Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)
3rd Edition
ISBN: 9781319019846
Author: Jon Rogawski, Colin Adams
Publisher: W. H. Freeman
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Question
Chapter 7.7, Problem 57E
To determine
To prove:
Use the comparison test to show that the given function converges
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1. Show that the vector field
F(x, y, z)
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satisfies the necessary conditions for a conservative vector field, and find a potential function for
F.
i need help please
6.
(i)
Sketch the trace of the following curve on R²,
(t) = (sin(t), 3 sin(t)),
tЄ [0, π].
[3 Marks]
Total marks 10
(ii)
Find the length of this curve.
[7 Marks]
Chapter 7 Solutions
Loose-leaf Version for Calculus: Early Transcendentals Combo 3e & WebAssign for Calculus: Early Transcendentals 3e (Life of Edition)
Ch. 7.1 - Prob. 1PQCh. 7.1 - Prob. 2PQCh. 7.1 - Prob. 3PQCh. 7.1 - Prob. 1ECh. 7.1 - Prob. 2ECh. 7.1 - Prob. 3ECh. 7.1 - Prob. 4ECh. 7.1 - Prob. 5ECh. 7.1 - Prob. 6ECh. 7.1 - Prob. 7E
Ch. 7.1 - Prob. 8ECh. 7.1 - Prob. 9ECh. 7.1 - Prob. 10ECh. 7.1 - Prob. 11ECh. 7.1 - Prob. 12ECh. 7.1 - Prob. 13ECh. 7.1 - Prob. 14ECh. 7.1 - Prob. 15ECh. 7.1 - Prob. 16ECh. 7.1 - Prob. 17ECh. 7.1 - Prob. 18ECh. 7.1 - Prob. 19ECh. 7.1 - Prob. 20ECh. 7.1 - Prob. 21ECh. 7.1 - Prob. 22ECh. 7.1 - Prob. 23ECh. 7.1 - Prob. 24ECh. 7.1 - Prob. 25ECh. 7.1 - Prob. 26ECh. 7.1 - Prob. 27ECh. 7.1 - Prob. 28ECh. 7.1 - Prob. 29ECh. 7.1 - Prob. 30ECh. 7.1 - Prob. 31ECh. 7.1 - Prob. 32ECh. 7.1 - Prob. 33ECh. 7.1 - Prob. 34ECh. 7.1 - Prob. 35ECh. 7.1 - Prob. 36ECh. 7.1 - Prob. 37ECh. 7.1 - Prob. 38ECh. 7.1 - Prob. 39ECh. 7.1 - Prob. 40ECh. 7.1 - Prob. 41ECh. 7.1 - Prob. 42ECh. 7.1 - Prob. 43ECh. 7.1 - Prob. 44ECh. 7.1 - Prob. 45ECh. 7.1 - Prob. 46ECh. 7.1 - Prob. 47ECh. 7.1 - Prob. 48ECh. 7.1 - Prob. 49ECh. 7.1 - Prob. 50ECh. 7.1 - Prob. 51ECh. 7.1 - Prob. 52ECh. 7.1 - Prob. 53ECh. 7.1 - Prob. 54ECh. 7.1 - Prob. 55ECh. 7.1 - Prob. 56ECh. 7.1 - Prob. 57ECh. 7.1 - Prob. 58ECh. 7.1 - Prob. 59ECh. 7.1 - Prob. 60ECh. 7.1 - 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Prob. 37ECh. 7.4 - Prob. 38ECh. 7.4 - Prob. 39ECh. 7.4 - Prob. 40ECh. 7.4 - Prob. 41ECh. 7.4 - Prob. 42ECh. 7.4 - Prob. 43ECh. 7.4 - Prob. 44ECh. 7.4 - Prob. 45ECh. 7.4 - Prob. 46ECh. 7.4 - Prob. 47ECh. 7.4 - Prob. 48ECh. 7.4 - Prob. 49ECh. 7.5 - Prob. 1PQCh. 7.5 - Prob. 2PQCh. 7.5 - Prob. 3PQCh. 7.5 - Prob. 4PQCh. 7.5 - Prob. 1ECh. 7.5 - Prob. 2ECh. 7.5 - Prob. 3ECh. 7.5 - Prob. 4ECh. 7.5 - Prob. 5ECh. 7.5 - Prob. 6ECh. 7.5 - Prob. 