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Thread: Application of Calculus to Analytic Geomtry Problem

  1. #1
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    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": $\displaystyle \frac{dy}{dx} = -\frac{x}{y}$.

    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:


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  2. #2
    MHF Contributor
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    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. $\displaystyle 49\cdot 3 = 147$
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  3. #3
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    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?
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  4. #4
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    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.
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