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Math Help - snell's law

  1. #1
    Member Veronica1999's Avatar
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    snell's law

    Jamie is at the point
    J = (0, 8) offshore, needing to
    reach the destination D = (12,3) on land as quickly
    as possible. The lake shore is the x-axis. Jamie is in a
    boat that moves at 10 uph, with a motor bike on board
    that will move 20 uph once the boat reaches land. Your
    job is to find the landing point P = (x, 0) that minimizes
    the total travel time from J to D. Assume that
    the trip from P to D is along a straight line.

    Could someone pls take a look at my work?

    Thanks

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  2. #2
    MHF Contributor MarkFL's Avatar
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    Re: snell's law

    You have correctly found, using the Pythagorean theorem and the relation between time, distance and constant velocity:

    f(x)=\frac{\sqrt{x^2+8^2}}{10}+\frac{\sqrt{(12-x)^2+3^2}}{20}

    where f measures time in hours. You then began the process of differentiating f with respect to x to find the critical values.

    f'(x)=\frac{2x}{20\sqrt{x^2+8^2}}+\frac{2(12-x)(-1)}{40\sqrt{(12-x)^2+3^2}}=

    \frac{x}{10\sqrt{x^2+8^2}}+\frac{x-12}{20\sqrt{(12-x)^2+3^2}}=

    \frac{2x\sqrt{(12-x)^2+3^2}+(x-12)\sqrt{x^2+8^2}}{20\sqrt{\left(x^2+8^2\right) \left((12-x)^2+3^2\right)}}

    Noting the denominator is always positive, we then consider:

    2x\sqrt{(12-x)^2+3^2}+(x-12)\sqrt{x^2+8^2}=0

    2x\sqrt{(12-x)^2+3^2}=(12-x)\sqrt{x^2+8^2}

    Observing both sides represent non-negative values, we may square:

    4x^2\left((12-x)^2+3^2\right)=(12-x)^2\left(x^2+8^2\right)

    (6x)^2=(12-x)^2\left(8^2-3x^2\right)

    36x^2=\left(144-24x+x^2\right) \left(64-3x^2\right)

    36x^2=-3x^4+72x^3-368x^2-1536x+9216

    3x^4-72x^3+404x^2+1536x-9216=0

    A graph of this polynomial shows only one real root for 0\le x\le 12.

    Using a root-finding technique such as Newton's method, we find:

    x\approx4.2194869338023206885

    The first derivative test confirms this is a minimum, i.e., f'(4) < 0 and f'(5) > 0.

    To apply Snell's Law, we could write:

    \frac{\sin\theta_2}{\sin\theta_1}=\frac{20}{10}=2

    \sin\theta_2=2\sin\theta_1

    \frac{12-x}{\sqrt{(12-x)^2+3^2}}=\frac{2x}{\sqrt{x^2+8^2}}

    This leads to the same equation we have above.
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: snell's law

    Let's generalize a bit to say we want to find the quickest path across two media:

    snell's law-qpath.jpg

    (sorry, still learning the idiosyncrasies of this board...don't yet know how to include an image directly into a post...)

    We are traveling from point A to point B then to point C by the quickest path possible. The question is, where should point B be placed?

    I have set up xy coordinate axes such that:

    O=\left(0,0\right)

    A=\left(0,-W_1\right)

    B=\left(B,0\right)

    C=\left(L,W_2\right)

    The distance from A to B is:

    AB=\sqrt{\left(B-0\right)^2+\left(0-\left(-W_1\right)\right)^2}=\sqrt{B^2+W_1^2}

    The distance from B to C is:

    BC=\sqrt{\left(L-B\right)^2+\left(W_2-0\right)^2}=\sqrt{\left(L-B\right)^2+W_2^2}

    The speed through medium 1 is v_1 and the speed through medium 2 is v_2. We assume both velocities are positive values. Thus, using the relationship between distance, constant velocity and time, we find the total time as a function of B is:

    t\left(B\right)=\frac{\sqrt{B^2+W_1^2}}{v_1}+\frac  {\sqrt{\left(L-B\right)^2+W_2^2}}{v_2}

    Differentiating with respect to B, we find:

    t'\left(B\right)=\frac{B}{v_1\sqrt{B^2+W_1^2}}+ \frac{\left(B-L\right)}{v_2\sqrt{\left(L-B\right)^2+W_2^2}}=

    \frac{Bv_2\sqrt{\left(L-B\right)^2+W_2^2}+\left(B-L\right)v_1\sqrt{B^2+W_1^2}}{v_1v_2\sqrt{B^2+W_1^2  }\sqrt{\left(L-B\right)^2+W_2^2}}

    The denominator will always be positive, so we need only consider:

    Bv_2\sqrt{\left(L-B\right)^2+W_2^2}+\left(B-L\right)v_1\sqrt{B^2+W_1^2}=0

    Bv_2\sqrt{\left(L-B\right)^2+W_2^2}=\left(L-B\right)v_1\sqrt{B^2+W_1^2}

    Let's take a moment to rewrite this:

    \frac{B}{\sqrt{B^2+W_1^2}}\cdot\frac{\sqrt{\left(L-B\right)^2+W_2^2}}{L-B}=\frac{v_1}{v_2}

    \frac{\frac{B}{\sqrt{B^2+W_1^2}}}{\frac{L-B}{\sqrt{\(L-B\)^2+W_2^2}}}=\frac{v_1}{v_2}

    \frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}

    Thus, we have shown that Snell's law (or Descartes' law) is satisfied when t'(B) = 0.
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