Bundle: Calculus, 10th + WebAssign Printed Access Card for Larson/Edwards' Calculus, 10th Edition, Multi-Term
10th Edition
ISBN: 9781305718661
Author: Ron Larson, Bruce H. Edwards
Publisher: Cengage Learning
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Textbook Question
Chapter 13.10, Problem 26E
Finding Minimum Distance In Exercises 19-28, use Lagrange multipliers to find the minimum distance from the curve or surface to the indicated point. (Hint: To simplify the computations, minimize the square of the distance.)
Curve Point
Cone:
(4, 0, 0)
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3.1 Limits
1. If lim f(x)=-6 and lim f(x)=5, then lim f(x). Explain your choice.
x+3°
x+3*
x+3
(a) Is 5
(c) Does not exist
(b) is 6
(d) is infinite
1 pts
Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and
G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is
Question 1
-0.246
0.072
-0.934
0.478
-0.914
-0.855
0.710
0.262
.
2. Answer the following questions.
(A) [50%] Given the vector field F(x, y, z) = (x²y, e", yz²), verify the differential identity
Vx (VF) V(V •F) - V²F
(B) [50%] Remark. You are confined to use the differential identities.
Let u and v be scalar fields, and F be a vector field given by
F = (Vu) x (Vv)
(i) Show that F is solenoidal (or incompressible).
(ii) Show that
G =
(uvv – vVu)
is a vector potential for F.
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- Question 2 Let F be a solenoidal vector field, suppose V × F = (-8xy + 12z², −9x² + 4y² + 9z², 6y²), and let (P,Q,R) = V²F(.725, —.283, 1.73). Then the value of sin(2P) + sin(3Q) + sin(4R) is -2.024 1.391 0.186 -0.994 -2.053 -0.647 -0.588 -1.851 1 ptsarrow_forward1 pts Let F and G be vector fields such that ▼ × F(0, 0, 0) = (0.76, -9.78, 3.29), G(0, 0, 0) = (−3.99, 6.15, 2.94), and G is irrotational. Then sin(5V (F × G)) at (0, 0, 0) is Question 1 -0.246 0.072 -0.934 0.478 -0.914 -0.855 0.710 0.262 .arrow_forwardanswerarrow_forward
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