Question

A street light is mounted at the top of a 11 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 7 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?

Answer 1

mancake, please don't contribute unless u can.. u're gonna get reported and banned if this keeps up

Answer 2

jinnie why did u leave the other question? nevermind lol let's just continue here.. i was just about to type anyway..

Answer 3

oh sorry for that.

Answer 4

similar triangles

Answer 5

btw how many more do u have left?

Answer 6

4 including this one

Answer 7

almost all of them are ones we started yesterday but im apparently too stupid to finish and get the right answer lol

Answer 8

drawing this... hold on

Answer 9

ok

Answer 10

try 42/5

Answer 11

that doesnt work

Answer 12

okay i messed up

Answer 13

heres one similar

A street light is hung 18 ft. above street level. A 6-foot tall man standing directly under the light walks away at a rate of 3 ft/sec. How fast is the tip of the man's shadow moving?

I know I would've to set up a proportion.

18 / 6 = x + y / y

x = distance of man from light
y = length of shadow
x + y = tip of shadow

Calculus - Damon, Sunday, February 6, 2011 at 6:24pm

You mean
18 / 6 = (x + y) / y
we know dx/dt, we need dy/dt
then the tip moves at dx/dt + dy/dt

18 y = 6x + 6 y
12 y = 6 x
12 dy/dt = 6 dx/dt
dy/dt = .5 dx/dt
so dy/dy = 3/2 = 1.5
and the sum
3+1.5 = 4.5 ft/sec

Answer 14

Here's how I did a problem like this on one of my tests

Answer 15

11/6 = (x+y)/y
11y = 6x+6y
5y = 6x
y = (6/5)x
dy/dt = (6/5)dx/dt = 36/5 ?

Answer 16

36/5 didnt work either.

Answer 17

hey rogue thanks for that i will see if i can follow it

Answer 18

42/5 should've worked.. hold on

Answer 19

11/6 = (x+y)/y
That's correct.. im sure

Answer 20

I came up with same ratio!

Answer 21

darn. webwork says its incorrect.

Answer 22

11y = 6x+6y
11y-6y = 6x
5y = 6x
y = (6/5)x
y' = (6/5)(dx/dt)

Answer 23

it provided me with this hint
This is a picture of a right triangle with the vertical side AB representing the pole, CD the woman, and the end E of the hypotenuse representing the tip of the woman's shadow.
Given: AB=11, and CD=6. Let BD=x, and BE=y. So dxdt=7.
The 2 triangles ABC and CDE are similar.
We can write the following proportion: CDAB=EDEB
So 611=y−xy
Cross multiply, 6y=11(y−x).
So (11−6)y=11x
Differentiate both sides: (11−6)dydt=11dxdt.
Solve for the derivative of y to get how fast the tip of her shadow is moving when x=35.

Answer 24

dx/dt = 7
y' = 42/5

Answer 25

jinnie try - 42/5

Answer 26

it rejected that as well

Answer 27

my response to that would be...: http://www.youtube.com/watch?v=cYplvwBvGA4 skip to 00:50

Answer 28

lolol

Answer 29

-15.4 ft/s?

Answer 30

it still says no

Answer 31

15.4?

Answer 32

yep

Answer 33

thanks you all

Answer 34

lol, wow.

Answer 35

lol rogue we make a good team!

Answer 36

lol. webwork is nuts

Answer 37

Yep, we do =)
Webworks... For me, I have Utexas quest, God I hate that so much.

Answer 38

Jinnie, where does the medal come from?

Answer 39

i never had to deal with any of this nonsense.. thank god!

Answer 40

the medal comes from having good answers

Answer 41

bahrom, you're so generous with medal :P

Answer 42

you guys want to help me just finish these 3 problems i have left

Answer 43

im generous with answers too lol

Answer 44

most definitely lol

Answer 45

I'll sure will stick with all your answer to copy and paste from it, bahrom :P

Answer 46

A particle is moving along the curve y=4 sqrt(4x+1) As the particle passes through the point (2,12), its x-coordinate increases at a rate of 3 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

Answer 47

Easy one WOOT!

