1
$\begingroup$

This is my conjecture:

If a line never intersects a differentiable curve then there exists at least one tangent to the curve whose slope is that of the line

I have tried proving it by maybe Lagrange's Mean Value Theorem but to no avail...

$\endgroup$
10
$\begingroup$

Not true. Consider the curve $y=e^x$. It never intersects the $x$-axis, but there is no point where the tangent has slope zero.

(No, there's nothing special about the slope being zero: Take the curve $y=ax+e^x$ and the line $y=ax$.)

$\endgroup$
3
$\begingroup$

As you have alredy been informed, the statement is false. However, if we are dealing with a loop or, to be more precise, with a differentiable function $c$ whose domain is $S^1$, then it is true. Just take the point $\theta\in S^1$ such that the distance from $c(\theta)$ to the line is the smallest (or the greatest) possible. The compactness of $S^1$ (and the continuity of $c$) assures us that such a $\theta$ exists. Then the line tangent to the curve at $c(\theta)$ will be parallel to the line. Therefore, they will have the same slope.

$\endgroup$
  • $\begingroup$ +1. Suitably phrased this would be a very nice exercise in Lagrange multipliers. $\endgroup$ – Jose27 Jun 4 '18 at 17:21
  • $\begingroup$ @Jose27 Indeed it would. $\endgroup$ – José Carlos Santos Jun 4 '18 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.