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Math Help - 2 Part Question: I've done the first bit but I don't understand the second bit

  1. #1
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    Post 2 Part Question: I've done the first bit but I don't understand the second bit

    a) Find the equation of the normal to y = x^2 / 3 at the point where x = 1

    b) This normal cuts the coordinate axis at P and Q. Given that P and Q are adjacent vertices of a rhombus, and that the diagonals of this rhombus intersect at the origin, determine the area of this rhombus
    [I don't understand this part and I'm not too sure on what to do can someone please help]

    My working out for part a [Please tell me if i made a mistake]

    y = x^2 / 3

    dy/dx = (x^2 x 0) - (3 x 2x)
    _________________

    9

    = x^2 - 6x
    ________

    9

    when x = 1

    m(tangent) = -5/9

    So m(normal) = 9/5

    Therefore Equation of normal = y = 9/5x - 22/15

    [Sorry I missed a few lines and skipped straight to the final answer]
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  2. #2
    Member Henderson's Avatar
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    Yeah, that's not right. Sorry.

    Looks like you're trying to use the quotient rule, which is making things difficult. Three problems with what you've done:
    • The top should be (3 \cdot 2x) - (x^2 \cdot 0). You've got it backwards.
    • x^2 \cdot 0 = 0, not x^2.
    • You've forgotten to square the bottom.
    An easier way to do this might be to look at the 'dividing by three' as a coefficient:
    \frac{1}{3}x^2
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  3. #3
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    Hello, sweetG!

    a) Find the equation of the normal to y \:= \:\frac{1}{3}x^2 where x = 1

    When x = 1,\;y = \frac{1}{3}

    We have: . y' \:=\:\frac{2}{3}x

    When x = 1, the slope of the tangent is: . m \:=\:\frac{2}{3}
    . . Hence, the slope of the normal is: -\frac{3}{2}

    The equation of the normal is: . y - \frac{1}{3}\:=\:-\frac{3}{2}(x-1) \quad\Rightarrow\quad \boxed{y \:=\:-\frac{3}{2}x + \frac{11}{6}}




    b) This normal cuts the coordinate axis at P and Q.
    Given that P and Q are adjacent vertices of a rhombus,
    and that the diagonals of this rhombus intersect at the origin,
    determine the area of this rhombus.

    The intercept of the normal are: . P\left(\frac{11}{9},\:0\right)\,\text{ and }\,Q\left(0,\:\frac{11}{6}\right)

    If the diagonals intersect at the origin, the rhombus looks like this:
    Code:
                          |
                        Q o (0,11/6)
                       *  |  *
                    *     |     *
              R  *        |        * P
        - - - o - - - - - + - - - - - o - - -
       (-11/9,0) *        |        * (11/9,0)
                    *     |     * 
                       *  |  *
                        S o (0,-11/6)
                          |

    The area of a rhomus is one-half the product of its diagonals.

    The horizontal diagonal has length: . 2 \times \frac{11}{9} \:=\:\frac{22}{9}

    The vertical diagonal has length: . 2 \times \frac{11}{6} \:=\:\frac{11}{3}


    The area is: . \frac{1}{2} \times \frac{22}{9} \times \frac{11}{3} \;=\;\boxed{\frac{121}{27}\text{ units}^2}



    *
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  4. #4
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    Thanks so much guys !
    =)
    that really helped and wow i made a lot of silly mistakes there
    and I just realized how easy this question really was I just wasn't thinking hard enough Thanks once again !
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