virtus
A 6m high vertical street lamp stands on horizontal ground. A 2 metre tall man runs away from the street lamp at a constant speed of 2.5m/s. When he is y metres from the street lamp his shadow has length x metres.
Find the rate at which his shadow lengthens
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virtus
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timo86m
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virtus
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i don't understand?
timo86m
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timo86m
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timo86m
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keep the light as the origin is 0
timo86m
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and the ground as zero also
timo86m
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there fore slope for the hypotonuse form is
dy/dx
(6-2) = dy which is fixed i think so 4
4/(0-2*x)
timo86m
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timo86m
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timo86m
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sfx x = -3.75*x1 by x i mean x+y
-3.75*x1=2.5x1+unkn
unkn = -6.25*x1
timo86m
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another way to look at it
timo86m
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where m1= -(6-4)/(2.5x) and m2= - 6/ (2.5x+b)
set them equal to eachothter find b
timo86m
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b=5x
so
s a b
1 2.5 5
2 5 10
3 7.5 15
It looks like everysecond the shadow gets 5 meters longer
so it is a constant growth of 5 m/s
i think
robtobey
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From the figure,\[\frac{y+x}{6}=\frac{x}{2} \]The Total Derivative of the above equation is:\[\frac{1}{6} (dx+dy)=\frac{dx}{2} \]Solve for dx:\[dx=\frac{dy}{2} \]Replace dy with 2.5 and simplify.\[dx=1.25 m/\sec \]