anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
it cut me off dx/dt = 3/2
amistre64
  • 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
  • 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

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anonymous
  • anonymous
so i did it wrong? lol
anonymous
  • anonymous
im guessing d is going to be moving at 32ft/s because the derivative of the distance equation is the velocity equation...
anonymous
  • anonymous
pellet I don't get it I'm over calc this is ridiculous.
anonymous
  • anonymous
I dont understand how my way doesn't work...
amistre64
  • 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
  • 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
  • amistre64
30 eh ....
anonymous
  • anonymous
actually I think you did this problem for me before
anonymous
  • anonymous
let me go look
anonymous
  • 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
  • anonymous
hmm I wonder if would be better to differentiate right away to see what things i need in respect to time
amistre64
  • amistre64
its best to play around with it and take it apart; break it, make mistakes with it; and learn from the experiences :)
amistre64
  • amistre64
|dw:1333469963580:dw|
anonymous
  • 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
  • amistre64
notice that the ratio is not 6 to d; but is a ration of the lower part 30-d
anonymous
  • anonymous
SEE those are the pictures i should draw
anonymous
  • anonymous
in the equation i just wrote d being the distant at time t=1
anonymous
  • anonymous
using -16t^2+30
amistre64
  • amistre64
yes
anonymous
  • anonymous
if it weren't for physics id be failing this class lol
anonymous
  • anonymous
I just have to do every situation of one of these and just hope one of them is on the test
amistre64
  • amistre64
physics do make the world go round :)
anonymous
  • anonymous
hahaha great pun
anonymous
  • 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
  • amistre64
some good practice problems can be found at interactmath.com i believe
amistre64
  • amistre64
in the similar tris, yes; but sometimes that information given makes us go into trig for an answer as well
anonymous
  • anonymous
I will check that out, out of curiosity does openstudy pay you? they should you and myn answer all of the questions
amistre64
  • 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
  • amistre64
this is volunteer stuff; something to pass the time with so no i dont get paid
anonymous
  • anonymous
drawing this sec
anonymous
  • anonymous
hmm I should keep the 15/2= 12/s+20 is what I have written so far
anonymous
  • anonymous
15/s*
anonymous
  • anonymous
gah 5/s*
amistre64
  • amistre64
is 20 a constant? or is it changing?
anonymous
  • anonymous
if my s+20 is changing at a rate shouldn't i leave that as a variable
anonymous
  • anonymous
like maybe set y=s+20?
amistre64
  • amistre64
|dw:1333470857325:dw|
anonymous
  • anonymous
if I do that and cross multiply to get 5d=12s can I just differentiate there? (i changed y to d)
amistre64
  • amistre64
i wouldnt add extra variables; work with what youve got
anonymous
  • anonymous
aww yeah now I remember...
anonymous
  • anonymous
(s+d)5=12s -> 5s+5d=12s -> 7s=5d-> 7s'=5d'? 15/7=s'
amistre64
  • 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
  • anonymous
sweet so I did it right?lol
amistre64
  • amistre64
as you can see; there is no place to include "d" or "s" in the rate of change now is there
amistre64
  • amistre64
so the rate of change at 20 feet is the same as at 5 feet is the same as at 150 feet
anonymous
  • 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
  • anonymous
I only have 16-20 more hours to study for this thing though idk how ill do
amistre64
  • 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
  • amistre64
|dw:1333471260552:dw|
anonymous
  • anonymous
ohhhh wow didn't even realize that
anonymous
  • anonymous
I gotta draw this stuff better instead of just trying to power through it
amistre64
  • amistre64
\[tip' = d'+s';\ but\ s'=\frac{5}{7}d'\] \[tip'=3+\frac{15}{7}\]
anonymous
  • anonymous
did you just make this question up?
amistre64
  • amistre64
the numbers yes; the format, no. its pretty standard
anonymous
  • anonymous
haha okay I was like that was a pretty good question to just have randomly made up
amistre64
  • 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
  • anonymous
don't you have to know the derivatives of inverse functions for that one though?
amistre64
  • amistre64
|dw:1333471559263:dw|
anonymous
  • 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
  • anonymous
like that?
