z3 fyx, y, z 2xyz3 fzx, y, z 3xy 2z2 Integrating 5 with respect to , we obtain x 8 f x, y, z xy 2z3 t y, z Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or ... |
df.com 97817_16_ch16_p1118-1127.qk_97817_16_ch16_p1118-1127 11/9/10 9:16 AM Page 1122 1122 CHAPTER 16 VECTOR CALCULUS 23–29 Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If is a scalar field and , and are vector fields, then are defined by F G F G f F , f GF , f Fx, y, ... |
f . So there are ua vb scalars u and such that A P0 P . (Figure 6 illustrates how this works, by v means of the Parallelogram Law, for the case where and are positive. See also Exercise 46 in Section 12.2.) If r is the position vector of P, then to by moving a certain P0 P b u v r OP0 P0P A A r0 u a vb So the vector eq... |
ch that meet at can be approximated v rv* by vectors. These vectors, in turn, can be approximated by the vectors because partial derivatives can be approximated by difference quotients. So we approxiu ru* . This parallelogram by the parallelogram determined by the vectors mate is shown in Figure 15(b) and lies in the t... |
surface planes x 0 x 1 z 0 , , y 4x z2 , and z 1 that lies between the 48. The helicoid (or spiral ramp) with vector equation , ru, v u cos v i u sin v j v k 0 u 1 0 v , 49. The surface with parametric equations 2v u2 y uv , , 50. The part of the sphere x 2 y 2 a 2 cylinder x 2 y2 z 2 b2 0 a b , where that lies inside ... |
uate xx S y dS S , where is the surface . , , SOLUTION Since y Formula 4 gives z x 1 and z y 2y y1 z x 2 z y 2 dA yy y dS yy S D y1 0 y2 0 ys1 1 4y 2 dy dx y1 0 dx s2 y2 0 ys1 2y 2 dy s2 ( 1 4) 2 3 1 2y 232]0 2 13s2 3 S S1 S2, . . . , is a piecewise-smooth surface, that is, a finite union of smooth surfaces , If S is de... |
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xx is the surface , where S , . 34. Find the exact value of z xy yz dS . 35. Find the value of S where lies above the x y -plane. S x 2 y 2z2 dS is the part of the paraboloid correct to four decimal places, z 3 2x 2 y 2 that , xx CAS 36. Find the flux of xy -plane and between the planes across the part of the cylinder ... |
the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_16_ch16_p1148-11... |
s could help to explain and predict physical phenomena in electricity and magnetism and fluid flow. The basic facts of the story are given in the margin notes on pages 1109 and 1147. Write a report on the historical origins of Green’s Theorem and Stokes’ Theorem. Explain the similarities and relationship between the theo... |
rves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_16_ch16_p1148-1157.qk_97817_16_ch16_p1148-1157 11/9/10 9:26 AM Page 1156 1156 CHAPTER 16 VECTOR CALCULUS n™ n¡ S™ S¡ _n¡ FIGURE 3 Although we have proved the Divergence Theorem only fo... |
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oid with upward orientation , where is the part of the paraboloid z x 2 y 2 z 4 , where S is the part of the plane that lies inside the cylinder x 2 y 2 4 , where x 2 y 2 z2 4 Fx, y, z x z i 2y j 3x k and with outward orientation S is Fx, y and z x 2 y 2 below the plane is the S z 1 31. Verify that Stokes’ Theorem is t... |
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erves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_17_ch17_p1165-1175.qk_97817_17_ch17_p1165-1175 11/9/10 10:28 AM Page 1171 SECTION 17.1 SECOND-ORDER LINEAR EQUATIONS 1171 The solution to Example 6 is graphed in Figure 5. It appears ... |
if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_17_ch17_p1165-1175.qk_97817_17_ch17_p1165-1175 11/9/10 10:28 AM Page 1175 SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS 1175 Then yp A sin x B cos x yp A cos x B sin x so substitution in the differential equation gives A cos x B sin x A sin... |
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ric circuits that contain a resistor and inductor (see Figure 5 in Section 9.3 or Figure 4 in Section 9.5) or a resistor and capacitor (see Exercise 29 in Section 9.5). Now that we know how to solve second-order linear equations, we are in a position to analyze (supplied by a battery the circuit shown in Figure 7. It c... |
