anonymous
  • anonymous
PLEASE HELP! PROBLEM ON RELATED RATES :) A ball is located at point Q on top of a flagpole 10m above the ground. a light is being raised vertically at the rate of 5 m/s casting the shadow of Q on the ground. how fast is the shadow moving along the ground when the light is 50m high?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
where is the light initially?
anonymous
  • anonymous
20 meters from the flagpole! coming from the top
amistre64
  • amistre64
dL/dt = 5 ....dh/dt = 0 dx/dt =? dx/dt = dx/dL * dx/dt find dx/dL ?? maybe not tho.....

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anonymous
  • anonymous
anonymous
  • anonymous
is there an initial height of the light?
amistre64
  • amistre64
ahhh...... the shadow moving problem in all the textbooks :)
anonymous
  • anonymous
well you need to cross multiply and divide. so it would be 20m/50m by 5m/x.... that would be 50x by 100... divide 50 by 100 to get x alone and you get x=2
amistre64
  • amistre64
thats a rough drawing by the way :)
amistre64
  • amistre64
h/(20+x) = 10/(x) if that helps :)
anonymous
  • anonymous
lol sorry thats what was given to me :p
amistre64
  • amistre64
y = (200+10x)/x
amistre64
  • amistre64
dx/dt = x^2(dy/dt)/-200 ?? maybe
amistre64
  • amistre64
dy/dy = 5
amistre64
  • amistre64
x=5
amistre64
  • amistre64
im gonna take a gander and say... 25(5) = 125 125/-200 = ?? -.0625 ?? if I did it right lol
amistre64
  • amistre64
i think I got the "equation" wrong at the start....
anonymous
  • anonymous
lol thank you! ..but i dont understand how can you have dy/dy? :s
anonymous
  • anonymous
lol so the whole thing is wrong??
amistre64
  • amistre64
:) fat fingers...little tiny keyboard :)
anonymous
  • anonymous
LMAO! ohhhh i see (:
amistre64
  • amistre64
it may be right, but I dont trust it yet.... need to find the equation to derive; then the rest is easy.
anonymous
  • anonymous
ya but thats the part i cant get! wahh :(
amistre64
  • amistre64
I am thinking it is the relation of congruent triangles; "y" being the height of the lamp: x = length of shadow. 10/x = y/(20+x) is what I come up with; there is no question about the distance of the light to the tip of the shadow...
amistre64
  • amistre64
10(20+x) = y(x) (200+10x)/x = y 200/x + 10 = y 200/x = y-10 200/(y-10) = x ??? sounds good
amistre64
  • amistre64
this relate y and x..... now implicit it with respect to time right?
anonymous
  • anonymous
yea thats what i gotta do
anonymous
  • anonymous
and thank youuu! for everything so far btw!
amistre64
  • amistre64
-200(dy/dt) / (y-10)^2 = dx/dt dy/dt = 5 y = 50;
amistre64
  • amistre64
no prob; I live to math things up lol
amistre64
  • amistre64
-0.625 is what i get, and I am almost sure its right... but i get that nagging feeling that I am just being stupid :)
anonymous
  • anonymous
hahaha but where did u get negative two hundred?
amistre64
  • amistre64
ok.... step it out.... 200 x = ------- (y-10) (y-10)(0) - (200)(dy/dt) dx/dt = ---------------------- (y-10)^2
anonymous
  • anonymous
lol im on the phone with my friend trying to figure this out and im reading her what u wrote and we both agreed your funny, just thought id let you know :p but yea we dont understand also where you got the 50 from? (: care to explain?
amistre64
  • amistre64
-200(dy/dt) dx/dt = ------------- (y-10)^2
amistre64
  • amistre64
when y=50...its right there :)
amistre64
  • amistre64
dy/dt is given already as the rate of change in "y"... = 5 m/s
amistre64
  • amistre64
as y rises; x shrinks.... so its (-) direction..
anonymous
  • anonymous
ohhh okay! thank you so so much honestly, it means a lot (: (:
amistre64
  • amistre64
just let me know if im right :)
anonymous
  • anonymous
i will haha but its the first time im on this site so how can i tell you?
anonymous
  • anonymous
because i have to check with my teacher first
amistre64
  • amistre64
shout really loud? :) AMISTRE WAS RIGHT!!!.... or if im wrong it would be: AMISTRE WAS WRONG .... ill feel it either way :)
anonymous
  • anonymous
LMAOOOOOOOOO!
amistre64
  • amistre64
that moving shadow thing is in all the calc textbooks If seen, and it always makes me nervous
anonymous
  • anonymous
lol im sure you got it right though, it seemed to make sense. i found it impossible!
amistre64
  • amistre64
good luck :)
anonymous
  • anonymous
thank you again!:)

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