# Rearranging to eliminate the R variable in volume equation

• FiveAlive
In summary: R, not 5/20. h should be multiplying R, not 5/20.In summary, at 5 seconds, the water glass is being filled at a rate of 50 cm^3/min. The height of the water is changing at a rate of 100 cm^3/min.

#### FiveAlive

A water glass (10 cm diameter at the top, 6 cm diameter at the bottom, 20 cm in height) is being filled at a rate of 50 cm^3/min. Find the rate of change of the height of the water after 5 seconds.

V=(∏* (h/3)) * (R^2 + r^2 +(R*r))

I've been messing around with this for a bit and made a bit of leeway but I'm unsure of how to eliminate R and r from my equation. If it was a cone I would use the r/h=5/20 and rearrange to 5h/20 and replace my 'r' then differentiate. However that ratio doesn't seem to work on a truncated cylinder and I have two radius-es. Any suggestions?

What are R, r, and h?

R=5 cm r= 3 cm h= 20 cm

FiveAlive said:
R=5 cm r= 3 cm h= 20 cm

Ok. So r is your lower radius. That's a constant. As the glass fills R and h will change. But there is a linear relation between them since the edge of the glass is a straight line. When h=0 cm then R=3 cm. When h=20 cm R=5 cm. What's the linear relation?

OK cool I think I get what you're going at. If I put those point on a graph the slope would be a slope of R/H=10/1. I can sub that in for R and put 3 in for r because it's a constant. Simplified it might look a little something like this

V= 100H^3(1/3)∏ + 90H^2(1/3)∏ + 9H(1/3)∏

I'm still a little uncertain what to do with the issue of time. Could I take the derivative of the height (20 cm) ,when the glass is full, get the rate of change and then attempt to find the rate algebraically?

FiveAlive said:
OK cool I think I get what you're going at. If I put those point on a graph the slope would be a slope of R/H=10/1. I can sub that in for R and put 3 in for r because it's a constant. Simplified it might look a little something like this

V= 100H^3(1/3)∏ + 90H^2(1/3)∏ + 9H(1/3)∏

I'm still a little uncertain what to do with the issue of time. Could I take the derivative of the height (20 cm) ,when the glass is full, get the rate of change and then attempt to find the rate algebraically?

But R=10h doesn't work. Remember, when h=0 cm then R=3 cm. When h=20 cm R=5 cm. If I put h=0 into R=10h I get 0, not 3. Can you fix that? For the other part of your question how much water is in the glass after 5s? Use that to find R and h at the time you want to find the rates.

Could I use R=5H/20 and plug 3 in as a constant? If you plug in 0, the answer will still be 0 however I am a little unsure what else to try since I can't think of a number that i can multiply times 0 and not get 0, unless I just add 3 on to the end like

R= 5h/20 + 3

FiveAlive said:
Could I use R=5H/20 and plug 3 in as a constant? If you plug in 0, the answer will still be 0 however I am a little unsure what else to try since I can't think of a number that i can multiply times 0 and not get 0, unless I just add 3 on to the end like

R= 5h/20 + 3

R=5h/20+3 is ok, for h=0. But at h=20 you want R=5, it's NOT ok there. Fix the slope. What should be multiplying h instead of 5/20?

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## 1. What is the volume equation and why is it important?

The volume equation is a mathematical formula used to calculate the amount of space occupied by an object. It is important because it allows us to determine the volume of various objects, which is crucial in many scientific fields such as physics and chemistry.

## 2. How do you rearrange the volume equation to eliminate the R variable?

To eliminate the R variable in the volume equation, you can use basic algebraic principles such as inverse operations and substitution. For example, if the volume equation is V = πr^2h, you can divide both sides by π and then substitute the value of r with another variable, such as x, to eliminate it.

## 3. Why would you want to eliminate the R variable in the volume equation?

Eliminating the R variable in the volume equation allows us to solve for other variables, such as the height or radius, without needing to know the value of the R variable. This can be useful in situations where we only have certain information and need to find the missing variable.

## 4. Are there any limitations to rearranging the volume equation to eliminate the R variable?

Yes, there are limitations to this method. It may not always be possible to eliminate the R variable, depending on the complexity of the volume equation and the number of variables involved. In some cases, it may be necessary to use other methods, such as integration, to solve for the volume.

## 5. Can you provide an example of rearranging the volume equation to eliminate the R variable?

Sure, let's use the volume equation for a cylinder, V = πr^2h. To eliminate the r variable, we can divide both sides by π, giving us V/π = r^2h. Then, we can substitute r with another variable, say x, which results in V/π = x^2h. Finally, we can solve for x by taking the square root of both sides, giving us x = √(V/πh). This eliminates the R variable from the equation and allows us to solve for the other variables.