7ECh. 7.5 - Prob. 8ECh. 7.5 - Prob. 9ECh. 7.5 - Prob. 10ECh. 7.5 - Prob. 11ECh. 7.5 - Prob. 12ECh. 7.5 - Prob. 13ECh. 7.5 - Prob. 14ECh. 7.5 - Prob. 15ECh. 7.5 - Prob. 16ECh. 7.5 - Prob. 17ECh. 7.5 - Prob. 18ECh. 7.5 - Prob. 19ECh. 7.5 - Prob. 20ECh. 7.5 - Prob. 21ECh. 7.5 - Prob. 22ECh. 7.5 - Prob. 23ECh. 7.5 - Prob. 24ECh. 7.5 - Prob. 25ECh. 7.5 - Prob. 26ECh. 7.5 - Prob. 27ECh. 7.5 - Prob. 28ECh. 7.5 - Prob. 29ECh. 7.5 - Prob. 30ECh. 7.5 - Prob. 31ECh. 7.5 - Prob. 32ECh. 7.5 - Prob. 33ECh. 7.5 - Prob. 34ECh. 7.5 - Prob. 35ECh. 7.5 - Prob. 36ECh. 7.5 - Prob. 37ECh. 7.5 - Prob. 38ECh. 7.5 - Prob. 39ECh. 7.5 - Prob. 40ECh. 7.5 - Prob. 41ECh. 7.5 - Prob. 42ECh. 7.5 - Prob. 43ECh. 7.5 - Prob. 44ECh. 7.5 - Prob. 45ECh. 7.5 - Prob. 46ECh. 7.5 - Prob. 47ECh. 7.5 - Prob. 48ECh. 7.5 - Prob. 49ECh. 7.5 - Prob. 50ECh. 7.5 - Prob. 51ECh. 7.5 - Prob. 52ECh. 7.5 - Prob. 53ECh. 7.5 - Prob. 54ECh. 7.5 - Prob. 55ECh. 7.5 - Prob. 56ECh. 7.5 - Prob. 57ECh. 7.5 - Prob. 58ECh. 7.5 - Prob. 59ECh. 7.5 - Prob. 60ECh. 7.5 - Prob. 61ECh. 7.5 - Prob. 62ECh. 7.6 - Prob. 1PQCh. 7.6 - Prob. 2PQCh. 7.6 - Prob. 3PQCh. 7.6 - Prob. 4PQCh. 7.6 - Prob. 5PQCh. 7.6 - Prob. 6PQCh. 7.6 - Prob. 7PQCh. 7.6 - Prob. 8PQCh. 7.6 - Prob. 9PQCh. 7.6 - Prob. 10PQCh. 7.6 - Prob. 11PQCh. 7.6 - Prob. 1ECh. 7.6 - Prob. 2ECh. 7.6 - Prob. 3ECh. 7.6 - Prob. 4ECh. 7.6 - Prob. 5ECh. 7.6 - Prob. 6ECh. 7.6 - Prob. 7ECh. 7.6 - Prob. 8ECh. 7.6 - Prob. 9ECh. 7.6 - Prob. 10ECh. 7.6 - Prob. 11ECh. 7.6 - Prob. 12ECh. 7.6 - Prob. 13ECh. 7.6 - Prob. 14ECh. 7.6 - Prob. 15ECh. 7.6 - Prob. 16ECh. 7.6 - Prob. 17ECh. 7.6 - Prob. 18ECh. 7.6 - Prob. 19ECh. 7.6 - Prob. 20ECh. 7.6 - Prob. 21ECh. 7.6 - Prob. 22ECh. 7.6 - Prob. 23ECh. 7.6 - Prob. 24ECh. 7.6 - Prob. 25ECh. 7.6 - Prob. 26ECh. 7.6 - Prob. 27ECh. 7.6 - Prob. 28ECh. 7.6 - Prob. 29ECh. 7.6 - Prob. 30ECh. 7.6 - Prob. 31ECh. 7.6 - Prob. 32ECh. 7.6 - Prob. 33ECh. 7.6 - Prob. 34ECh. 7.6 - Prob. 35ECh. 7.6 - Prob. 36ECh. 7.6 - Prob. 37ECh. 7.6 - Prob. 38ECh. 7.6 - Prob. 39ECh. 7.6 - Prob. 40ECh. 7.6 - Prob. 41ECh. 7.6 - Prob. 42ECh. 7.6 - Prob. 43ECh. 7.6 - Prob. 44ECh. 7.6 - Prob. 45ECh. 7.6 - Prob. 46ECh. 7.6 - Prob. 47ECh. 7.6 - Prob. 48ECh. 7.6 - Prob. 49ECh. 7.6 - Prob. 50ECh. 7.6 - Prob. 51ECh. 7.6 - Prob. 52ECh. 7.6 - Prob. 53ECh. 7.6 - Prob. 54ECh. 7.6 - Prob. 55ECh. 7.6 - Prob. 56ECh. 7.6 - Prob. 57ECh. 7.6 - Prob. 58ECh. 7.7 - Prob. 1PQCh. 7.7 - Prob. 2PQCh. 7.7 - Prob. 3PQCh. 7.7 - Prob. 4PQCh. 7.7 - Prob. 5PQCh. 7.7 - Prob. 1ECh. 7.7 - Prob. 2ECh. 7.7 - Prob. 3ECh. 7.7 - Prob. 4ECh. 7.7 - Prob. 5ECh. 7.7 - Prob. 6ECh. 7.7 - Prob. 7ECh. 7.7 - Prob. 8ECh. 