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Math Help - 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": \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:


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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?

    \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
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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.

    \tan(\theta) = 2h

    The obvious geometry is so glaring.
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