A street light is mounted at the top of a 15ft tall pole. A man 6ft tall walks away from the pole with a speed of 5ft/sec along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole?

Answer 1

#8.33(ft.)/(sec.)#

The street light is mounted at the top of a #15ft# tall pole. Let us consider the man #6ft# tall #xft# away from the pole. His shadow forms two ends - one end is at his feet and the shadow extends away from the pole till the tip of the shadow.

Let this be depicted by the figure shown below.

Here the distance of the man from lamp post be #xft.# and let his shadow be #yft# from man. Now as man is moving away from lamp post, #x# is a function of #t# and speed of man is #(dx)/(dt)#

Then, as they form similar triangle, we have

#15/(15-6)=(x+y)/x# i.e. #15x=9x+9y# or #9y=6x# and #y=2/3x#

and shadow is #x+2/3x=5/3x# from lamp post. And hence when man moves #deltax# feet, shadow moves #5/3deltax# feet

and hence, shadow moves with a speed of #5/3(dx)/(dt)# i.e. #5/3xx5=25/3=8.33(ft.)/(sec.)#

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Answer 2

To find the rate at which the tip of the man's shadow is moving, we can use related rates. Let ( x ) be the distance between the man and the base of the pole, and let ( y ) be the length of the shadow.

We are given that ( \frac{{dx}}{{dt}} = 5 ) ft/sec and we want to find ( \frac{{dy}}{{dt}} ) when ( x = 40 ) ft.

Using similar triangles, we have:

[ \frac{{y}}{{6}} = \frac{{y + 15}}{{x}} ]

Differentiate both sides with respect to time ( t ):

[ \frac{{d}}{{dt}}\left(\frac{{y}}{{6}}\right) = \frac{{d}}{{dt}}\left(\frac{{y + 15}}{{x}}\right) ]

[ \frac{1}{6}\frac{{dy}}{{dt}} = \frac{{x(y' + 0) - (y + 15)(5)}}{{x^2}} ]

Substitute ( x = 40 ), ( y = 40 + 15 = 55 ), and ( \frac{{dx}}{{dt}} = 5 ):

[ \frac{1}{6}\frac{{dy}}{{dt}} = \frac{{40(y' + 0) - (55)(5)}}{{40^2}} ]

[ \frac{1}{6}\frac{{dy}}{{dt}} = \frac{{40y' - 275}}{{1600}} ]

[ \frac{{dy}}{{dt}} = \frac{{240y' - 1650}}{{1600}} ]

Now, plug in ( y = 55 ) to solve for ( y' ):

[ \frac{{dy}}{{dt}} = \frac{{240y' - 1650}}{{1600}} ]

[ 5 = \frac{{240y' - 1650}}{{1600}} ]

[ 800 = 240y' - 1650 ]

[ 240y' = 2450 ]

[ y' = \frac{{2450}}{{240}} ]

[ y' = \frac{{49}}{{4}} ]

So, when the man is 40 ft from the pole, the tip of his shadow is moving at a rate of ( \frac{{49}}{{4}} ) ft/sec.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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