I'm looking at an example problem where $\sqrt{1-(x^2/36)}$ is changed to $\sqrt{36-x^2}$ with no explanation. How does that work?
2 Answers
You probably missed a factor of $\frac 16$. It is true that $\sqrt{1-\frac{x^2}{36}}=\frac 16\sqrt {36-x^2}$
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$\begingroup$ Thanks, that was it. I was using the chain rule and the 1/6 that factored out was actually negative, while the inner term was positive 1/6 so they cancelled each other out. Thanks :) $\endgroup$ Nov 12, 2013 at 21:18
It is just some basic algebra:
$$\begin{eqnarray*} 1-\frac{x^2}{36} &\equiv& \frac{36}{36}-\frac{x^2}{36} \\ \\ \\ &\equiv& \frac{36-x^2}{36}\end{eqnarray*}$$
It follows then that $$\begin{eqnarray*} \sqrt{1-\frac{x^2}{36}} &\equiv& \sqrt{\frac{36-x^2}{36}} \\ \\ \\ &\equiv& \frac{\sqrt{36-x^2}}{\sqrt{36}} \\ \\ \\ &\equiv& \frac{\sqrt{36-x^2}}{6} \\ \\ \\ &\equiv& \tfrac{1}{6}\sqrt{36-x^2} \end{eqnarray*}$$