Question

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?

Answer 1

where is the light initially?

Answer 2

20 meters from the flagpole!
coming from the top

Answer 3

dL/dt = 5 ....dh/dt = 0 dx/dt =?

dx/dt = dx/dL * dx/dt

find dx/dL ?? maybe not tho.....

Answer 4

is there an initial height of the light?

Answer 5

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

Answer 6

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

Answer 7

thats a rough drawing by the way :)

Answer 8

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

Answer 9

lol sorry thats what was given to me :p

Answer 10

y = (200+10x)/x

Answer 11

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

Answer 12

dy/dy = 5

Answer 13

x=5

Answer 14

im gonna take a gander and say... 25(5) = 125

125/-200 = ?? -.0625 ??

if I did it right lol

Answer 15

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

Answer 16

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

Answer 17

lol so the whole thing is wrong??

Answer 18

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

Answer 19

LMAO! ohhhh i see (:

Answer 20

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

Answer 21

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

Answer 22

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...

Answer 23

10(20+x) = y(x)

(200+10x)/x = y

200/x + 10 = y

200/x = y-10

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

Answer 24

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

Answer 25

yea thats what i gotta do

Answer 26

and thank youuu!
for everything so far btw!

Answer 27

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

dy/dt = 5

y = 50;

Answer 28

no prob; I live to math things up lol

Answer 29

-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 :)

Answer 30

hahaha but where did u get negative two hundred?

Answer 31

ok.... step it out....

200
x = -------
(y-10)


(y-10)(0) - (200)(dy/dt)
dx/dt = ----------------------
(y-10)^2

Answer 32

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?

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

just let me know if im right :)

Answer 39

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

Answer 40

because i have to check with my teacher first

Answer 41

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

Answer 42

LMAOOOOOOOOO!

Answer 43

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

Answer 44

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

Answer 45

good luck :)

Answer 46

thank you again!:)