Answer 48

a similar problem

A particle is moving along the curve y= 4sqrt{4 x + 9}. As the particle passes through the point (4, 20), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

dy/dt = 4(0.5)[(4x+9)^(-1/2)](4dx/dt)
so when x=4
dy/dt = (8/5)(dx/dt)=8 units/sec
D = (x^2+y^2)^0.5
dD/dt = 0.5(x^2+y^2)^(-0.5) (2dx/dt+2ydy/dt)
= (5 + 20*8) / sqrt(16+400) = 165/20.396=8.089 units/sec

Answer 49

Now, Jinnie suddenly become so greedy :)

Answer 50

dy/dt = 8*(dx/dt)/(sqrt(x+1)

Answer 51

x = 2, dx/dt = 3, dy/dt = 8*3/2 = 12

Answer 52

hahaha now that i have all these brilliant minds in one room, i just want to take advantage lol

Answer 53

x=2, y = 12, dx/dt = 3, dy/dt = 12
D = (x^2+y^2)^0.5
dD/dt = 0.5(x^2+y^2)^(-0.5) (2xdx/dt+2ydy/dt)
= (4+144)^(-0.5)*(2*3+12*12) = 150/sqrt(148)

Answer 54

webwork says no

Answer 55

FUUUUDGE!!!!

Answer 56

tell webwork that i hate it

Answer 57

i agree!!!

Answer 58

HELL WITH WEBWORK!!!!!!!!!!

Answer 59

writing it out on paper..

Answer 60

ok

Answer 61

dy/dt was 8/3

Answer 62

wait nevermind

Answer 63

dy/dt was 8, not twelve

Answer 64

copy paste this into webworks:
(6+12*8)/sqrt(148)

Answer 65

that works thans bah

Answer 66

thanks*

Answer 67

Do you lose points if you answer incorrectly on webworks?

Answer 68

no

Answer 69

rogue no

Answer 70

unlimited tries to get it right

Answer 71

on tests u get 3 trials though

Answer 72

You lucky bum.

Answer 73

I wish Utex was like that T_T

Answer 74

darn

Answer 75

hey bah even though you put me on the right track with this...i could never come up with an acceptable answer for webwork

If z^2=x^2+y^2,dx/dt=9, and dy/dt=6, find dz/dt when x=2 and y=4

Answer 76

that one has got to be the easiest question ever.. that u posted

Answer 77

z' = (xx'+yy')/z

Answer 78

z = 2sqrt(5)

Answer 79

my teacher so awk. the lvl of difficulties of his homework questions varies from easy to impossible lol

Answer 80

copy paste this:
(2*9+4*6)/2sqrt(5)

Answer 81

so the result is 8.4?

Answer 82

for 6 + 96/ sqrt (148)

Answer 83

neither worked

Answer 84

(2*9+4*6)/(2*sqrt(5))
try that

Answer 85

9.4 is the answer

Answer 86

that works. thanks alot bahrom. I have one last question but we have already looked at it and couldnt figure it out.

The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 4.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 90 square centimeters?

Answer 87

Jinnie, how about restart your PC! It's crazy !!!

Answer 88

ill do this one jinnie watch.. hahaha

Answer 89

lol

Answer 90

Chlorophyll if you guys are math majors and find these questions ridiculous. imagine people like me whom are just taking math to fulfill a requirement

Answer 91

i dont rly find them hard im annoyed at them.. ive had my share of RR problems

Answer 92

oh ok

Answer 93

Jinnie, consider today is your lucky day and hope that you're going have more and more days like this :P

Answer 94

I am very thankful of this day. I hope there are more like it lol

Answer 95

bah, have you gotten anything?

Answer 96

The altitude of a triangle is increasing at a rate of 1.5 centimeters/minute while the area of the triangle is increasing at a rate of 4.5 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 7 centimeters and the area is 90 square centimeters?
\[A = \frac {1}{2} bh\]\[\frac {dA}{dt} = \frac {1}{2} \left[ b \frac {dh}{dt} + h \frac {db}{dt} \right]\]\[A = 90, h = 7, b = \frac {180}{7}\]\[\frac {dh}{dt} = 1.5, \frac {dA}{dt} = 4.5\]\[9 = 1.5 (180/7) + 7 \frac {db}{dt}\]\[\frac {db}{dt} \approx -4.22449 cm/min\]