3 dimensional changes

[sndwatkins]

Analyzing the overall effects on a three-dimensional figure caused by a change in one of the figure's dimensions. I am looking for ways to describe these effects. For example if you double the sides of a rectangular prism what effect does it have on the total area and volume. I am looking for why for example the volume does not just double when you double the side lengths. Any other 3 dimensional figures such as cones, rectangular and square pyramids would be helpful as well. Thanks!

 

[Mrspi]

Analyzing the overall effects on a three-dimensional figure caused by a change in one of the figure's dimensions. I am looking for ways to describe these effects. For example if you double the sides of a rectangular prism what effect does it have on the total area and volume. I am looking for why for example the volume does not just double when you double the side lengths. Any other 3 dimensional figures such as cones, rectangular and square pyramids would be helpful as well. Thanks!

Let's just look at a rectangular prism....

Let a = length
Let b = width
Let c = height

The surface area is the sum of the areas of the faces of the prism. Each face is a rectangle. Two of the faces have area a*b. Two of the faces have area b*c. Two of the faces have area a*c. So, the surface area is

SA = ab + ab + bc + bc + ac + ac, or
SA = 2ab + 2bc + 2ac

Now, suppose you DOUBLE each dimension of the rectangular prism.

New length = 2a
New width = 2b
New height = 2c

New surface area = 2*(2a)*(2b) + 2*(2b)*(2c) + 2*(2a)*(2c)
New surface area = 8ab + 8bc + 8ac
New surface area = 4*2ab + 4*2bc + 4*2ac
How does this compare with the ORIGINAL surface area? Well, it is FOUR times as large:
New surface area = 4*[2ab + 2bc + 2ac)
Doubling each dimension results in a surface area which is 4 times as large.

How about the volume? For a rectangular prism,
volume = length * width * height
Original volume = a*b*c

After doubling the length, width and height, we have
New volume = 2a*2b*2c, or
New volume = 8*abc
New volume = 8 * original volume

Doubling each dimension results in a volume which is 8 times as large.

Note that 4 = 2[sup:31wmvlrt]2[/sup:31wmvlrt] and that 8 = 2[sup:31wmvlrt]3[/sup:31wmvlrt]

Surface area (even for a three-dimensional figure) is a 2-dimensional measure. If you multiply each dimension in the figure by n, then you multiply the area by n[sup:31wmvlrt]2[/sup:31wmvlrt].

Volume is a 3-dimensional measure. If you multiply each dimension in the figure by n, then you multiply the volume by n[sup:31wmvlrt]3[/sup:31wmvlrt]

You can apply similar reasoning to show that you'll get the same effect on the area and the volume of the other figures you mentioned when you double (for example) each of the dimensions.

 

[Denis]

"Get ready!" by looking at a TWO-dimensional figure ; take a square:
5 by 5 square = 5 * 5 = 25 area
double the side lengths:
10 by 10 square = 10 * 10 = 100 area (4 times larger)

 

[sndwatkins]

Thanks for the feedback. So following your information I am trying to make some assumptions as it applies to other 3-dimensional figures. So if I had a cylinder and I double the radius then the volume would be 4 times larger. Is my thinking correct on this? Would the volume of a cylinder then be 4 times smaller if I take half of the radius and then calculate the volume?

 

[mmm4444bot]

I am trying to make some assumptions There's no need for assuming.

if I had a cylinder and I double the radius then the volume would be 4 times larger. Is my thinking correct on this? Try it, and see!

Would the volume of a cylinder then be 4 times smaller if I take half of the radius and then calculate the volume? Again, experiment!

Do you know? You can pick 10 for the radius, as a test, and then let the radius also be 20 and 5, followed by comparing the corresponding volumes.

You can do the same thing with radii 1, 2, and 1/2, OR 3, 6, and 3/2.

You can conduct as many tests of actual radii as you please, but there is a way to test all of the possible radii in the universe at once.

Let r represent an arbitrary radius. You know the expression that represents the volume, when the radius is r, yes?

Now let the radius be twice as big. Calculate the volume when the radius is 2r.

Now let the radius be half as big. Calculate the volume when the radius is r/2.

Compare these volumes; in other words, analyze your results.

As far as extending your hypothesis to other three-dimensional objects, experiment!

\(\displaystyle \text{Volume of a Sphere} \ = \ \frac{4}{3} \cdot \pi \cdot r^{3}\)

Compare the volumes for spheres with radii 2r and r/2 to the volume above with radius r. Uh oh!

 

[sndwatkins]

Thanks a bunch. The light has finally come on thanks to your feedback! I am preparing for a test on analyzing the overall effects on a three-dimensional figure caused by a change in one of the figure's dimensions. If you could ask me some questions to see if I fully grasp it would be very helpful. Thanks again for this website it is the first time I have used it.