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OpenStudy (anonymous):
A woman 5 ft tall walks at the rate of 7.5 ft/sec away from a streetlight that is 10 ft above the ground.
At what rate is the tip of her shadow moving?
At what rate is her shadow lengthening?
11 years ago
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OpenStudy (anonymous):
What next?
11 years ago
OpenStudy (perl):
|dw:1426996061615:dw|
11 years ago
OpenStudy (perl):
by similar triangles you have
11 years ago
OpenStudy (perl):
$$
\Large{ \frac{10}{z} = \frac {5}{z-x}
\\10~(z-x) = 5z
\\ 10z - 10x = 5z
\\10z - 5z = 10x \\
\\ 5z = 10 x
\\ z = 2x \\ \therefore \\
\frac{dz}{dt} =2\frac{dx}{dt}
}
$$
11 years ago
OpenStudy (anonymous):
Is dz/dt 7.5?
11 years ago
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OpenStudy (anonymous):
Or dx/dt?
11 years ago
OpenStudy (perl):
dx/dt
11 years ago
OpenStudy (anonymous):
So dz/dt is the answer??
11 years ago
OpenStudy (anonymous):
Why is it 0.75, but not 7.5?
11 years ago
OpenStudy (perl):
typo
11 years ago
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OpenStudy (perl):
dz/dt = 2*7.5
11 years ago
OpenStudy (anonymous):
Is this the rate at which the tip of her shadow is moving?
11 years ago
OpenStudy (perl):
yes
11 years ago
OpenStudy (anonymous):
Then what about the rate at which the shadow is lengthening?
11 years ago
OpenStudy (perl):
in our diagram, z - x is the length of the shadow
11 years ago
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OpenStudy (anonymous):
So it's 7.5
11 years ago
OpenStudy (perl):
d/dt (z - x ) = dz/dt - dx/dt
= 2*7.5 - 7.5 = 7.5
11 years ago
OpenStudy (perl):
yes thats correct
11 years ago
OpenStudy (anonymous):
Ok awesome, it makes sense now. Thank you for your help!
11 years ago
OpenStudy (perl):
:)
11 years ago
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