# Numerical methods

• Jan 12th 2008, 08:43 AM
geton
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 $x = \frac {1}{3} e^{-x^2}$

How can I rearrange this?

2.
a) Sketch the graph of y = e^-x – 1. On the same axes sketch the graph of y = ½ (x - 1), for x ≥ 1, and y = -½ (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.

• Jan 12th 2008, 10:16 AM
earboth
Quote:

Originally Posted by geton
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 $x = \frac {1}{3} e^{-x^2}$

How can I rearrange this?

...

Hello,

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

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

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

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

5. Rearrange this equation:

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

6. All conditions are satisfied if $x=\frac13 \cdot e^{-x^2}$
• Jan 12th 2008, 01:08 PM
Soroban
Hello, geton!

Quote:

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

Show the coordinates of the points where the graph meets the axes.

Code:

                |         *      |             *  |           *    *|½                *           *    | *          *             *  |    *  1  *       ----------*-------*--------------               0|  *                 |        *                 |                *         - - - -1+ - - - - - - - - - -                 |

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

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

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

Quote:

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

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

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

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

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

Quote:

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

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

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

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

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

• Jan 12th 2008, 06:07 PM
geton
Thank you all.