Decrease in Diamter.

[richiesmasher]

What happens to the volume of certain shapes, for instance a cylinder or a sphere, as Diameter increases or decreases?

I deduced on my own that, for instance, if you reduced a diameter by 25% the radius will be halved, as 1/4 of a diameter is 1/2 of a radius, then I used some numbers and it worked out.

So by this logic the Area would decrease too, and the volume, but I wish to know if there are formulas or algebraic expressions I can use to represent these changes.

Also if 2r (radius) = D (diameter)

What would r^2 be in terms of D?

I know that D^2 4r^2

 

[Dr.Peterson]

What happens to the volume of certain shapes, for instance a cylinder or a sphere, as Diameter increases or decreases?

I deduced on my own that, for instance, if you reduced a diameter by 25% the radius will be halved, as 1/4 of a diameter is 1/2 of a radius, then I used some numbers and it worked out.

So by this logic the Area would decrease too, and the volume, but I wish to know if there are formulas or algebraic expressions I can use to represent these changes.

Also if 2R (radius) ^ D (diameter)

What would r^2 be in terms of D?

I know that D^2 = 4r^2

No, the radius is not halved when you reduce the diameter by 25%. Can you show us your reasoning, and your numerical check? There's some good stuff to learn here, and it will be best if you can discover it by yourself. I just want to guide you through the process. Algebra will definitely help in this.

I'm not sure what you mean by "2R ^ D". Can you put that into words? Did you just mean to say 2R = D? And are you taking r and R to be the same thing? Upper and lower case in algebra are commonly used for different variables, so it's important not to ignore case.

To find what r^2 is in terms of D, just solve D = 2r for r and square the resulting expression. You are right that D^2 = 4 r^2; you could just divide both sides by 4 to answer your question.

 

[richiesmasher]

No, the radius is not halved when you reduce the diameter by 25%. Can you show us your reasoning, and your numerical check? There's some good stuff to learn here, and it will be best if you can discover it by yourself. I just want to guide you through the process. Algebra will definitely help in this.

I'm not sure what you mean by "2R ^ D". Can you put that into words? Did you just mean to say 2R = D? And are you taking r and R to be the same thing? Upper and lower case in algebra are commonly used for different variables, so it's important not to ignore case.

To find what r^2 is in terms of D, just solve D = 2r for r and square the resulting expression. You are right that D^2 = 4 r^2; you could just divide both sides by 4 to answer your question.

HI, yes I did mean 2R = D and I did take r and R to be the same thing, but i edited it and will only use r from now on.

My reasoning was this, I figured if D = 2r, then 1/4D must be 1/2r, so I figured that 25% is a 1/4 and you're taking it away from D, so therefore I figured that if 1/4 is taken away from D, then 1/2 must be taken away from (r), so i used the numbers 5 for r, and 10 for D.

Now I found 1/4 of D is 2.5 and I subtracted it, so if D reduced by 1/4 it would become 7.5.

SO I then subtracted 2.5 from r, which is 5-2.5 = 2.5, which seemed to affirm my findings, as 5/2 = 2.5.

And then Yes I see that r^2 = D^2/4

 

[Dr.Peterson]

My reasoning was this, I figured if D = 2r, then 1/4D must be 1/2r, so I figured that 25% is a 1/4 and you're taking it away from D, so therefore I figured that if 1/4 is taken away from D, then 1/2 must be taken away from (r), so i used the numbers 5 for r, and 10 for D.

Now I found 1/4 of D is 2.5 and I subtracted it, so if D reduced by 1/4 it would become 7.5.

SO I then subtracted 2.5 from r, which is 5-2.5 = 2.5, which seemed to affirm my findings, as 5/2 = 2.5.

And then Yes I see that r^2 = D^2/4

Well, in the equation 1/4 D = 1/2 r, you are not taking away 1/4; you are multiplying by 1/4. Reducing by 25% means multiplying by 3/4, not by 1/4.

In your example, if you reduce D = 10 by 25%, you are right that the new D would be 7.5. And if you reduce r = 5 by 50%, the new r is 2.5. But these don't fit the relationship: 7.5 is not 2 times 2.5. If the diameter is 7.5, the radius is not 2.5.

Suppose the radius was 5, and you reduce it by 1/2 to make it 2.5. Then the original diameter was 10, and the new diameter is 5, right? By what percentage did you reduce the diameter?

You will find that working with percentage change doesn't work well here; it's better to think of multiplying or dividing by something, with no adding or subtracting involved.

Have you learned about proportions? The main idea here will be that the diameter is proportional to the radius. When we get into area and volume, we'll find that they are not proportional to the radius or diameter, but there is a nice relationship. But I'll hold off on that for the moment.