not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to ... |
ed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97817_17_ch17_p1186-1194.qk_97817_17_ch17_p1186-1194 11/9/10 10:29 AM Page 119... |
not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.www.EngineeringEBooksPdf.com 97879_Apdx7eMV_Apdx7eMV_pA02-A12.qk_97879_Apdx7eMV_Apdx7eMV_pA02-A12 11/10/10 1:04 PM Page A4 A4 APPENDIX F PROOF... |
ple 4 we have 1 i s2 cos s3 i 2cos and So, by Equation 1, i sin i sin 4 4 6 6 1 i(s3 i) 2s2 cos 2s2 cos 4 12 6 i sin i sin 4 12 6 FIGURE 8 This is illustrated in Figure 8. Copyright 2010 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some... |
ctangle described by and 25. 2 y 3 . y , starting and ending at to 27. y 1 (0, 1) t=1 (_1, 0) t=0 x (0, _1) t=_1 29. π _4 4 0, 2 1 x 4 t= 1 2 t=0 1 x _π x 2 5t, y 7 8t 0 t 1 x 2 cos t, y 1 2 sin t, 0 t 2 , x 2 cos t, y 1 2 sin t, 0 t 6 x 2 cos t, y 1 2 sin t, 2 t 32 31. (b) 33. (a) (b) (c) Copyright 2010 Cengage Learni... |
e 19 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES A19 43. All curves have the vertical asymptote curve bulges to the right. At For (0, 0). For , it bulges to the left. At , there is a loop. 1 c 0 c 0 c 1 , the curve is the line x 1 . For c 1 , the x 1 . there is a cusp at c 0 45. 1, 0, 3, 0 y 2œ„2 (1, 0) 3 x 3 0 2œ„2 4... |
nt rights restrictions require it.www.EngineeringEBooksPdf.com 97879_Ans7eMV_Ans7eMV_pA013-A023.qk_97879_Ans7eMV_Ans7eMV_pA013-A023 11/10/10 3:03 PM Page 23 APPENDIX H ANSWERS TO ODD-NUMBERED EXERCISES A23 y 8 25. A half-space consisting of all points to the left of the plane 27. All points on or between the horizontal... |
whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if su... |
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dy 47. xln 2 0 x2 e y f x, y dx dy y (0, 1) (1, 1) y=x 0 x y 1 0 y ln 2 0 y=cos x or x=cos_1y x π 2 y=ln x or x=e† x=2 y=0 1 2 x 49. 57. 65. 1 6 e 9 1 1 3 ln 9 51. 53. 16e116 xx Q ex 2y 2 2 dA 16 a 2b 3 2 ab 2 67. a 2b 1 3 (2 s2 1) 55. 1 59. 3 4 63. 9 EXERCISES 15.4 N PAGE 1026 x 32 0 x4 0 f r cos , r sin r dr d ¨= 3π ... |
tant 0 y 1 1 _1 0 x 1 0z √ constant _1 _1 0 y 1 1 u constant _1 0 x 9 2 17. III 13. IV 15. II x u, y v u, z v 19. y y, z z, x s1 y 2 1 21. x 2 sin cos y 2 sin sin , , 23. z 2 cos 0 4 0 2 [or x x, y y, z s4 x 2 y 2, x 2 y 2 2] 25 cos z 4 sin 0 x 5, 0 2 , x x, y ex cos , 29. z ex sin 01 2 x 31. (a) Direction reverses (b)... |
eeringEBooksPdf.com 97879_Index7eMV_Index7eMV_pA043-A050.qk_97879_Index7eMV_Index7eMV_pA043-A050 11/10/10 2:56 PM Page A44 A44 INDEX composition of functions, continuity of, 922 computer algebra system, 662 for integration, 775 computer algebra system, graphing with, function of two variables, 906 level curves, 910 par... |
ere, 1051 hypocycloid, 668 (standard basis vector), 820 i (imaginary number), A5 i ideal gas law, 938 image of a point, 1065 image of a region, 1065 implicit differentiation, 929, 952 Implicit Function Theorem, 953, 954 incompressible velocity field, 1119 increasing sequence, 720 increment, 945 independence of path, 110... |
dinates, 1052 conversion to spherical coordinates, 1057 recursion relation, 1189 reflection property of an ellipse, 697 of a hyperbola, 702 region connected, 1101 open, 1101 plane, of type I or II, 1013, 1014 simple plane, 1109 simple solid, 1153 simply-connected, 1102 solid (of type 1, 2, or 3), 1042, 1043, 1044 remain... |
x c 0 f x tx f x tx f x tx f x tx tx f x (Product Rule) f tx f tx tx (Chain Rule) Exponential and Logarithmic Functions 9. 11. d dx d dx e x e x ln x 1 x 2. 4. 6. 8. 10. 12. d dx d dx d dx d dx d dx d dx cf x c f x f x tx f x tx f x tx tx f x f x tx tx2 x n nx n1 (Power Rule) (Quotient Rule) a x a x ln a loga x 1 x ln ... |
normally would not call the result a box. If we were to allow these values, what would the corresponding volumes be? Does that make sense?) EXAMPLE 1.3.2 Circle of radius r centered at the origin The equation for this circle is usually given in the form x2 + y2 = r2. To write the equation in the form √ y = f (x) we so... |
