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Mathematics 15 Online
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?

OpenStudy (anonymous):

What next?

OpenStudy (perl):

|dw:1426996061615:dw|

OpenStudy (perl):

by similar triangles you have

OpenStudy (perl):

$$ \Large{ \frac{10}{z} = \frac {5}{z-x} \\10~(z-x) = 5z \\ 10z - 10x = 5z \\10z - 5z = 10x \\ \\ 5z = 10 x \\ z = 2x \\ \therefore \\ \frac{dz}{dt} =2\frac{dx}{dt} } $$

OpenStudy (anonymous):

Is dz/dt 7.5?

OpenStudy (anonymous):

Or dx/dt?

OpenStudy (perl):

dx/dt

OpenStudy (anonymous):

So dz/dt is the answer??

OpenStudy (anonymous):

Why is it 0.75, but not 7.5?

OpenStudy (perl):

typo

OpenStudy (perl):

dz/dt = 2*7.5

OpenStudy (anonymous):

Is this the rate at which the tip of her shadow is moving?

OpenStudy (perl):

yes

OpenStudy (anonymous):

Then what about the rate at which the shadow is lengthening?

OpenStudy (perl):

in our diagram, z - x is the length of the shadow

OpenStudy (anonymous):

So it's 7.5

OpenStudy (perl):

d/dt (z - x ) = dz/dt - dx/dt = 2*7.5 - 7.5 = 7.5

OpenStudy (perl):

yes thats correct

OpenStudy (anonymous):

Ok awesome, it makes sense now. Thank you for your help!

OpenStudy (perl):

:)

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