A person 6ft tall stands 10ft from point P directly beneath a lantern hanging 30ft above the ground. The lantern start to fall, thus causing the person shadow to lengthen(L). Given that the lantern falls 16t^2 ft in t seconds, how fast will the shadow be lengthening when t=1
*
l
l 30ft l 6ft s= the height f the falling lantern
l___10ft__l__L___
I got S/(x+10)=6/x solved for s=(6x+60)/x
put that in for -16t^2+30=s
found the x value for when the height has come down for a second (30-16) and got 7.5. took the derivative in respect to t
-32t=6x(-60/x^2)(dx/dt) d

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

it cut me off dx/dt = 3/2

- amistre64

if helps out greatly if you use variables that are more closely attuned to the tings you are trying to describe by them.
s for "s"hadow
d for "d"istance
b for "b"alloon
l for "l"ight
etc .......

- amistre64

in tis case youve got an application of similiar triangles:
6 is to "d"istance
as
"s"+10 is to "s"hadow
\[\frac{6}{s+10}=\frac{d}{s}\to\ 6s=d(s+10)\]
\[\frac{6}{s+10}=\frac{d}{s}\to\ 6s=sd+10d\]
take the derivative to pop out all the rates of change of the variables
\[6s' = s'd+sd' + 10d'\]
since you want to know how fast the shadow is moving, s', solve the eq for s'
\[6s'-s' =d'(s+10)\]
\[s'(6-1) =d'(s+10)\]
\[s' =\frac{d'(s+10)}{5}\]
now all we have to do is determine how fast d' is moving and how long s is, when t=1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

so i did it wrong? lol

- anonymous

im guessing d is going to be moving at 32ft/s because the derivative of the distance equation is the velocity equation...

- anonymous

pellet I don't get it I'm over calc this is ridiculous.

- anonymous

I dont understand how my way doesn't work...

- amistre64

*I got S/(x+10)=6/x solved for s=(6x+60)/x
S/(x+10)=6/x ; the variables are just for human convenience; so your write up of it is fine so far
s = 6(x+10)/x = (6x+60)/x is fine as well.
*put that in for -16t^2+30=s
your evaluating the height of the lamp at any given time value; thats fine
*found the x value for when the height has come down for a second (30-16) and got 7.5.
s = 30-16 = 14 , when t=1
14 = (6x+60)/x
14x = 6x+60
8x = 60
x = 60/8 = 30/4 = 15/2 = 7.5 , good
took the derivative in respect to t
-32t=6x(-60/x^2)(dx/dt)
this is a little obscure to me. we want to use a formula that gives us a relationship between the shadow the the distance the lamp is from the ground.
if i try to make sense of this tho ...
s = -16t^2 + 30; s' = 32t ; the negative is just a direction that provides
no relevant information
s = (6x+60)/x
sx = 6x+60
s'x+sx' = 6x'
s'x = x'(6-s)
x' = s'x/(6-s)
this looks fine by me; i see an error in my typing above; ive got my s+10 and my s under the wrong parts.
\[\frac{6}{s}=\frac{d}{s+10}\to\ 6(s+10)=ds\]
\[6s' = d's + ds'\]
\[s'(6-d)=d's\]
\[s'=\frac{d's}{6-d}\]
\[s'=\frac{32(7.5)}{6-(32-16)}\]
\[s'=\frac{32(7.5)}{-10}\]
\[s'=-16(1.5)=-24\]

- anonymous

hmm the answer is 30. I really appreciate taking the time to go through this. I just need to start figuring out the relationships a little more easily.

- amistre64

30 eh ....

- anonymous

actually I think you did this problem for me before

- anonymous

let me go look

- anonymous

http://openstudy.com/users/awstinf#/updates/4f78ee9ee4b0ddcbb89e7c5f yeah you did this for me before I just didn't get it lol we did something a little different

- anonymous

hmm I wonder if would be better to differentiate right away to see what things i need in respect to time

- amistre64

its best to play around with it and take it apart; break it, make mistakes with it; and learn from the experiences :)

- amistre64

|dw:1333469963580:dw|

- anonymous

wow i figured out a really easy way to do this lol so it turns out that -32 does do some magic for us if i just do implicit differentiation on this bad boy ds=6s+60 and fill in the blanks it woulds out really well

- amistre64

notice that the ratio is not 6 to d; but is a ration of the lower part 30-d

- anonymous

SEE those are the pictures i should draw

- anonymous

in the equation i just wrote d being the distant at time t=1

- anonymous

using -16t^2+30

- amistre64

yes

- anonymous

if it weren't for physics id be failing this class lol

- anonymous

I just have to do every situation of one of these and just hope one of them is on the test

- amistre64

physics do make the world go round :)

- anonymous

hahaha great pun

- anonymous

I wonder if that method will work for every similar triangles one
the key is just to find the two parts that are changing in respect with earh other and find a mathematical relationship in terms of 1 variable?

