## virtus Group Title 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 2 years ago 2 years ago

1. virtus Group Title

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2. timo86m Group Title

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3. virtus Group Title

i don't understand?

4. timo86m Group Title

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5. timo86m Group Title

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6. timo86m Group Title

keep the light as the origin is 0

7. timo86m Group Title

and the ground as zero also

8. timo86m Group Title

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 Group Title

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10. timo86m Group Title

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11. timo86m Group Title

sfx x = -3.75*x1 by x i mean x+y -3.75*x1=2.5x1+unkn unkn = -6.25*x1

12. timo86m Group Title

|dw:1342238127714:dw| another way to look at it

13. timo86m Group Title

where m1= -(6-4)/(2.5x) and m2= - 6/ (2.5x+b) set them equal to eachothter find b

14. timo86m Group Title

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 Group Title

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.$dx=1.25 m/\sec$