Heat flux The heat flow vector field for conducting objects is F = – k ▿ T , where T ( x , y , z ) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux of F across the following surfaces S for the given temperature distributions. Assume k = 1. 61. T ( x , y , z ) = 100 e − x − y ; S consists of the faces of the cube | x | ≤ 1 , | y | ≤ 1 , | z | ≤ 1 .
Heat flux The heat flow vector field for conducting objects is F = – k ▿ T , where T ( x , y , z ) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux of F across the following surfaces S for the given temperature distributions. Assume k = 1. 61. T ( x , y , z ) = 100 e − x − y ; S consists of the faces of the cube | x | ≤ 1 , | y | ≤ 1 , | z | ≤ 1 .
Solution Summary: The author explains how to compute the outward flux of F across the surface S.
Heat fluxThe heat flow vector field for conducting objects isF = –k▿T, where T(x, y, z) is the temperature in the object and k > 0 is a constant that depends on the material. Compute the outward flux ofFacross the following surfaces S for the given temperature distributions. Assume k = 1.
61.
T
(
x
,
y
,
z
)
=
100
e
−
x
−
y
; S consists of the faces of the cube
|
x
|
≤
1
,
|
y
|
≤
1
,
|
z
|
≤
1
.
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
Given lim x-4 f (x) = 1,limx-49 (x) = 10, and lim→-4 h (x) = -7 use the limit properties
to find lim→-4
1
[2h (x) — h(x) + 7 f(x)] :
-
h(x)+7f(x)
3
O DNE
17. Suppose we know that the graph below is the graph of a solution to dy/dt = f(t).
(a) How much of the slope field can
you sketch from this information?
[Hint: Note that the differential
equation depends only on t.]
(b) What can you say about the solu-
tion with y(0) = 2? (For example,
can you sketch the graph of this so-
lution?)
y(0) = 1
y
AN
(b) Find the (instantaneous) rate of change of y at x = 5.
In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of
change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the
following limit.
lim
h→0
-
f(x + h) − f(x)
h
The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule
defining f.
f(x + h) = (x + h)² - 5(x+ h)
=
2xh+h2_
x² + 2xh + h² 5✔
-
5
)x - 5h
Step 4
-
The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x).
-
f(x + h) f(x) =
= (x²
x² + 2xh + h² -
])-
=
2x
+ h² - 5h
])x-5h) - (x² - 5x)
=
]) (2x + h - 5)
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