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Thread: Another Stupid Differentiaion Problem

  1. #1
    Senior Member
    Jul 2008

    Exclamation Another Stupid Differentiaion Problem

    Hey guys

    Sorry's another differentiation problem i can't solve ><" ...i asked my friends but they have no clue and since it's the holz my teacher is inaccessible

    Although an example in the textbook showed a similar question, i can't seem to get the answer with the method shown.

    Problem: A man in a boat is 4 km from the nearest point O of a straight beach; his destination is 4 km along the beach from O. If he can row at 4 m/h and walk at 5 km/h, how should he proceed in order toreach his destination in the least possible time?

    The textbook answer is: Row direct to destination.

    Here's my solution:

    $\displaystyle 4^2 + x^2 = BC$(Pythagoras' Theorem)
    $\displaystyle BC = \sqrt{16 + x^2}$
    Time to travel BC $\displaystyle = \frac{\sqrt{16 + x^2}}{4}$
    Time to travel CD$\displaystyle = \frac{4-x}{5}$
    Total time T$\displaystyle = \frac{\sqrt{16 + x^2}}{4} + \frac{4-x}{5}$
    $\displaystyle \frac{dT}{dx} = \frac{x\sqrt{16 + x^2}}{4} - \frac{3}{50}$
    For least possible time, $\displaystyle \frac{dT}{dx} = 0$
    When I work out x, it equals approx. 0.06...which does not make sense at all.

    Please help?
    Last edited by xwrathbringerx; Apr 15th 2009 at 12:32 AM.
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, xwrathbringerx!

    Your work is excellent!

    Your error occured in the only part you didn't show us: solving for $\displaystyle x.$
    I'm sure it was a simple Algrbra mistake.

    The time function is: .$\displaystyle T \;=\;\frac{\sqrt{x^2+16}}{4} + \frac{4-x}{5}$

    Rewrite it: .$\displaystyle T \;=\;\frac{1}{4}(x^2+16)^{\frac{1}{2}} + \frac{4}{5} - \frac{1}{5}x$

    Then: .$\displaystyle \frac{dT}{dx} \;=\;\frac{1}{8}(x^2+16)^{\text{-}\frac{1}{2}}\!\cdot\!2x - \frac{1}{5} \;=\;0 $

    So we have: .$\displaystyle \frac{x}{4\sqrt{x^2+16}} \;=\;\frac{1}{5}$

    . . Then: .$\displaystyle 5x \;=\;4\sqrt{x^2+16}$

    . . Square: .$\displaystyle 25x^2 \:=\:16(x^2+16) \quad\Rightarrow\quad 25x^2 + 16x^2 + 256 \quad\Rightarrow\quad 9x^2 \:=\:256$

    Therefore:. $\displaystyle x^2 \:=\:\frac{256}{9} \quad\Rightarrow\quad x \:=\:\frac{16}{3} \:=\:5\tfrac{1}{3}$ km.

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