Student's Solution and Survival Manual for Calculus
7th Edition
ISBN: 9781524934040
Author: STRAUSS MONTY J, TODA MAGDALENA DANIELE, SMITH KARL J
Publisher: Kendall Hunt Publishing
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Chapter 11.2, Problem 56PS
To determine
To prove: Prove the statement.
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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.
A driver is traveling along a straight road when a buffalo runs into the street. This driver has a reaction time of 0.75 seconds. When the driver sees the buffalo he is traveling at 44 ft/s, his car can decelerate at 2 ft/s^2 when the brakes are applied. What is the stopping distance between when the driver first saw the buffalo, to when the car stops.
Topic 2
Evaluate S
x
dx, using u-substitution. Then find the integral using
1-x2
trigonometric substitution. Discuss the results!
Topic 3
Explain what an elementary anti-derivative is. Then consider the following
ex
integrals: fed dx
x
1
Sdx
In x
Joseph Liouville proved that the first integral does not have an elementary anti-
derivative Use this fact to prove that the second integral does not have an
elementary anti-derivative. (hint: use an appropriate u-substitution!)
Chapter 11 Solutions
Student's Solution and Survival Manual for Calculus
Ch. 11.1 - Prob. 1PSCh. 11.1 - Prob. 2PSCh. 11.1 - Prob. 3PSCh. 11.1 - Prob. 4PSCh. 11.1 - Prob. 5PSCh. 11.1 - Prob. 6PSCh. 11.1 - Prob. 7PSCh. 11.1 - Prob. 8PSCh. 11.1 - Prob. 9PSCh. 11.1 - Prob. 10PS
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- 1. Given the vector field F(x, y, z) = -xi, verify the relation 1 V.F(0,0,0) = lim 0+ volume inside Se ff F• Nds SE where SE is the surface enclosing a cube centred at the origin and having edges of length 2€. Then, determine if the origin is sink or source.arrow_forward4 3 2 -5 4-3 -2 -1 1 2 3 4 5 12 23 -4 The function graphed above is: Increasing on the interval(s) Decreasing on the interval(s)arrow_forwardQuestion 4 The plot below represents the function f(x) 8 7 3 pts O -4-3-2-1 6 5 4 3 2 + 1 2 3 5 -2+ Evaluate f(3) f(3) = Solve f(x) = 3 x= Question 5arrow_forward
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