Squaring and Cubing Equations: Explained!

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  • Thread starter Thread starter paulmdrdo1
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SUMMARY

The discussion focuses on the mathematical relationship between the volumes and areas of spheres, specifically how the ratios of their volumes and areas relate to the radii when squared and cubed. It establishes that the volume of a sphere is given by \( V = cr^3 \) and the area by \( A = dr^2 \). The derived equations show that \( \left(\frac{V_2}{V_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^6 \) and \( \left(\frac{A_2}{A_1}\right)^3 = \left(\frac{r_2}{r_1}\right)^6 \), demonstrating a consistent relationship between these geometric properties.

PREREQUISITES
  • Understanding of geometric formulas for volume and surface area of spheres
  • Knowledge of algebraic manipulation involving exponents
  • Familiarity with the constants \( c = \frac{4}{3}\pi \) and \( d = 4\pi \)
  • Basic comprehension of ratios and their properties
NEXT STEPS
  • Study the derivation of the volume formula for spheres
  • Explore the relationship between surface area and volume in higher dimensions
  • Learn about the implications of scaling in geometric figures
  • Investigate the properties of exponents and their applications in algebra
USEFUL FOR

Students of mathematics, educators teaching geometry, and anyone interested in the relationships between geometric properties of spheres.

paulmdrdo1
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Can you explain the part of the solution where v2/v1 = a2/a1were being squared and cubed repectively?

Thanks!
 
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$$(r^3)^2 = (r^2)^3$$
 
To provide a little more detail: It is known that the volume of a sphere of radius $r$ is $cr^3$ for some $c$ (in fact, $c=\frac43\pi$) and its area is $dr^2$ ($d=4\pi$). Therefore,
\[
\left(\frac{V_2}{V_1}\right)^2=\left(\frac{cr_2^3}{cr_1^2}\right)^2=\left(\frac{r_2}{r_1}\right)^6
\]
One can similarly show that $(A_2/A_1)^3=(r_2/r_1)^6$ regardless of $d$.
 

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