Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 76

Answer

(a) $V(d) = \frac{d^3 \pi} 6$ (b) \begin{matrix} d & V(d) \\ 0 & 0 \\ 1 & 0.524 \\ 2 & 4.189 \\ 3 & 14.137 \\ 4 & 33.51 \end{matrix} (c) Image below: ------

Work Step by Step

(a) $V(d) = ?$ - Take the cube of the diameter $V(d) = d^3 ...$ - Multiply by $\pi$: $V(d) = d^3 \pi...$ - Divide by 6: $V(d) = \frac{d^3 \pi} 6$ (b) Make a table of values of $V(d)$. You can choose any values for $d$. I will use 0, 1, 2, 3 and 4. - Calculate $V(0)$, $V(1)$, $V(2)$, $V(3)$ and $V(4)$: $$V(0) = \frac{(0)^3 \pi} 6 = 0$$ $$V(1) = \frac{(1)^3 \pi} 6 = 0.524$$ $$V(2) = \frac{(2)^3 \pi} 6 = 4.189$$ $$V(3) = \frac{(3)^3 \pi} 6 = 14.137$$ $$V(4) = \frac{(4)^3 \pi} 6 = 33.51$$ - Make the table: \begin{matrix} d & V(d) \\ 0 & 0 \\ 1 & 0.524 \\ 2 & 4.189 \\ 3 & 14.137 \\ 4 & 33.51 \end{matrix} (c) Use the table we made to plot the points with the respective coordinates. Then sketch a curve that passes through all these points.
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