Solving Cube's Volume Change: Related Rates Help

[Shiz Stain]

Homework Statement

This is the problem I am having...

All edges of a cube are expanding at a rate of 3 centimeters per second. How fast is the volume changing when each edge is

(a) 1 centimeter
(b) 30 centimeters

Homework Equations

The equation I am using is

V = S³

The Attempt at a Solution

This is what I did

dv/dt = 3s² ds/dt
dv/dt = 3 (1)² x 3
dv/dt = 3 x 3
dv/dt = 9

dv/dt = 3s² ds/dt
dv/dt = 3 (30)² x 3
dv/dt = 3 (900) x 3
dv/dt = 2700 x 3
dv/dt = 8100

I am just wondering wheter or not I did this equation right so I can look back over it and find the mistake.

 

[Mark44]

Technically speaking, your answers are numerically correct. Corrections I whould make are:

1. You started with V as a variable, but your derivative is dv/dt, which is often used as the derivative of velocity.
2. Distinguish between dv/dt and dv/dt at a particular time. Your two sets of equations make no distinction between dv/dt (which is a function) and dv/dt at a particular time.
   You have dv/dt = 3s^{2[\sup] ds/dt, and
   dv/dt = 9. The latter is dv/dt|t = t₀ (the time at which s = 1 cm.) Similar for the other value of s.
   [*]Add the appropriate units to your answers.}
   

 

[Shiz Stain]

ummm thanks but you kind of have me lost here could you explain it in "High Skool" student terms ?

I think i have an idea what your saying but I am not exactly sure

 

[Mark44]

1. V is typically used for volume, and was used this way in your original equation. Later on, you switched to v, and this letter is typically used for velocity. It's not a big thing, but if you get an equation like A = c*a + d*b, and you don't keep the A and a straight, you'll get lost.

2. There's a difference between dV/dt at any old time, and dV/dt at a particular time. The first dV/dt is a function (3s²*ds/dt in your problem), and dV/dt at a particular time is a number. Although the problem doesn't give a specific time, it hints at specific times when it asks for dV/dt when (at the specific times) each edge is 1 cm and when each edge is 30 cm. At those two times, dV/dt has specific and changing values.

3. Problems given in sentences should have answers that are also sentences. If the question is "How fast is the volume changing when each edge is 1 cm?" and you give only a number, many math teachers won't consider that you have answered the question. The problem gives you the length of each side in centimeters; what units will the time rate of change of the volume be in?

 

[HallsofIvy]

Technically speaking, your answers are numerically correct. Corrections I whould make are:

1. You started with V as a variable, but your derivative is dv/dt, which is often used as the derivative of velocity.

1. 
   You can call the quantities whatever letter you like but you used "V" (capital letter) in one case and "v" (small letter) in the others. In mathematics, those are different symbols and do not necessarily mean the same thing.
   
   

   [*]Distinguish between dv/dt and dv/dt at a particular time. Your two sets of equations make no distinction between dv/dt (which is a function) and dv/dt at a particular time.
   You have dv/dt = 3s^{2[\sup] ds/dt, and
   dv/dt = 9. The latter is dv/dt|t = t₀ (the time at which s = 1 cm.) Similar for the other value of s.
   [*]Add the appropriate units to your answers.}
   

 

[Shiz Stain]

ah ok thanks for the help