anonymous
  • anonymous
Calculus related rates problem. A street light is mounted at the top of a 15ft tall pole. A man 6ft tall walks away from the pole at with a speed of 5ft/s along a straight path. How fast is the tip of his shadow moving when he is 40ft from the pole.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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zepdrix
  • zepdrix
hmm
anonymous
  • anonymous
I can draw a diagram if you want to help you
zepdrix
  • zepdrix
lol

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zepdrix
  • zepdrix
Ya that might be helpful.
anonymous
  • anonymous
|dw:1446416743747:dw|
zepdrix
  • zepdrix
Aw I drew my triangle the other direction >.< Now I have to think backwards lol
zepdrix
  • zepdrix
|dw:1446417028893:dw|
zepdrix
  • zepdrix
Using similar triangles:\[\large\rm \frac{15}{x+a}=\frac{6}{a}\]Cross multiply and other stuff,\[\large\rm 15a=6x+6a\]\[\large\rm 9a=6x\]\[\large\rm a=\frac{2}{3}x\]And then just differentiate with respect to time I think?
anonymous
  • anonymous
ah, thats what I was forgetting to do, I was on the right track but, forgot to add the total distance in the x component
freckles
  • freckles
rate the tip of the shadow is moving is x'+a'
freckles
  • freckles
and yes I know that seems weird I honestly used to think it was just a'
anonymous
  • anonymous
ok ill tackle this in a bit, i gtg eat
freckles
  • freckles
@zepdrix do you know why it is x'+a' and not just a'? maybe I just don't understand shadows real well
freckles
  • freckles
or @IrishBoy123
IrishBoy123
  • IrishBoy123
|dw:1446476991782:dw| we want the velocity of the shadow relative to the ground or something else that is fixed to ground ... here, the base of the lamp, because that is convenient mathematically. so start with the distance \(y\) of the shadow tip from the lamp. \(\large y = x + s\) from sim triangles \(s = \dfrac{2}{3}x\) \(\large \dot y = \dot{(x + s)} = \dfrac{5}{3} \dot x\) which is the rate at which the distance between the lamp and the shadow's edge is increasing, aka the velocity of the shadow's tip this means that \(\large \dot s\) is the velocity at which the shadow's tip is racing away from the man because the man's velocity relative to the lamp is \(\large \dot x\). so if you glued the man to the street and moved the lamp away from him at a velocity of \(\large \dot x\) the shadow's tip should move in the other direction, relative to the may at \(\dot s\) and relative to the lamp at \(\dot y = \dot x + \dot s\)

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