7.7 - Prob. 9ECh. 7.7 - Prob. 10ECh. 7.7 - Prob. 11ECh. 7.7 - Prob. 12ECh. 7.7 - Prob. 13ECh. 7.7 - Prob. 14ECh. 7.7 - Prob. 15ECh. 7.7 - Prob. 16ECh. 7.7 - Prob. 17ECh. 7.7 - Prob. 18ECh. 7.7 - Prob. 19ECh. 7.7 - Prob. 20ECh. 7.7 - Prob. 21ECh. 7.7 - Prob. 22ECh. 7.7 - Prob. 23ECh. 7.7 - Prob. 24ECh. 7.7 - Prob. 25ECh. 7.7 - Prob. 26ECh. 7.7 - Prob. 27ECh. 7.7 - Prob. 28ECh. 7.7 - Prob. 29ECh. 7.7 - Prob. 30ECh. 7.7 - Prob. 31ECh. 7.7 - Prob. 32ECh. 7.7 - Prob. 33ECh. 7.7 - Prob. 34ECh. 7.7 - Prob. 35ECh. 7.7 - Prob. 36ECh. 7.7 - Prob. 37ECh. 7.7 - Prob. 38ECh. 7.7 - Prob. 39ECh. 7.7 - Prob. 40ECh. 7.7 - Prob. 41ECh. 7.7 - Prob. 42ECh. 7.7 - Prob. 43ECh. 7.7 - Prob. 44ECh. 7.7 - Prob. 45ECh. 7.7 - Prob. 46ECh. 7.7 - Prob. 47ECh. 7.7 - Prob. 48ECh. 7.7 - Prob. 49ECh. 7.7 - Prob. 50ECh. 7.7 - Prob. 51ECh. 7.7 - Prob. 52ECh. 7.7 - Prob. 53ECh. 7.7 - Prob. 54ECh. 7.7 - Prob. 55ECh. 7.7 - Prob. 56ECh. 7.7 - Prob. 57ECh. 7.7 - Prob. 58ECh. 7.7 - Prob. 59ECh. 7.7 - Prob. 60ECh. 7.7 - 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Prob. 9ECh. 7.8 - Prob. 10ECh. 7.8 - Prob. 11ECh. 7.8 - Prob. 12ECh. 7.8 - Prob. 13ECh. 7.8 - Prob. 14ECh. 7.8 - Prob. 15ECh. 7.8 - Prob. 16ECh. 7.8 - Prob. 17ECh. 7.8 - Prob. 18ECh. 7.8 - Prob. 19ECh. 7.8 - Prob. 20ECh. 7.8 - Prob. 21ECh. 7.8 - Prob. 22ECh. 7.8 - Prob. 23ECh. 7.8 - Prob. 24ECh. 7.8 - Prob. 25ECh. 7.8 - Prob. 26ECh. 7.8 - Prob. 27ECh. 7.8 - Prob. 28ECh. 7.8 - Prob. 29ECh. 7.8 - Prob. 30ECh. 7.9 - Prob. 1PQCh. 7.9 - Prob. 2PQCh. 7.9 - Prob. 3PQCh. 7.9 - Prob. 4PQCh. 7.9 - Prob. 5PQCh. 7.9 - Prob. 1ECh. 7.9 - Prob. 2ECh. 7.9 - Prob. 3ECh. 7.9 - Prob. 4ECh. 7.9 - Prob. 5ECh. 7.9 - Prob. 6ECh. 7.9 - Prob. 7ECh. 7.9 - Prob. 8ECh. 7.9 - Prob. 9ECh. 7.9 - Prob. 10ECh. 7.9 - Prob. 11ECh. 7.9 - Prob. 12ECh. 7.9 - Prob. 13ECh. 7.9 - Prob. 14ECh. 7.9 - Prob. 15ECh. 7.9 - Prob. 16ECh. 7.9 - Prob. 17ECh. 7.9 - Prob. 18ECh. 7.9 - Prob. 19ECh. 7.9 - Prob. 20ECh. 7.9 - Prob. 21ECh. 7.9 - Prob. 22ECh. 7.9 - Prob. 23ECh. 7.9 - Prob. 24ECh. 7.9 - Prob. 25ECh. 7.9 - Prob. 26ECh. 7.9 - Prob. 27ECh. 7.9 - 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- helppparrow_forward7. Let F(x1, x2) (F₁(x1, x2), F2(x1, x2)), where = X2 F1(x1, x2) X1 F2(x1, x2) x+x (i) Using the definition, calculate the integral LF.dy, where (t) = (cos(t), sin(t)) and t = [0,2]. [5 Marks] (ii) Explain why Green's Theorem cannot be used to find the integral in part (i). [5 Marks]arrow_forward6. Sketch the trace of the following curve on R², п 3п (t) = (t2 sin(t), t2 cos(t)), tЄ 22 [3 Marks] Find the length of this curve. [7 Marks]arrow_forward
