The function F of v if the actual frequency of sound emitted by the ambulance is 560 H z when an ambulance is moving toward an observer and the frequency of a sound relative to an observer is given by F v = f a s 0 s 0 − v , where f a is the actual frequency of the sound at the source, s 0 is the speed of the sound in air 772.4 mph , and v is the speed at which the source of sound is moving toward the observer.
The function F of v if the actual frequency of sound emitted by the ambulance is 560 H z when an ambulance is moving toward an observer and the frequency of a sound relative to an observer is given by F v = f a s 0 s 0 − v , where f a is the actual frequency of the sound at the source, s 0 is the speed of the sound in air 772.4 mph , and v is the speed at which the source of sound is moving toward the observer.
Solution Summary: The author explains the cost function F of v if the actual frequency of sound emitted by the ambulance is 560Hz.
The function F of v if the actual frequency of sound emitted by the ambulance is 560Hz when an ambulance is moving toward an observer and the frequency of a sound relative to an observer is given by Fv=fas0s0−v , where fa is the actual frequency of the sound at the source, s0 is the speed of the sound in air 772.4mph , and v is the speed at which the source of sound is moving toward the observer.
(b)
To determine
To graph: The sketch of a function given by Fv=560772.4772.4−v on the window 0,1000,100by0,5000,1000 .
(c)
To determine
The effect of the frequency of sound when the speed of the ambulance increases.
EXAMPLE 3
Find
S
X
√√2-2x2
dx.
SOLUTION Let u = 2 - 2x². Then du =
Χ
dx =
2- 2x²
=
信
du
dx, so x dx =
du and
u-1/2 du
(2√u) + C
+ C (in terms of x).
Let g(z) =
z-i
z+i'
(a) Evaluate g(i) and g(1).
(b) Evaluate the limits
lim g(z), and lim g(z).
2-12
(c) Find the image of the real axis under g.
(d) Find the image of the upper half plane {z: Iz > 0} under the function g.
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