Calculus made easy book7/6/2023 If any different power of $r$ was involved, such as $3$ for the volume of a sphere, then usually the difference in volumes at $0.5$ offsets would not be equal to the rate of change in volume at $r_0$. However, note this only works due to having terms using just $r^2$, $r$ and/or a constant in the original function, so the derivative is a linear function of the radius $r$. Calculus Made Easy: Being a Very Simplest Introduction to Those Beautiful Methods of Recknoning which are Generally Called by the Terrifying Names of the Differntial Calculus and the Integral Calculus: Author: Silvanus Phillips Thompson: Edition: 2: Publisher: Macmillan, 1914: Original from: Princeton University: Digitized: Apr 6, 2009: Length. Overall, this shows you can use $0.5$ at any value of $r_0$. Consider that, to check how the volume changes over a period of $1$ in. ![]() less radius.Īnother way to see is this is that it's due to the volume having a factor of the radius squared. If you want to ace your next exam, heres a software that you can legally use with your Ti89. ![]() This results in using the difference in volumes between $0.5$ in. Calculus Made Easy: Being A Very-Simplest Introduction to Those Beautiful Methods of Reckoning which are Generally Called by the Terrifying Names of the Differential Calculus and the Integral. Calculus Made Easy for TI89 - How to hack your next exam. As such, the change in volume, given by the integral wrt $r$, over any distance, such as $1$ inch, would be the same as this distance times the rate of change at the middle of the range. ![]() Perhaps the simplest, but not necessarily easiest, way to see the reason for using $r - 0.5$ for the lower value is to note that the derivative is a linear function of $r$.
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