Calculating Related Rates: Cone Fluid Flow

[Miike012]

Homework Statement

Fluid flows out of a cone at 3 cu ft/min. If the height of the cone is three times the radius how fast is the height of the fluid decreasing when the fluid is 6 inches deep in the middle?

The Attempt at a Solution

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

2. r = h/3 thus... V = ( (pi)(h^3) ) / 27

(Set up the problem) ...
3. dv/dt = (dv/dh)(dh/dt)

4. dv/dh = ( (pi)(h^2) ) / 9
5. dv/dt = 3 cu ft/min.
6.h = 6
7.(plug into problem from line 3.)

8. 3cu ft/min = ( ( (pi)(6^2) ) / 9 )(dh/dt)
9.Final answer: dh/dt = (3cu ft/min )/( ( (pi)(6^2) ) / 9 )

***in the back of the book it says... dh/dt = 1296/(pi) in/min***
Im assuming that I have to convert 3cu ft/min somehow? and I also did something else wrong... can some one please help me...
thank you.

 

[Char. Limit]

You have 3 \frac{ft^3}{min}. How can you convert that to x \frac{in^3}{min}?

 

[Miike012]

idk, but i did something wrong.. does anyone know

 

[Char. Limit]

Try multiplying it by \frac{12^3}{1} \frac{in^3}{ft^3}. You see why that would work, right?

Note: since 12^3 in^3 = 1 ft^3, you're multiplying by 1, which is allowed.

 

[Miike012]

Yeah, I tryed that but the answer comes out to I think 455... not the answer the book gives... Ill try again, maybe my algebra went wrong somewhere.

 

[Char. Limit]

Yeah, I tryed that but the answer comes out to I think 455... not the answer the book gives... Ill try again, maybe my algebra went wrong somewhere.

Yeah, because I get the right answer through your algebra. Make sure you simplify as much as possible, though...

 

[Mentallic]

How can you convert that to x \frac{in^3}{min}?

idk...

That would be your problem right there.

 

[Miike012]

Thank you, I see what I did wrong... thanks for the help.