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virtus

  • 3 years 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

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  1. virtus
    • 3 years ago
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    |dw:1342236117226:dw|

  2. timo86m
    • 3 years ago
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    |dw:1342236213718:dw|

  3. virtus
    • 3 years ago
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    i don't understand?

  4. timo86m
    • 3 years ago
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    |dw:1342236435888:dw|

  5. timo86m
    • 3 years ago
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    |dw:1342236535853:dw|

  6. timo86m
    • 3 years ago
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    keep the light as the origin is 0

  7. timo86m
    • 3 years ago
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    and the ground as zero also

  8. timo86m
    • 3 years ago
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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)

  9. timo86m
    • 3 years ago
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    |dw:1342237451450:dw|

  10. timo86m
    • 3 years ago
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    |dw:1342237577634:dw|

  11. timo86m
    • 3 years ago
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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

  12. timo86m
    • 3 years ago
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    |dw:1342238127714:dw| another way to look at it

  13. timo86m
    • 3 years ago
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    where m1= -(6-4)/(2.5x) and m2= - 6/ (2.5x+b) set them equal to eachothter find b

  14. timo86m
    • 3 years ago
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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

  15. robtobey
    • 3 years ago
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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.\[dx=1.25 m/\sec \]

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