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Mathematics 19 Online
OpenStudy (anonymous):

A circular conical reservoir, vertex down, has depth 20 ft and radius of the top 10 ft. Water is leaking out so that the surface is falling at a rate of 1/2 ft/hr. The rate, in cubic ft per hr, at which the water is leaving the reservoir when the water is 8 ft deep is...

OpenStudy (anonymous):

@tkhunny?

OpenStudy (anonymous):

cmon guys please?

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

what is the formula for volume of a cone?

OpenStudy (anonymous):

V=a^3

OpenStudy (anonymous):

whoopps

OpenStudy (amistre64):

thats a cube, we want a cone

OpenStudy (anonymous):

1/3pir^2h

OpenStudy (amistre64):

good, and we will also want to define the relationship between height of water and radius

OpenStudy (anonymous):

i don't know how to do that

OpenStudy (amistre64):

we know 2 points of reference: 0,0 and 10,20 for a given r,h

OpenStudy (amistre64):

when we have 2 points, we can make the equation of a line

OpenStudy (anonymous):

alright so slope=1/2, which we know

OpenStudy (anonymous):

so r=1/2h?

OpenStudy (amistre64):

h = 2r V = 1/3 pi r^2 h the rest is derivatives ...

OpenStudy (anonymous):

where do i start?

OpenStudy (amistre64):

with h or V we need the derivative of both of them to help us out

OpenStudy (amistre64):

use implicits

OpenStudy (anonymous):

h'=2, but what do i take the derivative of V=1/3pir^2h with respect to?

OpenStudy (amistre64):

h r and V are all functions of time, so h = 2r: h' = 2r' V = 1/3 pi r^2 h: V' = 1/3 pi (2rr' h + r^2 h')

OpenStudy (amistre64):

ask me why if that doesnt make sense

OpenStudy (anonymous):

no, i get it, product rule, dr/dt/dh/dt right?

OpenStudy (amistre64):

yep

OpenStudy (anonymous):

ok

OpenStudy (amistre64):

so, lets fill in the information

OpenStudy (amistre64):

h' is given as: 1/2, and h = 8 h = 2r: h' = 2r' 8 = 2r: 1/2 = 2r' fill in and solve for V' V' = 1/3 pi (2rr' h + r^2 h')

OpenStudy (amistre64):

h' might be better is -1/2 since it is falling

OpenStudy (anonymous):

ok i think i understand

OpenStudy (anonymous):

just gimme a min

OpenStudy (anonymous):

1/3pi(24) ?

OpenStudy (amistre64):

h' = -1/2 h = 8 r=4 r'=-1/4 fill in and solve for V' V' = 1/3 pi (2rr' h + r^2 h') V' = 1/3 pi (-2(4)(8)/4 - 4^2/2) V' = 1/3 pi (-16- 8) V' = -pi (8)/3

OpenStudy (anonymous):

yea i think i messed up a negative somewhere because i got the same answer, but positive

OpenStudy (anonymous):

think it had something to do with h=1/2

OpenStudy (amistre64):

you were right, i see my error

OpenStudy (anonymous):

rather than -1/2

OpenStudy (amistre64):

-16-8 = -24

OpenStudy (anonymous):

so it's -8pi?

OpenStudy (amistre64):

V' = -pi (24)/3 = -8pi the negative just tells us water is leaving the system

OpenStudy (anonymous):

yea, i think get it now , thanks a ton! I'm trying to teach myself calc but it's really hard, help is greatly appreciated

OpenStudy (amistre64):

i taught myself, so you should do fine

OpenStudy (tkhunny):

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