- amistre64

some good practice problems can be found at interactmath.com i believe

- amistre64

in the similar tris, yes; but sometimes that information given makes us go into trig for an answer as well

- anonymous

I will check that out, out of curiosity does openstudy pay you? they should you and myn answer all of the questions

- amistre64

lets say; man 5 foot tall; walking away from a lamppost that has a light 12 foot high at a constant rate of 3feet per sec.. how fast is the tip of his shadow moving when he is 20 feet away from the lamppost?

- amistre64

this is volunteer stuff; something to pass the time with so no i dont get paid

- anonymous

drawing this sec

- anonymous

hmm I should keep the 15/2= 12/s+20 is what I have written so far

- anonymous

15/s*

- anonymous

gah 5/s*

- amistre64

is 20 a constant? or is it changing?

- anonymous

if my s+20 is changing at a rate shouldn't i leave that as a variable

- anonymous

like maybe set y=s+20?

- amistre64

|dw:1333470857325:dw|

- anonymous

if I do that and cross multiply to get 5d=12s can I just differentiate there? (i changed y to d)

- amistre64

i wouldnt add extra variables; work with what youve got

- anonymous

aww yeah now I remember...

- anonymous

(s+d)5=12s -> 5s+5d=12s -> 7s=5d-> 7s'=5d'? 15/7=s'

- amistre64

\[\frac{5}{s}=\frac{12}{s+d}\]
\[5(s+d)=12s\]
\[D[5s+5d=12s]=5s' + 5d' = 12s'\]
\[5d' = 12s'-5s'\]
\[5d' = 7s'\]
\[\frac{5}{7}d' = s'\]

- anonymous

sweet so I did it right?lol

- amistre64

as you can see; there is no place to include "d" or "s" in the rate of change now is there

- amistre64

so the rate of change at 20 feet is the same as at 5 feet is the same as at 150 feet

- anonymous

ahhh this is kind of a cool I just wish I were better at it lol. I think i understand this stuff a little bit more

- anonymous

I only have 16-20 more hours to study for this thing though idk how ill do

- amistre64

now, the question i asked asks how fast the top of the shadow is moving; not the shadow itself; notice that the tip of the shadow moves at a rate of d' + s'

- amistre64

|dw:1333471260552:dw|

- anonymous

ohhhh wow didn't even realize that

- anonymous

I gotta draw this stuff better instead of just trying to power through it

- amistre64

\[tip' = d'+s';\ but\ s'=\frac{5}{7}d'\]
\[tip'=3+\frac{15}{7}\]

- anonymous

did you just make this question up?

- amistre64

the numbers yes; the format, no. its pretty standard

- anonymous

haha okay I was like that was a pretty good question to just have randomly made up

- amistre64

heres another basic format:
a ballon is rising at a rate of blah; you are standing blah distance from it watching it float up; how fast is the angle changing when the balloon is so many feet in the air

- anonymous

don't you have to know the derivatives of inverse functions for that one though?

- amistre64

|dw:1333471559263:dw|

- anonymous

a rocket is launched vertically from a point on level ground that is 3000 feet from the observer with binoculars. if the rocket is rising vertically at the rate of 750 ft/s at the instant it is 4000 above the ground, how fast must the observer change the angle of elevation of her line of sight in order to keep the rocket in sight at that instant?

- anonymous

like that?

- amistre64

yep

- anonymous

x/3000=tantheta

- amistre64

relate t and b ... tangent, good

- amistre64

how fast is the distance between you and the ballon changing?

- anonymous

1/(1+(x/3000)^2) i never got this part but 3000/((3000^2)+x^2) * dx/dt

- amistre64

|dw:1333471697722:dw|

- amistre64

for t'
tan(t) = b/d
sec^2(t) t' = b'/d
t' = b' cos^2(t)/d

- anonymous

gah i have no idea haha we can't use calculators :\ do I'm not sure learning it that way will be useful

- amistre64

pythag it for r'
d^2 + b^2 = r^2

- anonymous

so finding that you would take the d of r err'=2dd'+2bb'?

