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Math Help - differentiation question

  1. #1
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    differentiation question

    Hi, I'm new to this forum. Any help would be greatly appreciated. I'm actually tutoring a refugee in a volunteer program, but it's been so long since I've done this maths that I'm having real trouble with it so in order to help her, I myself need some help!

    A Queenland resort has a large swimming pool with AB=75. The pool is rectangular and has corners A,B,C,D, with A and C as opposites and B and D as opposites. P is a point somewhere between D and C.
    A boy can swim at 1 m/s and run at 1, 2/3 m/s. He starts at A, swims to a point P on DC, and runs from P to C. He takes 2 seconds to pull himself out of the pool.
    a, Let DP=x m and the total time be T s. Show that T=square root of (x squared +900)+3/5 (75-x)+2.
    b, find dT/dx
    c. i Find the value of x for which the time taken is a minimum.
    ii Find the minimum time.
    d. Find the time taken if the boy runs from A to D and then D to C.

    ----------

    An isosceles trapezoid is inscribed in the parabolas y=4-x squared as illustrated (the parabola intersects the x axis at (-2,0) and (2,0) and is symmetrical.
    a. show that the area of the trapezoid is :
    1/2(4-x squared) (2x+4)
    b, show that the trapezoid has its greatest area when x=2/3
    c, Repeat with the parabolas y = a squared - x squared
    i, show that the area, A, of the trapezoid = ( a squared - x squared) ( a + x)
    ii, use the product rule to find dA/ dx.
    iii, show that a maximum occurs when x = a/3.

    If anymore information is needed, please ask!
    Thanks so much.
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  2. #2
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    Quote Originally Posted by afan17 View Post
    ...

    An isosceles trapezoid is inscribed in the parabolas y=4-x squared as illustrated (the parabola intersects the x axis at (-2,0) and (2,0) and is symmetrical.
    a. show that the area of the trapezoid is :
    1/2(4-x squared) (2x+4)
    b, show that the trapezoid has its greatest area when x=2/3
    c, Repeat with the parabolas y = a squared - x squared
    i, show that the area, A, of the trapezoid = ( a squared - x squared) ( a + x)
    ii, use the product rule to find dA/ dx.
    iii, show that a maximum occurs when x = a/3.
    Let a and b denote the parallel sides of the trapezium and h it's height. Then the area of the trapezium is calculated by:

    A_t=\frac{a+b}2 \cdot h

    With your trapezium the base parallel has the length 4 and the upper parallel has the length 2x. The height correspond to the y-value. Plug in these terms into the formula given above:

    A_T(x)=\frac{2x+4}2 \cdot (4-x^2)

    which is exactly the given result.

    To get the maximum value of A_T calculate the derivative with respect to x:

    A_T(x)= (x+2)(4-x^2)=-x^3-2x^2+4x+8

    A'_T(x)=-3x^2-4x+4

    A'_T(x)=0~\implies~x=\frac23~\vee~x=-2

    The negative solution isn't very plausible here.

    Do the following questions in exactly the same manner.
    Attached Thumbnails Attached Thumbnails differentiation question-par_trapz.gif  
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  3. #3
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    Hello, afan17!

    I'm new to this forum . . . Welcome aboard!
    I'm actually tutoring a refugee in a volunteer program . . . Good for you!

    A resort has a rectangular swimming pool ABCD with AB = 75m, AD = 30m.
    P is a point somewhere between D and C.
    A boy can swim at 1 m/s and run at 1 2/3 m/s.
    He starts at A, swims to point P, and runs from P to C.
    He takes 2 seconds to pull himself out of the pool.

    a) Let DP = x meters and the total time be T seconds.
    Show that: . T\:=\:\sqrt{x^2+900}+ \frac{3}{5}(75-x)+2
    Code:
          A             75              B
          * - - - - - - - - - - - - - - *
          |  *                          |
          |     *                       |
       30 |        *                    |
          |           *                 |
          |              *              |
          * - - - - - - - - * - - - - - *
          D        x        P   75-x   C
    He will swim the diagonal distance AP.
    This is the hypotenuse of right triangle ADP.
    . . Hence:. . AP \:=\:\sqrt{x^2 + 30^2}
    At 1 m/sec, this takes him: . \frac{\sqrt{x^2+900}}{1} \:= \:{\color{blue}\sqrt{x^2+900}} seconds.

    He runs the distance PC \:=\:75-x\,\text{ at }\,\frac{5}{3} m/sec.
    This takes him: . \frac{75-x}{\frac{5}{3}} \:=\:{\color{blue}\frac{3}{5}(75-x)} seconds.

    He also used 2 seconds to leave the pool.

