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?

- anonymous

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- anonymous

where is the light initially?

- anonymous

20 meters from the flagpole!
coming from the top

- 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

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- anonymous

is there an initial height of the light?

- amistre64

ahhh...... the shadow moving problem in all the textbooks :)

- 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

thats a rough drawing by the way :)

- amistre64

h/(20+x) = 10/(x) if that helps :)

- anonymous

lol sorry thats what was given to me :p

- amistre64

y = (200+10x)/x

- amistre64

dx/dt = x^2(dy/dt)/-200 ?? maybe

- amistre64

dy/dy = 5

- amistre64

x=5

- amistre64

im gonna take a gander and say... 25(5) = 125
125/-200 = ?? -.0625 ??
if I did it right lol

- amistre64

i think I got the "equation" wrong at the start....

- anonymous

lol thank you! ..but i dont understand
how can you have dy/dy? :s

- anonymous

lol so the whole thing is wrong??

- amistre64

:) fat fingers...little tiny keyboard :)

- anonymous

LMAO! ohhhh i see (:

- amistre64

it may be right, but I dont trust it yet.... need to find the equation to derive; then the rest is easy.

- anonymous

ya but thats the part i cant get! wahh :(

- 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

10(20+x) = y(x)
(200+10x)/x = y
200/x + 10 = y
200/x = y-10
200/(y-10) = x ??? sounds good

- amistre64

this relate y and x..... now implicit it with respect to time right?

- anonymous

yea thats what i gotta do

- anonymous

and thank youuu!
for everything so far btw!

- amistre64

-200(dy/dt) / (y-10)^2 = dx/dt
dy/dt = 5
y = 50;

- amistre64

no prob; I live to math things up lol

- 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

hahaha but where did u get negative two hundred?

- amistre64

ok.... step it out....
200
x = -------
(y-10)
(y-10)(0) - (200)(dy/dt)
dx/dt = ----------------------
(y-10)^2

- 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

-200(dy/dt)
dx/dt = -------------
(y-10)^2

- amistre64

when y=50...its right there :)

- amistre64

dy/dt is given already as the rate of change in "y"... = 5 m/s

- amistre64

as y rises; x shrinks.... so its (-) direction..

- anonymous

ohhh okay!
thank you so so much
honestly, it means a lot (: (:

- amistre64

just let me know if im right :)

- anonymous

i will haha
but its the first time im on this site so how can i tell you?

- anonymous

because i have to check with my teacher first

- amistre64

shout really loud? :) AMISTRE WAS RIGHT!!!.... or if im wrong it would be: AMISTRE WAS WRONG .... ill feel it either way :)

- anonymous

LMAOOOOOOOOO!

- amistre64

that moving shadow thing is in all the calc textbooks If seen, and it always makes me nervous

- anonymous

lol im sure you got it right though, it seemed to make sense. i found it impossible!

- amistre64

good luck :)

- anonymous

thank you again!:)

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