• anonymous
A man 6ft tall walks at a rate of 6ft/s away from a lamppost that is 23 ft high. At what rate is the length of his shadow changing when he is 65 ft away from the lamppost?
  • Stacey Warren - Expert
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  • chestercat
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  • myname
Let x be the distance between the man and the lamppost. Let y be the length of the shadow. You can draw a sketch and you will see 2 similar triangles. So, from the relation of similar triangles you can get 23/(x+y)=6/y Now, you can just solve for y in terms of x 6x+6y=23y 6x=17y y=6x/17 Here, we already have the rate of change of the distance between the man and the lamppost given(6ft/sec), which is in algebrical terms refered as :dx/dt= 6ft/sec Then lets implicitly differentiate the function y=6x/17 dy/dt=(6/17)*6 ft/sec dy/dt= 36/17 (ft/se) Therefore the rate of change of the man’s shadow when he is 65 fts away from the lamppost is 36/17(ft/sec). I hope this helps.

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