Question

A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at is deepest point. If the pool is being filled at a rate of 1.4 ft3/min, how fast is the water level rising when the depth at the deepest point is 5 ft? (Round your answer to five decimal places.) Please Help! :P

Answer 1

what math r u in?

Answer 2

Calc 1

Answer 3

have u done integration yet?

Answer 4

Not yet, just finished learning about differentiation >< haha

Answer 5

ok ill do this by differentiation then

Answer 6

Its a related rates questions :(

Answer 7

Thank you!!

Answer 8

v=volume
h=height of water
l=length the water makes
w=width
if u look at it sideways it would make a trapezoid and until the water reached 6 ft the volume would be
v=1/2hlw

Answer 9

now we need to v in terms of 1 variable so we need another equation

Answer 10

Ok, do i differentiate V? or just plug in V=(1/2)[(40)(3+9)(20)]?

Answer 11

now the width is a constant because we are only changing the vertical level of the water

Answer 12

u have to differentiate v but we need to write the equation with only2 variables. v and h so the l is a problem

Answer 13

what we are trying to find is the rate of change of the h when the height equals 5

Answer 14

hmm ok :]

Answer 15

ok the answer to that is 4,800

Answer 16

now we need to find an equation that only uses l and h so that we can solve for l and write the volume equation in terms of h

Answer 17

now the easiest way to do this is to look at similar traingles.

Answer 18

if u drew it u would get a large triangle and a small triangle.
the length of the large trianglee would b 40 or the length of the pool
and the height of the large triangle would b 6(9-6) or how much deeper the pool is than the shallow end

Answer 19

the small triangle would b length l and height h

Answer 20

so we would have the formula:
40/6=l/h
and we would solvefor h
40h/6=l
so now we can substitue it back into our volume equation

Answer 21

so
v=(1/2)l*h*w
becomes
v=(1/2)(40/6)h^2*w

Answer 22

(1/2)[(l/40)(12)(20)]

Answer 23

oh nvm haha

Answer 24

udont wanna put in the values yet because u want variables. only variables can be differentiated

Answer 25

ok, so differentiate the (1/2)(40/6)h^2(w)?

Answer 26

yes but simplify it first will make it easier

Answer 27

5h^2*w

Answer 28

also w is a constant so treat it as u would any number like 6 or something

Answer 29

i got (10/3)h^2(w)

Answer 30

oh ok haha

Answer 31

my bad :)

Answer 32

wait how did u get 5?

Answer 33

i added 2 and 6 -_- instead of multiplying haha

Answer 34

sorry

Answer 35

dv/dt=(20/3)h?

Answer 36

almost. u r missing 2 things

Answer 37

dv/dt=dh/dt * dv/dh (20/3)h or something like that? ><

Answer 38

well u got 1 of them. u r differentiating in terms of time or dt so that should b the onlydenominator. so the dv/dh is wrong but u r still missing 1 more thing

Answer 39

umm...haha wait i have no idea :(

Answer 40

well look at the equation for V that i gave u. what letter is in there thats not in ur derivative?

Answer 41

w! but when you derive that wouldnt it become 1?

Answer 42

haha

Answer 43

w is a constant.
when u derive 6x in terms of it becomes 6 right?
similarly if u derived wx in terms of x and w is a constan it would become w

Answer 44

oh haha ok (20/3)hw then

Answer 45

and then plug in the 5 so 100/3 (w)?

Answer 46

(20/3)hw(dh/dt)

Answer 47

oh then 1.4/ [(100/3)]w =dh/dt

Answer 48

? hopefully ...

Answer 49

yes

Answer 50

isnt the w= 20?

Answer 51

u r dividing the walso right?

Answer 52

yup :)

Answer 53

yes it is

Answer 54

I got 0.0021?

Answer 55

(1.4)(3/100)(1/w) will give u ur answer

Answer 56

ok thank you!! :D