# Tangent will not meet the curve again

1. May 2, 2014

### delsoo

1. The problem statement, all variables and given/known data

can anyone give me hint on how to show the tangent will not meet the curve again?

2. Relevant equations

3. The attempt at a solution

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2. May 2, 2014

### Staff: Mentor

Just plug your coordinates in the equation for the tangent and look for all solutions?

3. May 2, 2014

### Ray Vickson

You can try to check the convexity (or concavity) properties of the curve $y = Y(x)$, by checking the second derivative $d^2 y / dx^2$.

4. May 5, 2014

### delsoo

what's the relationship between checking the concavity and show the curve will not meet the tangent again?

5. May 6, 2014

### Curious3141

Have you tried sketching the curve?

6. May 6, 2014

### Ray Vickson

Google "convex function" or "concave function". See, eg., http://ece.tamu.edu/~cui/ECEN629/lecture2.pdf [Broken] slide 4.

Last edited by a moderator: May 6, 2017
7. May 6, 2014

### HallsofIvy

It might be simplest to eliminate the parameter and write a Cartesian equation for this curve:
$x= ln(cos(\theta)$ so $e^x= cos(\theta)$ and $y= ln(sin(\theta)$ so $e^y= sin(\theta)$. Then $e^{2x}+ e^{2y}= cos^2(\theta)+ sin^2(\theta)= 1$. Of course, for $0< \theta< \pi/2$, $cos(\theta)$ goes from 1 to 0 so x goes from 0 to $-\infty$ and y goes from $-\infty$ to 0. The graph is in the third quadrant.

At $\theta= \pi/4$, $y= x= -(1/2)ln(2)$ so that $e^{2y}= e^{2x}= 2^{-1/2}= 1/2$. Further, differentiating $e^{2x}+ e^{2y}= 1$, $2e^{2x}+ 2e^{2y}y'= 0$ so, at $(1/2, 1/2)$, $y'= -1$. The tangent line is $y= -1(x+ (1/2)ln(2))- (1/2)ln(2)= -x- ln(2)$ and the question becomes solving $e^{2x}+ e^{2(-x- ln(2)}= e^{2x}- (1/2)e^{-2x}= 1$.