Heat flux in a plate A square plate R = {( x , y ): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} has a temperature distribution T ( x , y ) = 100 – 50 x – 25 y . a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature ▿ T ( x, y ) . c. Assume that the flow of heat is given by the vector field F = –▿ T ( x , y ). Compute F . d. Find the outward heat flux across the boundary {( x , y ): x = 1, 0 ≤ y ≤ 1}. e. Find the outward heat flux across the boundary {( x , y ): 0 ≤ x ≤ 1, y = 1}.
Heat flux in a plate A square plate R = {( x , y ): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} has a temperature distribution T ( x , y ) = 100 – 50 x – 25 y . a. Sketch two level curves of the temperature in the plate. b. Find the gradient of the temperature ▿ T ( x, y ) . c. Assume that the flow of heat is given by the vector field F = –▿ T ( x , y ). Compute F . d. Find the outward heat flux across the boundary {( x , y ): x = 1, 0 ≤ y ≤ 1}. e. Find the outward heat flux across the boundary {( x , y ): 0 ≤ x ≤ 1, y = 1}.
Solution Summary: The author illustrates the two level curves of the temperature in the square plate.
Heat flux in a plate A square plate R = {(x, y): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} has a temperature distribution T(x, y) = 100 – 50x – 25y.
a. Sketch two level curves of the temperature in the plate.
b. Find the gradient of the temperature ▿ T(x, y).
c. Assume that the flow of heat is given by the vector field F = –▿ T(x, y). Compute F.
d. Find the outward heat flux across the boundary {(x, y): x= 1, 0 ≤ y ≤ 1}.
e. Find the outward heat flux across the boundary {(x, y): 0 ≤ x ≤ 1, y = 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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