eplacing x by x/0.5 = x/(1/2) = 2x has the effect of contracting toward the y-axis by a factor of 2. If A is negative, we dilate by a factor of |A| and then flip about the y-axis. Thus, replacing x by −x has the effect of taking the mirror image of the −x, which has domain graph with respect to the y-axis. For example, th... |
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ond from t = 2 to t = 2.5 the ball dropped 80.4 − 69.375 = 11.025 meters, at an average speed of 11.025/(1/2) = 22.05 meters per second; this should be a better estimate of the speed at t = 2. So it’s clear now how to get better and better approximations: compute average speeds over shorter and shorter time intervals. ... |
guarantee that −19.6 − 4.9∆x is in (−19.6 − ϵ, −19.6 + ϵ)? If ∆x is positive, we need: −19.6 − 4.9∆x > −19.6 − ϵ −4.9∆x > −ϵ ∆x < −ϵ/ − 4.9 ∆x < ϵ/4.9 So if I pick any number δ that is less than ϵ/4.9, the algebra tells me that whenever ∆x < δ then ∆x < ϵ/4.9 and so −19.6 − 4.9∆x is within ϵ of −19.6. (This is exactly... |
From this point of view it makes sense to ask what happens to the height of the function as x approaches 1. Of course, as we’ve already seen, we’re primarily interested in limits that aren’t so easy, namely, limits in which a denominator approaches zero. There are a handful of algebraic tricks that work on many of the... |
hat come together at the origin. We can summarize this as ( y′ = if x > 0; 1 −1 if x < 0; undefined if x = 0. EXAMPLE 2.4.5 Discuss the derivative of the function y = x2/3, shown in figure 2.4.1. We will later see how to compute this derivative; for now we use the fact that y′ = (2/3)x−1/3. Visually this looks much like ... |
is differentiable if is differentiable at every point (excluding endpoints and isolated points in the domain of f ) in the domain of f . Take note that, for technical reasons not discussed here, both of these definitions exclude endpoints and isolated points in the domain from consideration. We close with a useful theore... |
t at time t is given by f (t) = −49t2/10 + 5t + 10. Find a function giving the speed of the object at time t. The acceleration of an object is the rate at which its speed is changing, which means it is given by the derivative of the speed function. Find the acceleration of the object at time t. ⇒ 10. Let f (x) = x3 and... |
more complicated, but it too looks something like a derivative. The denominator, g(x + ∆x) − g(x), is a change in the value of g, so let’s abbreviate it as ∆g = g(x + ∆x) − g(x), which also means g(x + ∆x) = g(x) + ∆g. This gives us lim ∆x→0 f (g(x) + ∆g) − f (g(x)) ∆g . As ∆x goes to 0, it is also true that ∆g goes t... |
n the figure, x is the standard location of the angle π/6, that is, the length of the arc from (1, 0) to A is π/6. The angle y in the picture is −π/6, because the distance from (1, 0) to B along the circle is also π/6, but in a clockwise direction. Now the fundamental trigonometric definitions are: the cosine of x and th... |
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tail, like “Which exponent?” and “Exponent of what?”) EXAMPLE 4.6.1 What is the value of log10(1000)? The “10” tells us the appropriate number to use for the base of the exponential function. The logarithm is the exponent, so the question is, what exponent E makes 10E = 1000? If we can find such an E, then log10(1000) =... |
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stance between two points, we can write down an implicit equation for the ellipse: p (x − x1)2 + (y − y1)2 + p (x − x2)2 + (y − y2)2 = 2a. Then we can use implicit differentiation to find the slope of the ellipse at any point, though the computation is rather messy. EXAMPLE 4.8.4 We have already justified the power rule b... |
.............................................................................................................................................................................................................................................................. −π/2 π/2 ........................................................... |
