College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.5 - Variation - 2.5 Assess Your Understanding: 15

Answer

$\color{blue}{T^2=\dfrac{8a^3}{d^2}}$

Work Step by Step

RECALL: (1) If $y$ varies directly as $x$, then $y=kx$ where $k$ is the constant of proportionality. (2) If $y$ varies inversely as $x$, then $y=\dfrac{k}{x}$ where $k$ is the constant of proportionality. Notice that when the variation is direct, the variable is on the numerator while if the variation is inverse, the variable is in the denominator. $T^2$ varies directly with the cube of $a$ and inversely with the square of $d$. Thus, the equation of the variation is: $T^2=k\cdot \dfrac{a^3}{d^2}$ Since $T=2$ when $a=2$ and $d=4$, then substituting these into the tentative equation above gives: $\require{cancel} T^2=k \cdot \frac{a^3}{d^2} \\2^2=k \cdot \frac{2^3}{4^2} \\4=k \cdot \frac{8}{16} \\4=k \cdot \frac{1}{2} \\2 \cdot 4=k \cdot \frac{1}{2} \cdot 2 \\8=k$ Thus, the equation of the inverse variation is: $T^2=8\cdot \dfrac{a^3}{d^2} \\\color{blue}{T^2=\dfrac{8a^3}{d^2}}$
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