Question

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

Answer 1

it cut me off dx/dt = 3/2

Answer 2

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 .......

Answer 3

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

Answer 4

so i did it wrong? lol

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

*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.

\[\frac{6}{s}=\frac{d}{s+10}\to\ 6(s+10)=ds\]
\[6s' = d's + ds'\]
\[s'(6-d)=d's\]
\[s'=\frac{d's}{6-d}\]

\[s'=\frac{32(7.5)}{6-(32-16)}\]
\[s'=\frac{32(7.5)}{-10}\]
\[s'=-16(1.5)=-24\]

Answer 9

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.

Answer 10

30 eh ....

Answer 11

actually I think you did this problem for me before

Answer 12

let me go look

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

SEE those are the pictures i should draw

Answer 19

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

Answer 20

using -16t^2+30