Calculus Problem?

Question

I don't need an answer to the problem, but I was wondering if anyone could tell me how to set up a problem to solve. "The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 feet above the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at the rate of 3 cubic feet per minute. Find the rate of change of the depth of water in the tank." I've tried everything, but can't figure this one out. Any help is appreciated.

Thanks for the help. My teacher actually showed us how to do the problem before turning in the worksheet and dh/dt was -.220 ft/min (or -.022, I don't have the answer in front of me). Pretty complicated!

Answer 1

get the function for the depth of the tank based on the parameters like height of the tank, volume etc etc.

get the relation for linking the parameters.

you have dV/dt = 3 cubic feets/min.

diffrentiate the eqn wrt time and you will get the solution

Answer 2

The volume of a cone is:

v = (1/3)πr²h

h = height of the cone
r = radius of the cone

This is a related rates problem I believe......yes?

You know:

R= 5'
H = 12'
h = 7'
dv/dt = 3 ft³/min

By similar triangles you can get a relation between the radius of the cone (R) to its height (H) to the radius of the cone at any time t (r) and the height at any time t (h).

H/R = h/r

H/R = 12/5 = h/r

==> r = (5/12)h

Sub this in to 'v' and you have:

v = (1/3)π[(5/12)h]²h

Simplify this a bit...take the derivatives of this with respect to time (inplicitly of course), solve it for "dh/dt" which you will get on differentiation and you should have it.

Did that make sense?

Answer 3

f (x) = x^3 - 12x + a million . . . the 1st by-product set to 0 unearths turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2nd by-product evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== beneficial fee exhibits x=2 is a close by minimum f ' ' (-2) = 6*(-2) = -12 <== damaging fee exhibits x=-2 is a close by optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is expanding x = -2 to +2 is lowering x = +2 to + infinity is expanding b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2nd by-product set to 0 unearths inflection factors, or the place concavity differences 6x = 0 x = 0 <=== inflection factor x = - 2 is a optimal, so ought to be concave down concavity differences on the inflection factor(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 4

Let r = the radius of the cone, h = the height of water in the cone, and v = the volume of the water.

The volume of a cone is
v = (1/3) pi (r^2)h......(1)

Since you have two variables, you need to find a restriction, which is
r/h = 5/12 or r = 5h/12......(2)

Plug in (2) into (1),
v = (1/3) pi (5/12)^2 h^3......(3)

Differentiate (3) with respect to time,
v' = (25/432) pi (3h^2)h'......(4)

Plugging in v' = -3 and h = 7 into (4) gives the solution of h',
h' = -0.11225 ft/min

The negative sign of h' means the water level drops.

Answer 5

This is indeed very compicated.
the volume change of the tank is 3 cunic feet per minute

first write the equation of the watervolume as function of the
height
say this is V(h(t)); h height at time t. t = 0 h = 7, NOTE h is a function of time t.
each minute 3 cube is gone
thus dV(h(t))/dt = 3
dV(h(t))/dt = dV/dh*dh/dt = 3
dV/dh you can calculate,sat this is S threat h as a variable.
the function h(t) can now be solved as follows

S*h' = 3

thus h' = 3/S
and h(t) is the integral from s=0 to s=t; and h(0) = 7

something like h(t) = h(0) + Integral 3/S , boundery of integral 0 and t


DOES THIS HELP ?

Answer 6

V=1/3*Pi*r^2.h
v=volume
r=radius
h=height
r[t]=radius at the height h[t] at any time t
h[t]=height of water at any time t[below the apex of cone]
V[t]=volume of water at any time t when
height of water is h[t] below the cone
=1/3*Pi*25*12-1/3*Pi*r[t]^2*h[t]
=310.86-1.037*r[t]^2*h[t]
now r/h=cotz=5/12
r[t]=5/12*h[t]
Vt=310.86-1.037*25/144*[h[t]]^3
Vt=310.86-.18*h[t]^3
dVt/dt=3ft^3/min
=-.18*3*ht^2*dh/dt
dh/dt=1/[.18*h[t]^2]
Here we observe that the variation
of height of water with respect to
time is a inverse squre function of
the height itself. At each instant the
rate of change of height will depend
upon inverse squre of the height itself.