 

[richiesmasher]

Well, in the equation 1/4 D = 1/2 r, you are not taking away 1/4; you are multiplying by 1/4. Reducing by 25% means multiplying by 3/4, not by 1/4.

In your example, if you reduce D = 10 by 25%, you are right that the new D would be 7.5. And if you reduce r = 5 by 50%, the new r is 2.5. But these don't fit the relationship: 7.5 is not 2 times 2.5. If the diameter is 7.5, the radius is not 2.5.

Suppose the radius was 5, and you reduce it by 1/2 to make it 2.5. Then the original diameter was 10, and the new diameter is 5, right? By what percentage did you reduce the diameter?

You will find that working with percentage change doesn't work well here; it's better to think of multiplying or dividing by something, with no adding or subtracting involved.

Have you learned about proportions? The main idea here will be that the diameter is proportional to the radius. When we get into area and volume, we'll find that they are not proportional to the radius or diameter, but there is a nice relationship. But I'll hold off on that for the moment.

Ah, I feel like a bulb has gone off in my brain. I have learned about proportions and I see that you are absolutely right about multiplying or dividing, it's much easier, I'm following this now, radius is proportional to diameter, so if the diameter decreases by 1/4 then the radius will as well, and this can be verified with the relationship D= 2r .

But what will the next steps be, in terms of volume and area? These seem a bit more complex.

 

[Deleted member 4993]

Ah, I feel like a bulb has gone off in my brain. I have learned about proportions and I see that you are absolutely right about multiplying or dividing, it's much easier, I'm following this now, radius is proportional to diameter, so if the diameter decreases by 1/4 then the radius will as well, and this can be verified with the relationship D= 2r .

But what will the next steps be, in terms of volume and area? These seem a bit more complex.

When you are considering volume - you have to consider a cylinder (as opposed to a circle for area). The volume of a cylinder:

Volume (of cylinder) = Area (of base of cylinder) * height (of cylinder)

So if you reduce (base area of the circular cylinder) by 25% (by reducing radius by certain amount) and kept the height constant - then the the volume of the cylinder will be reduced by 25%.

From previous discussion you may have found that to reduce the area of the circle by 25% - you have to reduce the radius by (~13.398% =) 13%

 

[Dr.Peterson]

Ah, I feel like a bulb has gone off in my brain. I have learned about proportions and I see that you are absolutely right about multiplying or dividing, it's much easier, I'm following this now, radius is proportional to diameter, so if the diameter decreases by 1/4 then the radius will as well, and this can be verified with the relationship D= 2r .

But what will the next steps be, in terms of volume and area? These seem a bit more complex.

Good thinking!

There are several things to consider.

First, staying in two dimensions, it makes a difference whether you are talking about keeping figures similar while increasing or decreasing size, or just change one dimension. For example, if you start with a square, one thing will happen if you increase both width and height, making a larger (similar) square, or just change the height and keep the same width, making a taller rectangle. Try experimenting with both. (I recommend starting with rectangles, which are a lot easier than other figures to work with; the conclusion you end up with will be independent of the shape.) Working with the area of a circle will be like the square, rather than the circle, because the shape remains similar.

When you move to three dimensions, you have the same issue, but more possibilities. Subhotosh assumed you are changing only one dimension; and he stuck with the "increase by a percent" approach, which makes the results less obvious. Suppose you multiply the height of the cylinder by k; what will happen to the area? What if you multiply the radius by k (which amounts to changing two dimensions at once, the length and width like my square)?

Keep thinking, and let us know what you find.

 

[richiesmasher]

Good thinking!

There are several things to consider.

First, staying in two dimensions, it makes a difference whether you are talking about keeping figures similar while increasing or decreasing size, or just change one dimension. For example, if you start with a square, one thing will happen if you increase both width and height, making a larger (similar) square, or just change the height and keep the same width, making a taller rectangle. Try experimenting with both. (I recommend starting with rectangles, which are a lot easier than other figures to work with; the conclusion you end up with will be independent of the shape.) Working with the area of a circle will be like the square, rather than the circle, because the shape remains similar.

When you move to three dimensions, you have the same issue, but more possibilities. Subhotosh assumed you are changing only one dimension; and he stuck with the "increase by a percent" approach, which makes the results less obvious. Suppose you multiply the height of the cylinder by k; what will happen to the area? What if you multiply the radius by k (which amounts to changing two dimensions at once, the length and width like my square)?

Keep thinking, and let us know what you find.

Ok, I will think hard about this and send a reply later or tomorrow.

