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Say we have some positive value K. How can we know what K is if the function $f(x)=K tan(x)$ intersects its derivative perpendicularly on a graph?

So I know how to do this using some form of a software but don't know how to do it theoretically, or by paper and pencil, to make sure I am getting the same value. How I tried to approach the problem: $f(x)=K tan(x)$. $f(x)'=K sec^2(x)$. Then we have $-K/sec^2(x)$ since we are trying to find the derivative perpendicularly. Then to find K, I set the the two equal and that is where I am stuck. Any help with shown steps would be appreciated.

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You are fine with $f'(x)=K\sec^2 x$. When you say "then we have $-K/\sec^2 x$" it is not clear what you mean. To have a perpendicular intersection of two curves, you need their derivatives to be negative reciprocals, so now you need to find $f''(x)$. Having done that, you require that $f'(x)f''(x)=-1$ for some $x$ and $K$.

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  • $\begingroup$ What would "intersects the graph of its derivative perpendicularly" mean? Would it be where f(x) and -1/f'(x) form a 90 degree angle? $\endgroup$ – hherklj kljkljklj Oct 9 '13 at 22:18
  • $\begingroup$ It would be where $f(x)$ and $f'(x)$ make a 90 degree angle. But since the slope of $f(x)$ is $f'(x)$ and the slope of $f'(x)$ is $f''(x)$ we get the condition I gave at the end. $\endgroup$ – Ross Millikan Oct 9 '13 at 22:19

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