# Solving Cube's Volume Change: Related Rates Help

• Shiz Stain
In summary, the conversation was about finding the rate of change of a cube's volume when its edges are expanding at a rate of 3 centimeters per second. The equations used were V = s³ and dv/dt = 3s² ds/dt. The expert provided corrections to the solution, such as distinguishing between dv/dt and dv/dt at a particular time and adding appropriate units to the answers.
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.

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 = 3s2[\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.

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

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?

Mark44 said:
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 = 3s2[\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.

ah ok thanks for the help

## 1. How do I calculate the volume of a cube?

To calculate the volume of a cube, you need to know the length of one side. Then, simply use the formula V = s^3, where V is the volume and s is the length of one side.

## 2. What is the formula for related rates in a cube's volume change?

The formula for related rates in a cube's volume change is dV/dt = 3s^2 * ds/dt, where dV/dt represents the rate of change of volume, s is the length of one side, and ds/dt represents the rate of change of the side length.

## 3. How can I solve for the rate of change of a cube's volume?

To solve for the rate of change of a cube's volume, you will need to use the formula dV/dt = 3s^2 * ds/dt, where dV/dt is the rate of change of volume and ds/dt is the rate of change of the side length. You will also need to know the values of both dV/dt and ds/dt.

## 4. What are some real-life applications of related rates for a cube's volume change?

Related rates for a cube's volume change can be used in many real-life situations, such as calculating the rate at which a melting ice cube is losing volume, or determining how quickly a balloon is expanding as it is filled with air.

## 5. How can I check my solution for solving a cube's volume change using related rates?

To check your solution for solving a cube's volume change using related rates, you can use the original formula V = s^3 and plug in the values for the rate of change of volume and the side length. If the two sides of the equation are equal, then your solution is correct.

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