very hard calc question, in need of serious help the question is below, if any1 can help id be very grateful?

Question

Water is draining at the rate of 48pi ft^3/ minute from a conical tank whose diameter at its base is 40 ft and whose height is 60 ft.

a) find an expression for the volume of water in the tank in terms of its radius
b) at what rate is the radius of the water in the tank shrinking when the radius is 16ft?
c) how fast is the height of the water in the tank dropping at the instant that the radius is 16ft?

Answer 1

(a) Recall that the volume of the water inside the cone is given by the following formula:

V = (1/3) pi (r^2) h

One thing to note is that the height of the entire conical tank is 60 and the diameter is 40. That would make the radius of the tank equal to 20.

Notice that we have a cone within a cone. We have (1) the tank itself, which is a cone, and (2) the water in the tank, which has to be a cone. This means the cones are proportionate. More specifically,

[height of water] / [radius of water] =
[height of tank] / [radius of tank]

Height and radius of the water are h and r respectively. The radius of the tank is given to be 20, and the height is given to be 60. Therefore,

h/r = 60/20. All we have to do is solve this for h. We get

h/r = 3
h = 3r

Now that we know h = 3r, we can plug it into our volume formula,

V = (1/3)pi(r^2)h. Plugging this in gives us our revised formula,

V = (1/3)pi(r^2) [3r], which, if we simplify, becomes

V = pi(r^3)

This is the expression for the volume of water in terms of its radius. Note that we had to use the method of similar triangles to eliminate the h.

(b) In order to find the radius of the water shrinking when the radius is 16 feet, we must first declare our unknowns.

Water is draining at a rate of 48 pi ft^3/minute means
dV/dt = -48 pi [Note that water is draining, which is why it is negative].

What we want to find is dr/dt. So dr/dt = ? {our unknown}.

V = pi(r^3), so implicitly differentiating with respect to t,

dV/dt = pi [3r^2] [dr/dt]
dV/dt = 3pi(r^2) [dr/dt]

Now, we substitute dV/dt = - 48pi

-48pi = 3pi(r^2) [dr/dt]

Dividing both sides by 3pi, we get

-16 = (r^2) [dr/dt]

It is now at this point that we plug in our numerical value. In any related rates question, we require a "WHEN" statement. Here is our "when" statement in this case:

WHEN THE RADIUS OF THE WATER IS 16 FT ....

It is at this point we plug in r = 16. BEWARE OF PLUGGING IN r = 16 TOO EARLY! You want to do is AFTER you substitute your rates.

So now, we plug in r = 16, to get

-16 = (16)^2 [dr/dt], which means
-1/16 = dr/dt.

So dr/dt is -1/16. Here's where you want to properly make your concluding statement:

"The radius of the water in the tank is shrinking at a rate of 1/16 ft/minute."

Note the significance of 1/16 being positive. Had we said "shrinking at a rate of -1/16 ft/minute", that would be a double negative, indicating that it's growing. You could alternatively say,

"The radius of the water is growing at a rate of -1/16 ft/minute."

(c) To solve this one, note the original equation

V = (1/3)pi(r^2)h

We can solve via our proportion equation that r = h/3, so

V = (1/3) pi ([h/3]^2)h
V = (1/3) pi ([h^2 / 9])h
V = (1/3) pi (h^3 / 9)
V = (pi/27) h^3

Differentiating, we have

dV/dt = (pi/27) [3h^2] [dh/dt], which means
-48pi = (pi/9) (h^2) [dh/dt], and simplified, gets us
-384 = (h^2) [dh/dt]

Now for our "when" statement:

WHEN THE RADIUS IS 16 FT ...
Remember our relation r = h/3. That means h = 3r, which means
h = 3(16) = 48. So "when the radius is 16 feet, the height is 48", and h = 48. We plug in this numerical value for h.

-384 = (48)^2 [dh/dt]
-1/6 = dh/dt

Concluding statement:

"The height of the water in the tank is dropping at a rate of -1/6 ft per minute."

