Application of Calculus to Analytic Geomtry Problem

Hello everyone,

I have two questions on which I would greatly appreciate help. I have also included the given solution below.

**Problem:**

http://img193.imageshack.us/img193/2...ithdiagram.png

**Question #1: **For part (iii), is it possible to solve it using calculus?

By Implicit Differentiation, the circle has "slope": $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$.

The required line has slope $\displaystyle m = -h$ and the equation: $\displaystyle y = h(3 - x)$

$\displaystyle \Rightarrow$ The line is tangent to the circle when $\displaystyle -\frac{x}{y} = -h \Rightarrow \frac{x}{y} = h $.

Substitution of the equation of the line at the point of tangency gives:

$\displaystyle \frac{x}{h(3 - x)} = h $.

But this does not appear to help too much.

**Question #2: This question is based on part (v), but I am trying to find something else that is not asked by the question.**

Since $\displaystyle h = \frac{6}{7}$, the vertex of the triangle $\displaystyle \text{ie} \left(1, \frac{12}{7}\right)$ is inside the circle.

This means that we are in the case where $\displaystyle h < \frac{\sqrt{3}}{2}$.

The question does not ask for this, but how would I show that

$\displaystyle h = \frac{6}{7} < \frac{\sqrt{3}}{2} $

By basic algebra:

$\displaystyle \frac{\sqrt{3}}{7} < \frac{\sqrt{3}}{2} $ but $\displaystyle \frac{6}{7} > \frac{\sqrt{3}}{7}$.

Thank you very much for your help.

**Given solution:**

http://img833.imageshack.us/img833/9359/solutionf.png

Re: Application of Calculus to Analytic Geomtry Problem

Have you considered common denominators?

$\displaystyle \frac{6}{7}$ vs. $\displaystyle \frac{\sqrt{3}}{2}$

or

$\displaystyle \frac{12}{14}$ vs. $\displaystyle \frac{7\sqrt{3}}{14}$

so

$\displaystyle 12$ vs. $\displaystyle 7\sqrt{3}$

Or, since everyone is positive

$\displaystyle 144$ vs. $\displaystyle 49\cdot 3 = 147$

Re: Application of Calculus to Analytic Geomtry Problem

Thanks TKHunny, for answering Question #2. I should have thought of that!

May I ask if that has been any insight into **Question #1**?

Re: Application of Calculus to Analytic Geomtry Problem

I'm a little puzzled why you would want to think of another solution.

$\displaystyle \tan(\theta) = 2h$

The obvious geometry is so glaring.