 

[richiesmasher]

Ok, I've thought about it, and I see that if you increase one dimension of a rectangle, it will either get longer or wider, one of the 2, and when you increase both dimensions, the entire shape becomes bigger my multiples if what you increase it by?

And I'm picturing the cylinder in image form, not numerical it helps me, I see if you increase the height of the cylinder, it will increase the volume, and if you increase area, it will also increase the volume, so for both properties the volume is increased

 

[Dr.Peterson]

Ok, I've thought about it, and I see that if you increase one dimension of a rectangle, it will either get longer or wider, one of the 2, and when you increase both dimensions, the entire shape becomes bigger my multiples if what you increase it by?

And I'm picturing the cylinder in image form, not numerical it helps me, I see if you increase the height of the cylinder, it will increase the volume, and if you increase area, it will also increase the volume, so for both properties the volume is increased

In particular, if you change one dimension (e.g. the height of a rectangle), the area is PROPORTIONAL to that dimension, so that doubling the height doubles the area, for example; if you change both dimensions by the same multiple, then you are multiplying TWICE, so the area is proportional to the SQUARE of the linear dimensions. If you double both the length and the width, then you double the area, and then double it again, multiplying by 4 (2 squared).

Algebraically, if the original rectangle is L by W, its area is LW; if you double only the height, the new area is (2L)W = 2(LW), so it is doubled. If you double both L and W, the new area is (2L)(2W) = 4(LW).

Something similar is true for volumes.

 

[richiesmasher]

In particular, if you change one dimension (e.g. the height of a rectangle), the area is PROPORTIONAL to that dimension, so that doubling the height doubles the area, for example; if you change both dimensions by the same multiple, then you are multiplying TWICE, so the area is proportional to the SQUARE of the linear dimensions. If you double both the length and the width, then you double the area, and then double it again, multiplying by 4 (2 squared).

Algebraically, if the original rectangle is L by W, its area is LW; if you double only the height, the new area is (2L)W = 2(LW), so it is doubled. If you double both L and W, the new area is (2L)(2W) = 4(LW).

Something similar is true for volumes.

\

I see when I do the algebra for Volume of a cylinder, It gives me this: Pi x rsquared x H
when I double the height,it gives me this : Pi x rsquared x 2(H)
which I can rearrange to see this : 2(Pi x rsqaured x H)

So I conclude, doubling the height would double the volume?

But I assume when it comes to the radius it's different.

 

[richiesmasher]

In particular, if you change one dimension (e.g. the height of a rectangle), the area is PROPORTIONAL to that dimension, so that doubling the height doubles the area, for example; if you change both dimensions by the same multiple, then you are multiplying TWICE, so the area is proportional to the SQUARE of the linear dimensions. If you double both the length and the width, then you double the area, and then double it again, multiplying by 4 (2 squared).

Algebraically, if the original rectangle is L by W, its area is LW; if you double only the height, the new area is (2L)W = 2(LW), so it is doubled. If you double both L and W, the new area is (2L)(2W) = 4(LW).

Something similar is true for volumes.

So I assume that the volume of a cylinder would double as you double the height?
Because

V=πr^2 h
so V = π xr^2 x2(H)



which you can rearrange to V=2(π x r^2 xH) ?

 

[Dr.Peterson]

So I assume that the volume of a cylinder would double as you double the height?
Because

V=πr^2 h
so V = π xr^2 x2(H)

which you can rearrange to V=2(π x r^2 xH) ?

Yes, excellent.

A suggestion: when you write formulas, you should avoid using "x" for multiplication, as it looks like a variable and can be confusing. Traditionally we use "*" for multiplication when we type; or just leave it out:

V = π r^2 h, so if we replace h with 2h, we have V = π r^2 (2h) = 2(π r^2 h)
​

Now, what happens if you double the radius?

 

[richiesmasher]

Yes, excellent.

A suggestion: when you write formulas, you should avoid using "x" for multiplication, as it looks like a variable and can be confusing. Traditionally we use "*" for multiplication when we type; or just leave it out:

V = π r^2 h, so if we replace h with 2h, we have V = π r^2 (2h) = 2(π r^2 h)
​

Now, what happens if you double the radius?

I think I've solved it.

When you double the radius the volume quadruples, because:

V = π r^2 h
SO : V = π (2r)^2 h
V = π 4r^2 h
V = 4(π r^2 h)

 

[Dr.Peterson]

When you double the radius the volume quadruples, because:

V = π r^2 h
SO : V = π (2r)^2 h
V = π 4r^2 h
V = 4(π r^2 h)

You've got the technique down.

The area of a plane figure is proportional to the square of its dimension (in this case, the radius); and the volume is proportional to that.