Thread: Application of Calculus to Analytic Geomtry Problem

1. 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:

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

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

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

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

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

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

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

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

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

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

By basic algebra:

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

Thank you very much for your help.

Given solution:

2. Re: Application of Calculus to Analytic Geomtry Problem

Have you considered common denominators?

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

or

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

so

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

Or, since everyone is positive

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

3. 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?

4. Re: Application of Calculus to Analytic Geomtry Problem

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

$\tan(\theta) = 2h$

The obvious geometry is so glaring.