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Thread: Numerical methods

  1. #1
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    Numerical methods

    1.
    The curve with equation y = ln (3x) crosses the x-axis at the point P (1/3, 0).

    The normal to the curve at the point Q, with x-coordinate q, passes through the origin.

    Show that the equation can be rearranged in the form $\displaystyle x = \frac {1}{3} e^{-x^2}$


    How can I rearrange this?

    2.
    a) Sketch the graph of y = e^-x – 1. On the same axes sketch the graph of y = (x - 1), for x ≥ 1, and y = - (x - 1), for x < 1. Show the coordinates of the points where the graph meets the axes.

    The x-coordinate of the point of intersection of the graphs of α.
    b) Show that x = α is a root of the equation x + 2e^-x – 3 = 0.
    c) Show that -1<α<0.


    I’ve done part a. But how can I do part b & c. Please help me.
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  2. #2
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    Quote Originally Posted by geton View Post
    1.
    The curve with equation y = ln (3x) crosses the x-axis at the point P (1/3, 0).

    The normal to the curve at the point Q, with x-coordinate q, passes through the origin.

    Show that the equation can be rearranged in the form $\displaystyle x = \frac {1}{3} e^{-x^2}$


    How can I rearrange this?

    ...
    Hello,

    1. Calculate the 1st derivative of the function: $\displaystyle y'=\frac1{3x} \cdot 3=\frac1x$

    2. The slope of the tangent in Q(q, ln(q)) is $\displaystyle m_t=\frac1q$. Therefore the perpendicular direction has the slope: $\displaystyle m_{perp}=-\frac1{m_t}=-q$

    3. The point-slope-form of the normal straight line is:
    $\displaystyle y-\ln(3q)=-q\cdot(x-q)~\implies~y-\ln(3q)=-p \cdot x + q^2$

    4. This line passes through O(0, 0) that means $\displaystyle x = 0~\wedge~y=0$ . Plug in these values into the equation of the normal line:
    $\displaystyle -\ln(3q)=q^2~\implies~\frac1{3q}=e^{q^2}$

    5. Rearrange this equation:

    $\displaystyle q=\frac1{3e^{q^2}}~\implies~q=\frac13 \cdot e^{-q^2}$

    6. All conditions are satisfied if $\displaystyle x=\frac13 \cdot e^{-x^2}$
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  3. #3
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    Hello, geton!

    2. a) Sketch the graph of: $\displaystyle y \:=\:e^{\text{-}x} - 1$
    On the same axes, sketch the graphs of: .$\displaystyle \begin{array}{cc}y \:= \:\frac{1}{2}(x-1) & \text{for }x \geq 1 \\ y \:= \text{-}\frac{1}{2}(x - 1)&\text{for }x < 1\end{array}$

    Show the coordinates of the points where the graph meets the axes.
    Code:
                    |
             *      |
                *   |
              *    *|                *
               *    | *           *
                 *  |    *  1  *
          ----------*-------*-------------- 
                   0|   *
                    |         *
                    |                 *
            - - - -1+ - - - - - - - - - -
                    |

    We're expected to be familiar with the graph of $\displaystyle (a)\;y \:=\:e^x$
    The graph of $\displaystyle (b)\;y \:=\:e^{\text{-}x}$ is a reflection of (a) over the y-axis.
    The graph of $\displaystyle y\:=\:e^{\text{-}x} - 1$ is the graph of (b) lowered one unit.

    The two lines intersect at (1,0) and exist above the x-axis.

    The intercepts are: .$\displaystyle \0,\,0),\;(1,\,0),\;\left(0,\,\frac{1}{2}\right)$



    The x-coordinate of the point of intersection of the graphs is $\displaystyle \alpha.$

    b) Show that $\displaystyle x = \alpha$ is a root of the equation $\displaystyle x + 2e^{\text{-}x} - 3 \:= \:0$

    The only intersection is that of: .$\displaystyle y \:=\:e^{\text{-}x} - 1\text{ and }y \:=\:-\frac{1}{2}(x-1)$

    We have: .$\displaystyle e^{\text{-}x} - 1 \:=\:-\frac{1}{2}(x-1)\quad\Rightarrow\quad 2e^{\text{-}x} - 2 \:=\:-x + 1$

    Therefore, $\displaystyle \alpha$ is a root of: .$\displaystyle x + 2e^{\text{-}x} - 3 \:=\:0$




    c) Show that: .$\displaystyle -1 \,<\,\alpha \,<\,0$

    Since both the exponential function and the line are continuous,
    . . the most elementary method is to test the endpoints.

    At $\displaystyle x = \text{-}1:\;\begin{array}{ccccc}y & = & e^{\text{-}(\text{-}1)} - 1 & \approx & 1.7 \\ y & = & \text{-}\frac{1}{2}(\text{-}1-1) & = & 1\end{array}$ . . . the exponential is above the line

    At $\displaystyle x = 0:\;\begin{array}{ccccc}y & = & e^0-1 &=&0 \\ y&=&\text{-}\frac{1}{2}(0-1) & = & \frac{1}{2}\end{array}$ . . . the exponential is below the line

    Therefore, they intersect somewhere on the interval $\displaystyle (\text{-}1,\,0).$


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  4. #4
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    Thank you all.
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