    His total time is: . \boxed{T \;=\;\sqrt{x^2+900} + \frac{3}{5}(75-x) + 2} seconds.




    b) Find \frac{dT}{dx}
    We have: . T \;=\;\left(x^2+900\right)^{\frac{1}{2}} - \frac{3}{5}x + 47

    Then: . \frac{dT}{dx} \;=\;\frac{1}{2}\left(x^2+900\right)^{-\frac{1}{2}}(2x) - \frac{3}{5} \;=\;\boxed{\frac{x}{\sqrt{x^2+900}} - \frac{3}{5}}




    c) (i) Find the value of x for which T is a minimum.
    (ii) Find the minimum time.
    (i) Solve \frac{dT}{dx} \:=\:0

    We have: . \frac{x}{\sqrt{x^2+900}} \:=\:\frac{3}{5} \quad\Rightarrow\quad 5x\:=\:3\sqrt{x^2+900}

    Square both sides: . 25x^2 \:=\:9(x^2 + 900)\quad\Rightarrow\quad 25x^2 \:=\:9x^2 + 8100

    . . 16x^2 \:=\:8100\quad\Rightarrow\quad x^2 \:=\:\frac{2025}{4}\quad\Rightarrow\quad x \:=\:\frac{45}{2} \:=\:\boxed{22.5\text{ meters}}


    (ii) Substitute into the formula in part (a).

    T \;=\;\sqrt{22.5^2 + 900} + \frac{3}{5}(75-22.5) + 2 \;\;=\;\;\boxed{71\text{ seconds}}




    d) Find the time taken if the boy runs from A to D and then D to C.
    He would run: . 30 + 75 meters at \frac{5}{3} m/sec

    This will take him: . \frac{105}{\frac{5}{3}} \:=\:\boxed{63\text{ seconds}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Very funny!

    If the boy's only concern is getting from A to C as fast as possible,
    . . he should all the way (and skip the math).

    However, if this some sport where some swimming is required,
    . . then our solution provides the shortest time.
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, afan17!

    Code:
          A             75              B
          * - - - - - - - - - - - - - - *
          |  *                          |
          |     *                       |
       30 |        *                    |
          |           *                 |
          |              *              |
          * - - - - - - - - * - - - - - *
          D        x        P   75-x   C
    He will swim the diagonal distance AP.
    This is the hypotenuse of right triangle ADP.
    . . Hence:. . AP \:=\:\sqrt{x^2 + 30^2}
    At 1 m/sec, this takes him: . \frac{\sqrt{x^2+900}}{1} \:= \:{\color{blue}\sqrt{x^2+900}} seconds.

    He runs the distance PC \:=\:75-x\,\text{ at }\,\frac{5}{3} m/sec.
    This takes him: . \frac{75-x}{\frac{5}{3}} \:=\:{\color{blue}\frac{3}{5}(75-x)} seconds.

    He also used 2 seconds to leave the pool.

    His total time is: . \boxed{T \;=\;\sqrt{x^2+900} + \frac{3}{5}(75-x) + 2} seconds.



    We have: . T \;=\;\left(x^2+900\right)^{\frac{1}{2}} - \frac{3}{5}x + 47

    Then: . \frac{dT}{dx} \;=\;\frac{1}{2}\left(x^2+900\right)^{-\frac{1}{2}}(2x) - \frac{3}{5} \;=\;\boxed{\frac{x}{\sqrt{x^2+900}} - \frac{3}{5}}



    (i) Solve \frac{dT}{dx} \:=\:0

    We have: . \frac{x}{\sqrt{x^2+900}} \:=\:\frac{3}{5} \quad\Rightarrow\quad 5x\:=\:3\sqrt{x^2+900}

    Square both sides: . 25x^2 \:=\:9(x^2 + 900)\quad\Rightarrow\quad 25x^2 \:=\:9x^2 + 8100

    . . 16x^2 \:=\:8100\quad\Rightarrow\quad x^2 \:=\:\frac{2025}{4}\quad\Rightarrow\quad x \:=\:\frac{45}{2} \:=\:\boxed{22.5\text{ meters}}


    (ii) Substitute into the formula in part (a).

    T \;=\;\sqrt{22.5^2 + 900} + \frac{3}{5}(75-22.5) + 2 \;\;=\;\;\boxed{71\text{ seconds}}



    He would run: . 30 + 75 meters at \frac{5}{3} m/sec

    This will take him: . \frac{105}{\frac{5}{3}} \:=\:\boxed{63\text{ seconds}}

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Very funny!

    If the boy's only concern is getting from A to C as fast as possible,
    . . he should all the way (and skip the math).

    However, if this some sport where some swimming is required,
    . . then our solution provides the shortest time.
    Thank you so much! Honestly, such a lifesaver and it really is appreciated. Now I can go through this with the student tomorrow
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