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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
 one year ago
 one year ago
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
 one year ago
 one year ago

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timo86mBest ResponseYou've already chosen the best response.3
keep the light as the origin is 0
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
and the ground as zero also
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
there fore slope for the hypotonuse form is dy/dx (62) = dy which is fixed i think so 4 4/(02*x)
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
sfx x = 3.75*x1 by x i mean x+y 3.75*x1=2.5x1+unkn unkn = 6.25*x1
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
dw:1342238127714:dw another way to look at it
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
where m1= (64)/(2.5x) and m2=  6/ (2.5x+b) set them equal to eachothter find b
 one year ago

timo86mBest ResponseYou've already chosen the best response.3
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
 one year ago

robtobeyBest ResponseYou've already chosen the best response.0
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.\[dx=1.25 m/\sec \]
 one year ago
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