Question

A floodlight illuminates a tall vertical wall that is 6m from it. A man, 2m tall, walks towards the wall in front of the light. HELP! I need to find a function that gives the height of his shadow as a function of his distance from the light???

Answer 1

application of derivatives :)

Answer 2

how fast is he walking?

Answer 3

nice art!

Answer 4

i got a tattoo like that lol

Answer 5

\[\frac{ds}{dt}=\frac{ds}{}\frac{}{dt}\]
we need something to relate this to; most likely the speed of the walk

\[\frac{ds}{dt}=\frac{ds}{dw}\frac{dw}{dt}\]
or some combo if chains

Answer 6

6-d = x
h = 2

s^2 = (6-d)^2 + 2^2

Answer 7

im thinking of this as how fast the shadow is moving; wrong question altogether

Answer 8

the shadow itself as a function

Answer 9

you have similar triangles;

6 is to 6-(distance from light)
as
2 is to (height of shadow)

Answer 10

ya, I'm confused. There is no rate given?!?! That literally is what the question says.....I'm doing a Calculus course by correspondence....frustrating to say the least! The next question says "what is his shadow height when he is 1m from the light?"...and i| know the answer is 12?!?!? does that help??

Answer 11

but 6 is a horizontal measure as well as x...I was thinking hte same thing but having a mental block...can't make up the triangle, as I dont' know the height of the light?

Answer 12

the ratio would be:
\[\frac{6}{2}=\frac{6-d}{sh}\]

Answer 13

the light is at ground level; its a flood light

Answer 14

my 6-d is wrong; it should simply be d

Answer 15

\[\frac{6}{2}=\frac{d}{sh}\frac{\text{ distance from light}}{\text{ height of shadow}}\]

the ratios are good since we are ddealing with similar triangles right?

Answer 16

height of his shadow as a function of his distance from the light
got to revise it still

\[\frac{6}{sh}=\frac{d}{2}\]
thats better

Answer 17

shadow height = 12/d

Answer 18

does that make sense to you?

Answer 19

similar triangles produce a scalar effect; so we ratio the measurements

Answer 20

sort of....it makes sense because I understand that I needed to set up ratios, and I know the answer when d=1 is 12, but why match the 6 with the s and the d with his height? how will I know for future questions which ones match up?

Answer 21

the wall never moves; so its always at a distance of 6; the shadow is always on the wall, so its gots to be part of that specific triangle

the height of the man never changes; so it has to be part of the triangle that moves as the man moves

Answer 22

man height (2) is to shadow height
as
distance is to wall distance (6)

\[\frac{2}{d}=\frac{sh}{6}\]
solve for sh: sh = 12/d

Answer 23

ok, got it, thanks! next questions is to find a function that gives the rate of change of the shadow, so, I guess differentiate it,,,,,correct?

Answer 24

rate of change with respect to what? time or distance?

Answer 25

distance

Answer 26

might as well do it both ways :)

Answer 27

wrt distance we get
\[\frac{d}{dd}(sh=12/d)\implies\frac{d\ sh}{dd}=\frac{-12}{d^2}*\frac{dd}{dd}\]

Answer 28

wrt time its pretty much the same:

\[\frac{d}{dt}(sh=12/d)\implies\frac{d\ sh}{dt}=\frac{-12}{d^2}*\frac{dd}{dt}\]

notice the denominator "dd" is now "dt"

Answer 29

dd/dt is the speed at which the guy walks

Answer 30

what's dd/dd in above answer? I got -12/d^2 but didn't write dd/dd...is that wrong?

Answer 31

its a part of derivatives that they never teach you, hoping to confuse you later on ...

recall that the chain rule always pops out a derived bit. for example:

\[\frac{d}{dx}(5y^3)=15y^2*\frac{dy}{dx}\text{ ;the chain rule poped out a derived bit of dy/dx}\]

\[\frac{d}{dx}(6x^2)=12x*\frac{dx}{dx}\text{ ; since dx/dx = 1 they tend to omit it}\]

Answer 32

dd/dd is just the derived bit of distance with respect to distance

Answer 33

it would do you good to keep all your derived bit intact until the end so that you can sort thru them afterwards. make sense?

Answer 34

yup. thanks so much! probably be asking for more help shortly....we'll see! lol!