- amistre64

2dd' + 2bb' = 2rr' yes

- amistre64

d is constant so d' = 0

- anonymous

shouldn't we eliminate the one that isn't changing which is our distance between the ballon

- anonymous

yeah haha

- amistre64

dont think eliminate; think; d' is what i will use for the rate of change of d
since d'=0 we can then insert it and clean thngs up in order

- anonymous

ahhh okay that does make more sense mathematically

- anonymous

now take my test for me amistre!

- amistre64

lol, but i might fail ;)

- anonymous

the first test was a joke somehow myself and one other person walked away with 100s and this test is going to be the death of me

- amistre64

a conic tank has a leak and is being filled up at some rate by a hose; how fast is the radius changing when the tank is half full?

- amistre64

this one a made up, but its an amalgamation of others ive come across

- anonymous

r/h=r/2/h/2?

- amistre64

oh theres a bit of work to do to it; determine the h and r when at V/2; relate in and out to determine V' to determine such and such

- anonymous

gah hahaha hmm so this is a volume problem

- amistre64

it volume related yes; so we might need a formula for cone volume

- anonymous

1/3pir^2h?

- amistre64

notice that the rate of change in volume can be simplified at rate in - rate out as well

- anonymous

what do you want it in terms of?

- amistre64

yes
\[V = \frac{pi}{3}r^2h\]
\[V' = \frac{2rh\ pi}{3}r'+\frac{r^2pi}{3}r^2h'\]
\[V' = in' - out';\ also\]

- amistre64

if we solve for r' ; our intended target ... and fill in the rest; that should do it

- anonymous

oh this is way more complex than I had originally though

- amistre64

pfft, we only have 2 variable to concern ourselves with at this point and the rest of the information would be supplied

- anonymous

can I set V'/2 = that over 2? after taking the derivative? or sub in for h or something?

- amistre64

maybe, but sincce i dont see what your seeing i would be leary of it

- amistre64

we want to know the h and r at V/2 ; not rates of change of V' at V/2

- anonymous

hmmmm

- amistre64

\[V' = \frac{2rh\ pi}{3}r'+\frac{r^2pi}{3}h'\]
\[V'-\frac{r^2\ pi}{3}h' = \frac{2rh\ pi}{3}r'\]
\[3V'-r^2 pi\ h' = 2rh\ pir'\]
\[\frac{3V'-r^2 pi}{2rh\ pi}\ h' = r'\]
\[\frac{3(in'-out')-r^2 pi}{2rh\ pi}\ h' = r'\]

- amistre64

\[\frac{V}{2}=\frac{1}{2*3}pi\ r^2h\]
\[\frac{V}{2}=\frac{1}{6}pi\ r^2h\]

- anonymous

wow this is quit complex I don't know if I would have ever gotten this lol

- amistre64

as is, we would have to be given some specifics to determine a proper value; but im sure this is all the stuff that would help us find the answer :)

- amistre64

this wont be on the test

- anonymous

you never know! I think I'm really starting to understand what all of these damn word problems mean

- amistre64

the idea tho; is to draw a diagram and label all the parts that are of interest; and give them the info provided; place arrows in teh directions of change and relate the given info to the desired results

- anonymous

I would never want to be a mathematician...

- amistre64

lol, it does get better. you find out that they tend to teach math backwards for some reason

- anonymous

hahaha what do you mean? teach backwards?

- anonymous

calc II sounds like hell I'm opting out and taking basic stats

- amistre64

|dw:1333473490855:dw|
the math we learn today is built upon a whole lot of past stuff that gets us to today.
we learn math today; and as we progress higher in math we learn how we got to today.

- amistre64

so we are learning math backwards

- anonymous

OH i get it hahaha I was like wtf

- anonymous

damn you isaac newton....

- anonymous

/everyone other person that contributed to calculus

- anonymous

everyother*

- amistre64

and leibnez, and wallis, and Laplace, and Euler, and and and lol

- anonymous

guass!

- anonymous

gauss*

- amistre64

not so much guass, he just figured out how to count really :)

- anonymous

hahahaha this is true

- anonymous

i loved matrices in precalc

- amistre64

oh those get more complicated

- amistre64

linear algebra is all about the matrixes

- anonymous

I wish I could tough it out for that but I can't lol

- amistre64

gotta get to class, good luck and all ;)

- anonymous

same! thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.