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...
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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
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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
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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 ...
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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')
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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
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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) ?
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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
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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
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OpenStudy (tkhunny):
General Rule: Post YOUR work, and you get better answers, sooner.