ncy in applications, and are quite similar in many respects to the trigonometric functions. This is a bit surprising given our initial definitions. DEFINITION 4.11.1 The hyperbolic cosine is the function cosh x = ex + e−x 2 , and the hyperbolic sine is the function sinh x = ex − e−x 2 . Notice that cosh is even (that is... |
maximum or minimum can occur where the derivative does not exist, and so forget to check whether the derivative exists everywhere. You might also assume that any place that the derivative is zero is a local maximum or minimum point, but this is not true. A portion of the graph of f (x) = x3 is shown in figure 5.1.2. Th... |
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lt to identify and we will ignore them. p If the domain of the function does not extend out to infinity, we should also ask what happens as x approaches the boundary of the domain. For example, the function r2 − x2 has domain −r < x < r, and y becomes infinite as x approaches y = f (x) = 1/ either r or −r. In this case w... |
.................................................................. −1 1 2 −2 Figure 6.1.2 f (x) = x3 − x √ √ 3/3 and a global minimum at x = √ sidered. Is the lone local maximum a global maximum? Here we must look more closely at the graph. We know that on the closed interval [− 3/3] there is a global maximum at x = − ... |
“really” there is only one variable. So our next step is to find the relationship and use it to solve for one of the variables in terms of the other, so as to have a function of only one variable to maximize. In this problem, the condition is apparent in the figure: the upper corner of the triangle, whose coordinates are... |
similar triangles to get an equation relating h and r.) ⇒ 17. In example 6.1.12, what happens if w ≥ v (i.e., your speed on sand is at least your speed on the road)? ⇒ 18. A container holding a fixed volume is being made in the shape of a cylinder with a hemispherical top. (The hemispherical top has the same radius as ... |
entation of the situation, as in figure 6.2.1. Because the plane is in level flight directly away from you, the rate at which x changes is the speed of the plane, dx/dt = 500. The distance between you and the plane is y; it is dy/dt that we wish to know. By the Pythagorean Theorem we know that x2 + 9 = y2. 6.2 Related Ra... |
ate of 0.6 m/sec. How fast is the top sliding down the wall when the foot of the ladder is 5 m from the wall? ⇒ 4. A ladder 13 meters long rests on horizontal ground and leans against a vertical wall. The top of the ladder is being pulled up the wall at 0.1 meters per second. How fast is the foot of the ladder approach... |
As the blades are closed (i.e., the angle θ in the diagram is decreased), the distance x between A and C increases, cutting the paper. a. Express x in terms of a, θ, and β. b. Express dx/dt in terms of a, θ, β, and dθ/dt. c. Suppose that the distance a is 20 cm, and the angle β is 5 ◦ is decreasing at 50 deg/sec. At t... |
....................................................................................................................................... 1 1.5 Figure 6.3.2 y = tan x and y = 2x on the left, y = tan x − 2x on the right. Exercises 6.3. You may want to use this Sage worksheet. 1. Approximate the fifth root of 7, using x0 = ... |
(b) − f (a) b − a , and consider a new function g(x) = f (x)−m(x−a)−f (a). Proof. Let m = We know that g(x) has a derivative everywhere, since g′(x) = f ′(x) − m. We can compute g(a) = f (a) − m(a − a) − f (a) = 0 and g(b) = f (b) − m(b − a) − f (a) = f (b) − f (b) − f (a) b − a (b − a) − f (a) = f (b) − (f (b) − f (a)... |
t distance traveled during one second, and the final position is 11.5. So for t = 1, at least, this rather cumbersome approach gives the same answer as the first approach. But really there’s nothing special about t = 1; let’s just call it t instead. In this case the approximate distance traveled during time interval numb... |
amental Theorem of Calculus. To avoid confusion, some people call the two versions of the theorem “The Fundamental Theorem of Calculus, part I” and “The Fundamental Theorem of Calculus, part II”, although unfortunately there is no universal agreement as to which is part I and which part II. Since it really is the same ... |