Answer 2

a) Volume of water in the tank at any moment V=1/3{ pi R^2 H}
R is the radius'
H is the height =60/20 =3R
V =1/3{ pi * 3R^3} = [ pi R^3 }

b) Rate shrinking of the water volume = -48 pi ft^3 / min
that is= differentiation of V wrt time T = dV / dT
= (3 pi R^2 ) (dR /dT ) = - 48 pi
there fore d R /dT = -16/(R^2)
when R =16 , rate of shrinkage of the radius = 1/16ft /min


c) H=3 R
dH / dT = 3 dR / dT =3/16 ft /min

Answer 3

This is not a hard task, you only need to use the expression for volume of a cone as one third of its height multiplied by the area of its base.
V=(1/3)HS, S=PiR^2, and while the water goes down, R of water surface depends on its height. When its height is 60ft, its radius is 20ft (one half of the diameter 40ft). So, the height of partially filled tank H will depend on the radius of its water surface as H=3R.
Hence
(a) V = (1/3)3RPiR^2 = PiR^3.
As the rate of water draining is fixed, it means that the rate of changing of the remaining water volume is fixed also.
Each minute the volume changes for 48Pi(ft^3/minute) = -dV/dT, where T is time
And as we take the derivative from the expression,
dV/dT = 3PiR^2dR/dT = - 48Pi(ft^3/minute)
When the radius is 16ft, the volume is V=Pi(16^3)ft^3, and the
(b) -dR/dT = (48/3)/16^2 (ft/minute) = 1/16 ft/minute.
(c) as H=3R, so -dH/dT = -3dR/dT = 3/16 ft/minute
-dH/dT = 3/16 ft/minute

Answer 4

Which way is the cone facing??? If the vertex of the cone is down, the water will always be in the shape of a cone similar to the tank. If the vertex is up and the base is down, which is the way the problem sounds, the water is NOT in the shape of a cone, but in the shape of a truncated cone.

I will assume that the vertex of the cone is up. In that case, to make the math easier, I will compute the rate at which the cone of air above the water grows.

Given

Diameter at base = 40 ft
Radius at base = 40/2 = 20 ft
Height of cone = 60 ft
dV/dt = -48π ft³/min

Let

h = height of water in tank
a = height of air in tank
r = radius of the surface of the water in tank
V = volume of water in tank

a) find an expression for the volume of water in the tank in terms of its radius

a = 60 - h

Ratio of height of air to radius = 60/20 = 3
a = 3r
60 - h = 3r
h = 60 - 3r

V = (Volume of full tank) - (Volume of air in tank)
V = ⅓π(20²)(60) - ⅓πr²a
V = ⅓π(400)(60) - ⅓πr²a
V = 8000π - ⅓πr²a

b) at what rate is the radius of the water in the tank shrinking when the radius is 16ft?

Find dr/dt when r = 16

a = 3r

V = 8000π - ⅓πr²a = 8000π - ⅓πr²(3r) = 8000π - πr³
dV/dr = -3πr² = -3π(16²) = -768π ft²

dr/dt = (dV/dt)(dV/dr) = (-48π ft³/min)/(-768π ft²)
dr/dt = 1/16 ft/min

c) how fast is the height of the water in the tank dropping at the instant that the radius is 16ft?

Find dh/dt when r = 16
a = 3r = 3(16) = 48
h = 60 - a = 60 - 48 = 12
a = 60 - h
da/dh = -1

r = a/3
V = 8000π - ⅓πr²a = 8000π - ⅓π(a/3)²a
V = 8000π - (1/27)πa³
dV/da = (1/9)πa²
dV/dh = (dV/da)(da/dh) = (dV/da)(-1) = -(dV/da)

dh/dt = (dV/dt)/(-dV/dh)
dh/dt = (-48π ft³/min)/[-(1/9)πa² ft²]
dh/dt = (-48π ft³/min)/[-(1/9)π(48²) ft²]
dh/dt = 9/48 ft/min = 3/16 ft/min

Answer 5

Better get out the ol' Calculus book.

First, define the height in terms of radius., h(r) = 60 - 3r

Then you can define

For a cone, V = pi r^2 h/3, so the total volume is 25,120 ft^3

so fo a) the volume is going to be the total volume - the volume above whatever h you want to use. This is the trick.

Find the volume above h, where h' = 60 - h where h = 60 - 3r
and thus h' = 60 - (60 - 3r) = 3r

so the volume above h = pi r^2 h'/3 = pi r^2 3r/3 = pi r^3

===>> V(r) = 25,120 - pi r^3

Since dV/dt = -48 pi ft^3/min = -150.7 ft^3/min

V(t) = Vo - 150.7 t
dV/dt = -150.7 ft^3/min

dV/dr = - 9.425r^2
when r = 16', dV/dr = -2412.8 ft^3/ft

b) dr/dV * dV/dt = dr/dt so -150.7/2412.8 = -0.062 ft/min


h = 60 - 3r
dh/dr = -3

c) so dr/dt * dh/dr = dh/dt = -0.186 ft/min

I think this is right. I don't have my book to check.