How Fast Does Water Depth Change in a Conical Tank?

[lim]

Homework Statement

A conical tank has a base radius of 6 feet and a height of 10 feet. Initially the tank is empty. Water is poured into the tank at a rate of 75 ft/min. How fast is the depth of the water in the tank changing when the water in the tank reaches a height of 5 ft? 8 ft? when the tank is half full?

Homework Equations

V= 1/3πr^2h
(π=pi)

The Attempt at a Solution

r/h=6/10
6h= 10r
r=3/5h.
My teacher told us to set up a radius and height relation to get ride of differentiating the radius later.

V= 1/3πr^2h
dv/dt= 1/3π(9/25h^2)h =1/3π(9/25)h^3
dv/dt= 1/3π(9/25)3h^2
dv/dt= π(9/25)100(5)= 180π ft/s @ 5 ft
π(9/25)100(8)= 280π ft/s @ 8 ft
π(9/25)100(5)= 180π ft/s @ 1/2 tank (since half of the tank is also 5 ft)
Are these correct?

2. Homework Statement
A rectangle has a constant area of 200 sq meters and length, L, is increasing at 4 meters/s.
a. Width, W, at instant the width is decreasing at .5 m/s?
b. At what rate is the diagonal, D, of the rectangle changing at the instant the width is 10 m?

Homework Equations

A=lw

The Attempt at a Solution

A=lw
Da/dt= l(dw/dt) + w (dl/dt)
200= l (-.5) + 4w
I set it up, but couldn’t figure out how to substitute for l or w. Is there another way?

3. Homework Statement
Water is draining from a conical tank with height 12 feet and diameter 8 feet into a cylindrical tank that has a base with area square feet. The depth, h, in feet, of the water in the conical tank is changing at the rate of (h-12) feet per minute.

Equations:
V= 1/3πr^2h
A) Write an expression for the volume of the water in the conical tank as a function of h.
r/h = 4/12
4h= 12r
r=1/3h
V= 1/3π1/9 h^3
correct?

B) At what rate is the volume of the water in the conical tank changing when h=3?
Dv/dt= 1/3 π 3h^2 dh/dt
Π/3 (1/9)3 (12)^2 dh/dt
Π(144)3/27
432 π/27= 16 π ft/min
correct?

C) Let y be the depth, in feet, of the water in the cylindrical tank. At what rate is y changing when h = 3?

Homework Equations

v=bh

attempt

y=bh, letting y be the depth
dy/dt= b dh/dt + h db/dt
dy/dt= 3 db/dt + 400 π dh/dt, ? Not sure what to do here.

 

[mighty2000]

Thank you for your post. I am a scientist and I would be happy to assist you with your questions.

1. For the conical tank problem, your approach is correct. You have set up the relation between the radius and the height and then differentiated the volume equation with respect to time. Your final answers for the rates of change of depth at 5 ft, 8 ft, and when the tank is half full are all correct.

2. For the rectangle problem, you have correctly set up the equation for the area of the rectangle and used the product rule to find the rate of change of the diagonal. To solve for the rate of change of width when the width is decreasing at 0.5 m/s, you can substitute the given values into the equation and solve for the rate of change of length, which is the same as the rate of change of width.

For part b, you can use the Pythagorean theorem to find the relation between the diagonal and the length and width of the rectangle. Then, you can differentiate this equation with respect to time and solve for the rate of change of the diagonal at the given width of 10 m.

3. For the conical tank and cylindrical tank problem, your approach is correct for parts A and B. For part C, you can use the fact that the volume of water in the conical tank is equal to the volume of water in the cylindrical tank to set up an equation relating the rates of change of the depth in the two tanks. Then, you can solve for the rate of change of depth in the cylindrical tank when the depth in the conical tank is 3 ft.

I hope this helps. Let me know if you have any further questions.