Green’s Theorem for line integrals Use either form of Green’s Theorem to evaluate the following line integrals. 30. ∮ C ( − 3 y + x 3 / 2 ) d x + ( x − y 2 / 3 ) d y ; C is the boundary of the half disk {( x , y ): x 2 + y 2 ≤ 2, y ≥ 0} with counterclockwise orientation.
Green’s Theorem for line integrals Use either form of Green’s Theorem to evaluate the following line integrals. 30. ∮ C ( − 3 y + x 3 / 2 ) d x + ( x − y 2 / 3 ) d y ; C is the boundary of the half disk {( x , y ): x 2 + y 2 ≤ 2, y ≥ 0} with counterclockwise orientation.
Solution Summary: The author evaluates the value of the line integral displaystyleundersetCoint.
Green’s Theorem for line integralsUse either form of Green’s Theorem to evaluate the following line integrals.
30.
∮
C
(
−
3
y
+
x
3
/
2
)
d
x
+
(
x
−
y
2
/
3
)
d
y
;
C is the boundary of the half disk {(x, y): x2 + y2 ≤ 2, y ≥ 0} with counterclockwise orientation.
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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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