- Total marks 10 Total marks on naner: 80 7. Let DCR2 be a bounded domain with the boundary OD which can be represented as a smooth closed curve : [a, b] R2, oriented in the anticlock- wise direction. Use Green's Theorem to justify that the area of the domain D can be computed by the formula 1 Area(D) = ½ (−y, x) · dy. [5 Marks] (ii) Use the area formula in (i) to find the area of the domain D enclosed by the ellipse y(t) = (10 cos(t), 5 sin(t)), t = [0,2π]. [5 Marks]arrow_forwardTotal marks 15 Total marks on paper: 80 6. Let DCR2 be a bounded domain with the boundary ǝD which can be represented as a smooth closed curve : [a, b] → R², oriented in the anticlockwise direction. (i) Use Green's Theorem to justify that the area of the domain D can be computed by the formula 1 Area(D) = . [5 Marks] (ii) Use the area formula in (i) to find the area of the domain D enclosed by the ellipse (t) = (5 cos(t), 10 sin(t)), t = [0,2π]. [5 Marks] (iii) Explain in your own words why Green's Theorem can not be applied to the vector field У x F(x,y) = ( - x² + y²²x² + y² ). [5 Marks]arrow_forwardTotal marks 15 པ་ (i) Sketch the trace of the following curve on R2, (t) = (t2 cos(t), t² sin(t)), t = [0,2π]. [3 Marks] (ii) Find the length of this curve. (iii) [7 Marks] Give a parametric representation of a curve : [0, that has initial point (1,0), final point (0, 1) and the length √2. → R² [5 Marks] Turn over. MA-201: Page 4 of 5arrow_forward
- Total marks 15 5. (i) Let f R2 R be defined by f(x1, x2) = x² - 4x1x2 + 2x3. Find all local minima of f on R². (ii) [10 Marks] Give an example of a function f: R2 R which is not bounded above and has exactly one critical point, which is a minimum. Justify briefly your answer. [5 Marks] 6. (i) Sketch the trace of the following curve on R2, y(t) = (sin(t), 3 sin(t)), t = [0,π]. [3 Marks]arrow_forwardA ladder 25 feet long is leaning against the wall of a building. Initially, the foot of the ladder is 7 feet from the wall. The foot of the ladder begins to slide at a rate of 2 ft/sec, causing the top of the ladder to slide down the wall. The location of the foot of the ladder, its x coordinate, at time t seconds is given by x(t)=7+2t. wall y(1) 25 ft. ladder x(1) ground (a) Find the formula for the location of the top of the ladder, the y coordinate, as a function of time t. The formula for y(t)= √ 25² - (7+2t)² (b) The domain of t values for y(t) ranges from 0 (c) Calculate the average velocity of the top of the ladder on each of these time intervals (correct to three decimal places): . (Put your cursor in the box, click and a palette will