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(05-16-2022, 09:11 PM)LonePredator Wrote: I'll tell you in one sentence what I did in my calculation and then you answer me with yes or no.
If the volume of a cube (or any particular shape) increases by 2 times then it's surface area will increase by 2^2/3 times. YES OR NO? YOU SAID NO which means you never studied maths in your entire life.
Absolutely a big 'yes'. Yes! Are you satisfied now? That isnt even my argument in the first place, and you dont understand even what is the issue, and for the same reason you perceive everything wrong.
Look at yourself, my friend! You arent still getting the blunder you have made.
I respect you a lot and atleast now, try to understand the thing where you are going wrong! Take your own time to read this and I positively hope you understand what I mean to say.
You are right! If you increase the length by two times on every sides, you increase the surface area by 4 times and volume by 8 times. But, this isnt even the case here to the least. Let me say you how.
Now, you could notice that the cuboid 2, after increasing the sides of the cuboid 1 two times on each side. As you said, when we increase the sides by two times everywhere, the surface area increases 4 times and the volume increases by 8 times.
Since you said about the square-cube law here, you are saying about increasing all the sides by the equal multiple, which is basically multiplying all the sides by, for example, two and getting 4 times the surface area and 8 times the volume.
Its all right that you get 4 times the surface area, 8 times the volume if you increase the side by 2 times. Now, how do you find a cuboid with twice the volume of cuboid A which is isometric to cuboid A? Thats what we have to find. Lets look at it.
What was your task? You had to find the bite force of a jaguar after isometrically scaling it to twice its volume, right? So, you are scaling the jaguar to twice the volume isometrically, and then finding the surface area which you got after doing it.
Likewise, we are gonna scale the cuboid A to twice its volume isometrically (without changing the ratios between the vertics) and then find the surface area which we get form it. Isnt this the exact same way?
If you look at cuboid A and Cuboid B, cuboid B has twice the volume of cuboid A. But the proportions arent the same. So, how do you find the measures of a cuboid with twice the volume of A and which maintains the same proportions of A? Lets do it!
Now, lets scale the cuboid A which has volume 1, to volume 2 without changing its proportions.
Ratio of vertices of cuboid A : 1:8:8
Now, the equation of this cube be: x*8x*8x = 1 (where 'x' = measure of the smallest vertice)
Now, we are gonna keep the ratio the same and equate the volume (product of the three vertices) to twice the initial volume. By this, we could find the vertices of that cuboid which has twice the volume of cuboid A and stays with the same ratio. By doing this: we get:
x*8x*8x = 2,
64x^3 = 2,
x = (2/64)^ 1/3
= (1/32)^3
Therefore, x = 0.314980262473361
As we said, x is the value of the smallest vertice.
Now, lets find out the other two vertices. For that, we must multiply x by 8 (since ratio is 1:8:8).
Lets call this cuboid the 'cuboid awesome'. By doing so, the vertices of cuboid awesome are:
Now, check for the proportion. 2.519842099786888/ 0.314980262473361 = 8.
Thus, the ratios are kept the same, and the volume of "cuboid awesome" is twice the volume of cuboid A, as we were intending to get.
This is the exact same thing I did in the comparison. If you go back and see, the measure of the cuboid I took too was the same. I just elaborated my steps with reasoning. Thats all I did.
Now, can you say this isnt isometric variation? If you say so, I am happy to hear from you about why this isnt isometric.