amistre64
  • amistre64
yep
anonymous
  • anonymous
x/3000=tantheta
amistre64
  • amistre64
relate t and b ... tangent, good
amistre64
  • amistre64
how fast is the distance between you and the ballon changing?
anonymous
  • anonymous
1/(1+(x/3000)^2) i never got this part but 3000/((3000^2)+x^2) * dx/dt
amistre64
  • amistre64
|dw:1333471697722:dw|
amistre64
  • amistre64
for t' tan(t) = b/d sec^2(t) t' = b'/d t' = b' cos^2(t)/d
anonymous
  • 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
  • amistre64
pythag it for r' d^2 + b^2 = r^2
anonymous
  • anonymous
so finding that you would take the d of r err'=2dd'+2bb'?
amistre64
  • amistre64
2dd' + 2bb' = 2rr' yes
amistre64
  • amistre64
d is constant so d' = 0
anonymous
  • anonymous
shouldn't we eliminate the one that isn't changing which is our distance between the ballon
anonymous
  • anonymous
yeah haha
amistre64
  • 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
  • anonymous
ahhh okay that does make more sense mathematically
anonymous
  • anonymous
now take my test for me amistre!
amistre64
  • amistre64
lol, but i might fail ;)
anonymous
  • 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
  • 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
  • amistre64
this one a made up, but its an amalgamation of others ive come across
anonymous
  • anonymous
r/h=r/2/h/2?
amistre64
  • 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
  • anonymous
gah hahaha hmm so this is a volume problem
amistre64
  • amistre64
it volume related yes; so we might need a formula for cone volume
anonymous
  • anonymous
1/3pir^2h?
amistre64
  • amistre64
notice that the rate of change in volume can be simplified at rate in - rate out as well
anonymous
  • anonymous
what do you want it in terms of?
amistre64
  • 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
  • amistre64
if we solve for r' ; our intended target ... and fill in the rest; that should do it
anonymous
  • anonymous
oh this is way more complex than I had originally though
amistre64
  • amistre64
pfft, we only have 2 variable to concern ourselves with at this point and the rest of the information would be supplied
anonymous
  • anonymous
can I set V'/2 = that over 2? after taking the derivative? or sub in for h or something?
amistre64
  • amistre64
maybe, but sincce i dont see what your seeing i would be leary of it
amistre64
  • amistre64
we want to know the h and r at V/2 ; not rates of change of V' at V/2
anonymous
  • anonymous
hmmmm
amistre64
  • 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
  • amistre64
\[\frac{V}{2}=\frac{1}{2*3}pi\ r^2h\] \[\frac{V}{2}=\frac{1}{6}pi\ r^2h\]
anonymous
  • anonymous
wow this is quit complex I don't know if I would have ever gotten this lol
amistre64
  • 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
  • amistre64
this wont be on the test
anonymous
  • anonymous
you never know! I think I'm really starting to understand what all of these damn word problems mean
amistre64
  • 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
  • anonymous
I would never want to be a mathematician...
amistre64
  • amistre64
lol, it does get better. you find out that they tend to teach math backwards for some reason
anonymous
  • anonymous
hahaha what do you mean? teach backwards?
anonymous
  • anonymous
calc II sounds like hell I'm opting out and taking basic stats
amistre64
  • 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
  • amistre64
so we are learning math backwards
anonymous
  • anonymous
OH i get it hahaha I was like wtf
anonymous
  • anonymous
damn you isaac newton....
anonymous
  • anonymous
/everyone other person that contributed to calculus
anonymous
  • anonymous
everyother*
amistre64
  • amistre64
and leibnez, and wallis, and Laplace, and Euler, and and and lol
anonymous
  • anonymous
guass!
anonymous
  • anonymous
gauss*
amistre64
  • amistre64
not so much guass, he just figured out how to count really :)
anonymous
  • anonymous
hahahaha this is true
anonymous
  • anonymous
i loved matrices in precalc
amistre64
  • amistre64
oh those get more complicated
amistre64
  • amistre64
linear algebra is all about the matrixes
anonymous
  • anonymous
I wish I could tough it out for that but I can't lol
amistre64
  • amistre64
gotta get to class, good luck and all ;)
anonymous
  • anonymous
same! thanks!

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