come up to help you enter your symbolic answer.) time interval ave velocity [0,2] -0.766 [6,8] -3.225 time interval ave velocity -1.224 -9.798 [2,4] [8,9] (d) Find a time interval [a,9] so that the average velocity of the top of the ladder on this…arrow_forwardTotal marks 15 3. (i) Let FRN Rm be a mapping and x = RN is a given point. Which of the following statements are true? Construct counterex- amples for any that are false. (a) If F is continuous at x then F is differentiable at x. (b) If F is differentiable at x then F is continuous at x. If F is differentiable at x then F has all 1st order partial (c) derivatives at x. (d) If all 1st order partial derivatives of F exist and are con- tinuous on RN then F is differentiable at x. [5 Marks] (ii) Let mappings F= (F1, F2) R³ → R² and G=(G1, G2) R² → R² : be defined by F₁ (x1, x2, x3) = x1 + x², G1(1, 2) = 31, F2(x1, x2, x3) = x² + x3, G2(1, 2)=sin(1+ y2). By using the chain rule, calculate the Jacobian matrix of the mapping GoF R3 R², i.e., JGoF(x1, x2, x3). What is JGOF(0, 0, 0)? (iii) [7 Marks] Give reasons why the mapping Go F is differentiable at (0, 0, 0) R³ and determine the derivative matrix D(GF)(0, 0, 0). [3 Marks]arrow_forward
- 5. (i) Let f R2 R be defined by f(x1, x2) = x² - 4x1x2 + 2x3. Find all local minima of f on R². (ii) [10 Marks] Give an example of a function f: R2 R which is not bounded above and has exactly one critical point, which is a minimum. Justify briefly Total marks 15 your answer. [5 Marks]arrow_forwardTotal marks 15 4. : Let f R2 R be defined by f(x1, x2) = 2x²- 8x1x2+4x+2. Find all local minima of f on R². [10 Marks] (ii) Give an example of a function f R2 R which is neither bounded below nor bounded above, and has no critical point. Justify briefly your answer. [5 Marks]arrow_forward4. Let F RNR be a mapping. (i) x ЄRN ? (ii) : What does it mean to say that F is differentiable at a point [1 Mark] In Theorem 5.4 in the Lecture Notes we proved that if F is differentiable at a point x E RN then F is continuous at x. Proof. Let (n) CRN be a sequence such that xn → x ЄERN as n → ∞. We want to show that F(xn) F(x), which means F is continuous at x. Denote hnxn - x, so that ||hn|| 0. Thus we find ||F(xn) − F(x)|| = ||F(x + hn) − F(x)|| * ||DF (x)hn + R(hn) || (**) ||DF(x)hn||+||R(hn)||| → 0, because the linear mapping DF(x) is continuous and for all large nЄ N, (***) ||R(hn) || ||R(hn) || ≤ → 0. ||hn|| (a) Explain in details why ||hn|| → 0. [3 Marks] (b) Explain the steps labelled (*), (**), (***). [6 Marks]arrow_forward
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