Partial Review for Test 3 I. Study the WebAssign homework questions for Sections 14.6, 14.7, and Chapter 15. II. Study the lecture notes for Sections 14.6, 14.7, and Chapter 15. III. In addition, study the following questions: 80 1. Suppose that the temperature at a point (x, y, z) in space is given by T(x, y, z) = where T is 1+ x +2y +3z measured in °C and x, y, z in meters. (a) Find the rate of change of T with respect to distance at the point (1, 1, -2) in the direction of the vector v = 2î + 2j-k. (b) In what direction dose the temperature increase fastest at the point (1, 1, -2)? What is the maximum rate of increase? Solutions 1. The gradient of T is aT aT j+k VT= i+ ax az 160x 320y 480z (1 + x + 2y2 + 3z2) (1 + x? + 2y + 3z2) - -k (1 + x + 2y? + 3z2) 160 (-xi – 2yj - 3zk) %3D (1 + x + 2y2 + 3z?)? At the point (1, 1, -2) the gradient vector is V(1, 1, -2) =(-i - 2j + 6k) ={-i- 2j + 6k) (a) The unit vector in the direction of v=2î + 2j–k is û == (2i + 2j – k) The rate of change of T with respect to distance at the point (1, 1,-2) is DaT(1,1, –2) = VT(1,1,–2)û = (-î - 2j + 6k). 1 (2î + 2j – k) (b) The temperature increases fastest in the directiongradient vector VT(1, 1,-2) =(-i-2j + 6k) or, equivalently, in the direction of -i - 2j + 6k or the unit vector (-i- 2j + 6 k)//41. The maximum rate of increase is the length of the gradient vector: |V7L. 1.-2)|3히-i- 2j + 6k| = BV4T
Partial Review for Test 3 I. Study the WebAssign homework questions for Sections 14.6, 14.7, and Chapter 15. II. Study the lecture notes for Sections 14.6, 14.7, and Chapter 15. III. In addition, study the following questions: 80 1. Suppose that the temperature at a point (x, y, z) in space is given by T(x, y, z) = where T is 1+ x +2y +3z measured in °C and x, y, z in meters. (a) Find the rate of change of T with respect to distance at the point (1, 1, -2) in the direction of the vector v = 2î + 2j-k. (b) In what direction dose the temperature increase fastest at the point (1, 1, -2)? What is the maximum rate of increase? Solutions 1. The gradient of T is aT aT j+k VT= i+ ax az 160x 320y 480z (1 + x + 2y2 + 3z2) (1 + x? + 2y + 3z2) - -k (1 + x + 2y? + 3z2) 160 (-xi – 2yj - 3zk) %3D (1 + x + 2y2 + 3z?)? At the point (1, 1, -2) the gradient vector is V(1, 1, -2) =(-i - 2j + 6k) ={-i- 2j + 6k) (a) The unit vector in the direction of v=2î + 2j–k is û == (2i + 2j – k) The rate of change of T with respect to distance at the point (1, 1,-2) is DaT(1,1, –2) = VT(1,1,–2)û = (-î - 2j + 6k). 1 (2î + 2j – k) (b) The temperature increases fastest in the directiongradient vector VT(1, 1,-2) =(-i-2j + 6k) or, equivalently, in the direction of -i - 2j + 6k or the unit vector (-i- 2j + 6 k)//41. The maximum rate of increase is the length of the gradient vector: |V7L. 1.-2)|3히-i- 2j + 6k| = BV4T
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
This question is already solved. I just need you to explain it more simpler for me since I not understand part and a and b fully. I am mostly curious as to where the 160/256 comes from and where the 1/3 comes from. I put question marks next to them. Explain the best way for this problem. thank you.
Transcribed Image Text:Partial Review for Test 3
I.
Study the WebAssign homework questions for Sections 14.6, 14.7, and Chapter 15.
II. Study the lecture notes for Sections 14.6, 14.7, and Chapter 15.
III. In addition, study the following questions:
80
1. Suppose that the temperature at a point (x, y, z) in space is given by T(x, y, z) =
where T is
1+ x' +2y² +3z?
measured in °C and x, y, z in meters.
(a) Find the rate of change of T with respect to distance at the point (1, 1, -2) in the direction of the vector
v = 2î + 2j-k.
(b) In what direction dose the temperature increase fastest at the point (1, 1, -2)? What is the maximum rate of
increase?
Solutions
1.
The gradient of T is
aT
aT
VT=
ayj+k
i+
ax
az
320y
(1 + x + 2y² + 3z?)² J
160x
480z
(1 + x² + 2y² + 3z?)
(1 + x + 2y? + 3z²)?
160
(-xi – 2yj - 3zk)
%3D
(1+ x² + 2y² + 3z2)?
At the point (1, 1, -2) the gradient vector is
V71, 1, –2) = (-i - 2j + 6k) =(-i - 2j + 6k)
(a) The unit vector in the direction of v=2i +2j–k is û ==" (2î + 2j – k)
The rate of change of T with respect to distance at the point (1, 1,–2) is
DaT(1,1, –2) = VT(1,1,–2)û =
(-î – 2j + 6k).
1
(2î + 2ĵ –
(b)
The temperature increases fastest in the directiongradient vector
VT(1, 1, -2) =(-i – 2j + 6k) or, equivalently, in the direction of -i – 2j + 6k or
the unit vector (-i – 2j + 6 k)//41. The maximum rate of increase is the length of the
gradient vector:
|V7(1, 1, –2)| = |-i – 2j + 6k| =VAT
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