Question

Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant?
Round your answer to three decimal places.

Answer 1

Okay, you can find the volume first since you have h and r , so :

\[V = \pi r^2 h\]

\[= ( 3)^2 (6) \pi\]
= 169.65 ft^3

Answer 2

sstarica, isn't this the same thing we were tackling

Answer 3

now find V' which is :\[V' = 2\pi rr'h'\]

Answer 4

not really @iam , somehow similar :)

Answer 5

now we have:

h' = 0.2 ft/sec
we need to find r'

Answer 6

do i just plug in

Answer 7

yes :)

but we need to find r' now

Answer 8

\[V' = 2 \pi rr' h'\]

we have h' , but we need to find r' to calculate V' = rate of volume increase

Answer 9

how do i do that? :(

Answer 10

hold on :)

Answer 11

thx

Answer 12

you need to find a relationship between r and h

so r/h = r'/h'

Answer 13

thats where im having problem for all the problems

Answer 14

cross mulitply and you'll get:

r' = rh'/r

Answer 15

sorry! wait you'll get:

r' = rh'/h :)

Answer 16

are you with me soo far?

Answer 17

yes :)

Answer 18

now plug :

r =3 ft
h = 6ft
h' = 0.2ft/sec

r' = (3)(0.2)/6
= 0.1 ft/sec

Answer 19

we have found r' now, let's go back to V'\[V' = 2 \pi rr'h'\]

so,

V' = 2(3)(0.1)(0.2) pi
= 0.377 ft^3/sec

Answer 20

Correct me if I'm wrong please

Answer 21

OMG this makes so much sense now

Answer 22

but the website marked it wrong :(

Answer 23

why?

Answer 24

I guess there's a mistake, but that's the idea, maybe in calculation?

Answer 25

ok

Answer 26

did you get the idea?

Answer 27

one qquestion

Answer 28

sure :)

Answer 29

where did u get \[V \prime = 2\pi r r \prime h \prime \]

Answer 30

you derive with respect to r and h :)

Answer 31

you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem

Answer 32

oooooh...im sorry can u show me one more problem

Answer 33

hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?

Answer 34

I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec

Answer 35

so what is the given?

Answer 36

we have :

- R = 2 m
- R' = 0.5 m/sec

and we know the formula of the Area of the circle which is:

A = pi r^2, right?

Answer 37

uhhuh

Answer 38

Now , you want to know how much is the area increasing with respect to R, since R is increasing :)

so you find the derivative of it which will be :

A' = 2 pi rr'

Answer 39

so when exactly do i start pluging in numbers?

Answer 40

right now, after you find the derived formula, start pluggingthe numbers :)

Answer 41

okay ...these are pretty simple compared to the ones i have on hw :'(

Answer 42

Like this problem, they want to know how much is the area increasing , we got the formula so the answer is:

A' = 2(2)(0.5) pi

= 2 pi = 2.68 m^2/sec :)

So the area is increasing at 2.68 m^2/sec

Answer 43

Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)

Answer 44

did you understand my problem?

Answer 45

yes but how would i approach towards a hemisphere problem

Answer 46

think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH!

To test your understanding , you've got to sketch what they give you to be fully understand what they want :)

Answer 47

to fully*

Answer 48

1) write the given and what you need to find
2) sketch
3) write down the general formula that you're going to use :)

Answer 49

^_^

Answer 50

ok :)

Answer 51

good luck !

You're a smart student, you can figure it out :)

Answer 52

you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!

Answer 53

lol..thx

Answer 54

np :)

Give it a try once more. ^_^