ed, that is, the difference between the starting and ending point. As we have already seen, Z 6 Z 5 Z 6 v(t) dt = v(t) dt + v(t) dt. 0 0 5 Computing the two integrals on the right (do it!) gives 125/6 and −17/6, and the sum of these is indeed 18. But what does that negative sign mean? It means precisely what you might t... |
hen Z Z (ax + b)n dx = 1 a un du = 1 a(n + 1) un+1 + C = 1 a(n + 1) (ax + b)n+1 + C. Z EXAMPLE 8.1.2 Evaluate sin(ax + b) dx, assuming that a and b are constants and Z a ̸= 0. Again we let u = ax + b so du = a dx or dx = du/a. Then (− cos u) + C = − 1 1 a a sin(ax + b) dx = sin u du = 1 a Z cos(ax + b) + C. EXAMPLE 8.1... |
n x dx. Let u = ln x so du = 1/x dx. Then we must Z let dv = x dx so v = x2/2 and Z x ln x dx = Z x2 ln x 2 − Z x2 2 1 x dx = x2 ln x 2 − x 2 dx = x2 ln x 2 − x2 4 + C. Z EXAMPLE 8.4.2 Evaluate x sin x dx. Let u = x so du = dx. Then we must let dv = sin x dx so v = − cos x and Z Z Z x sin x dx = −x cos x − − cos x dx =... |
dx ⇒ x3 4 + x2 dx ⇒ 1 2x2 − x − 3 dx ⇒ Z Z Z Z Z 2. 4. 6. 8. 10. x4 4 − x2 dx ⇒ x2 4 − x2 dx ⇒ 1 x2 + 10x + 29 1 x2 + 10x + 21 1 x2 + 3x dx ⇒ dx ⇒ dx ⇒ 8.6 Numerical Integration We have now seen some of the most generally useful methods for discovering antiderivatives, and there are others. Unfortunately, some functio... |
d 0.746855 + 0.0003 = 0.7471555, both of which round to 0.75. Exercises 8.6. 8.7 Additional exercises 189 In the following problems, compute the trapezoid and Simpson approximations using 4 subintervals, and compute the error estimate for each. (Finding the maximum values of the second and fourth derivatives can be cha... |
xpand on the geometric interpretation of the hyperbolic functions. Refer to section 4.11 and particularly to figure 4.11.2 and exercise 6 in section 4.11. Z p 13. Compute x2 − 1 dx using the substitution u = arccosh x, or x = cosh u; use exercise 6 in section 4.11. 14. Fix t > 0. Sketch the region R in the right half pl... |
base 2(1 − x2 i ) and height 3(1 − x2 i ), so the area of the cross-section is (base)(height) = (1 − x2 i ) √ 3(1 − x2 i ), 1 2 9.3 Volume 201 and the volume of a thin “slab” is then Thus the total volume is √ (1 − x2 i ) 3(1 − x2 i )∆x. Z 1 √ −1 3(1 − x2)2 dx = √ 3. 16 15 One easy way to get “nice” cross-sections is b... |
ume of the solid. ⇒ 206 Chapter 9 Applications of Integration 9.4 Average value of a function The average of some finite set of values is a familiar concept. If, for example, the class scores on a quiz are 10, 9, 10, 8, 7, 5, 7, 6, 3, 2, 7, 8, then the average score is the sum of these numbers divided by the size of the... |
top? Here we have a large number of atoms of water that must be lifted different distances to get to the top of the tank. Fortunately, we don’t really have to deal with individual atoms—we can consider all the atoms at a given depth together. To approximate the work, we can divide the water in the tank into horizontal s... |
ero and solve for ¯x we get an approximation to the balance point of the beam: xi(1 + xi)∆x − ¯x n−1X i=0 (1 + xi)∆x 0 = n−1X i=0 n−1X n−1X ¯x (1 + xi)∆x = xi(1 + xi)∆x i=0 i=0 n−1X xi(1 + xi)∆x ¯x = i=0 n−1X . (1 + xi)∆x i=0 9.6 Center of Mass 215 The denominator of this fraction has a very familiar interpretation. Co... |
ergy. We know that the work required to move an object from the surface of the earth to infinity is W = Z ∞ r0 k r2 dr = k r0 . At the surface of the earth the acceleration due to gravity is approximately 9.8 meters per second squared, so the force on an object of mass m is F = 9.8m. The radius of the earth is approxima... |
ely so Z ∞ 1 e−x/2 dx = 2√ e , Z ∞ e−x2/2 dx is some finite number smaller than 2/ 1 √ e. Because f is symmetric around the y-axis, Z −1 −∞ e−x2/2 dx = Z ∞ 1 e−x2/2 dx. This means that Z ∞ −∞ e−x2/2 dx = Z −1 −∞ e−x2/2 dx + Z 1 −1 e−x2/2 dx + Z ∞ 1 e−x2/2 dx = A for some finite positive number A. Now if we let g(x) = f (... |
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