[SOLVED] Differential Equation Geometry/Word Problem Help!

**horan** · September 22nd 2007, 08:43 PM

Hi everyone, I'm currently having trouble mentally grasping this question I have:

Find the equation of the curve that goes through (1,3) for which any tangent and the line from the origin to the point of contact make with the ordinate a triangle having area A.

I have been trying several approaches to this, but to no avail. I assume it can be set up as a first-order differential equation of some sort. I've defined the curve as a, the tangent line as b, and the line from the point of tangency to the origin as c. I also defined a random point along the curve (alpha, beta). Thus,

$b(x)=beta+a'(alpha)*(x-alpha)$
$c(x)=beta/alpha * x$

EDIT: The area of the triangle, A, could also be found with one of the following:
$A_1=int(b(x))-int(c(x))$
$A_2=(1/2)(b)(h)=(1/2)(b(0))(alpha)=(1/2)(beta+a'(alpha)*(x-alpha))$

I have verified that, in fact, $A_1=A_2$ .

Any insight would be greatly appreciated!

**Opalg** · September 23rd 2007, 02:38 AM

[According to Wikipedia, the ordinate is another name for the y-axis. I never came across that terminology before.]

I find that a, b, c, alpha, beta, ... notation hard to follow, so I'll use other letters. Suppose the curve is given by y=f(x), and let $(x_0,y_0)$ be a point on the curve. The tangent at this point has equation $y-y_0=f'(x_0)(x-x_0)$ . It meets the y-axis at the point $(0,y_1)$ , where y_1 is given by $y_1-y_0 = -f'(x_0)x_0$ .

If we take the triangle to have its base along the y-axis then its base is $\pm y_1$ ( the +/- is needed because we don't know whether y_1 is positive or negative) and its height is x_0. So its area is given by $\textstyle A=\pm\frac12x_0y_1 = \pm\frac12x_0(y_0-x_0f'(x_0))$ .

At this stage we need to change the notation. Up until now, $(x_0,y_0)$ has been fixed. But in order to get a differential equation for f(x), we need to see what happens as $(x_0,y_0)$ varies. The best thing to do is to drop the zero subscripts and to replace f'(x_0) by dy/dx (or just y'). Then the previous equation becomes $x(y-xy') = \pm2A$ . At last, we have a differential equation!

You'll need to use an integrating factor to solve the equation, and then to use the initial condition that y=3 when x=1. There are two solutions, corresponding to the +/- sign in the equation.

**Soroban** · September 23rd 2007, 08:02 AM

Hello, bherila!

Find the equation of the curve that goes through (1,3)
for which any tangent and the line from the origin to the point of contact
make with the ordinate a triangle having area $A$ .

Code:

        |
        *                 * y = f(x)
        |    *
        |        *       *
        |             *  P(p,q)
        |           o /
        |       o   /
        |   o     /
      B o       /
        |     /
        |   /
        | /
    - - * - - - - - - - - - - -
        |O
        |

We have a function: $y \,=\,f(x)$

Let $P(p,q)$ be a point on $y \,=\,f(x).$

$OP$ is the segment from the origin to point $P.$

$BP$ is the tangent at $P$ .
. . The slope of this tangent is $y'$ and it contains $P(p,q).$
. . Its equation is: . $y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q$
. . Its y-intercept is: . $B(0,\,q - py')$

The base of $\Delta OBP$ is: . $OB \:=\:q - py'$
The height of the triangle is: . $p$

The area of $\Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A$

\. . which simplifies to: . $p^2y' - pq \:=\:-2A$

Divide by $p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}$

Hence, we have: . $\frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}$

Integrating factor: . $I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}$

So we have: . $\frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}$

Then: . $\frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}$

Integrate: . $\frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C$

Multiply by $x\!:\;\;y \;=\;-\frac{A}{x} + Cx$

Since $(1,3)$ is on the curve,
. . $3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3$

Therefore: . $\boxed{y \;=\;-\frac{A}{x} + (A + 3)x}$

**Opalg** · September 23rd 2007, 12:22 PM

The other solution is $\boxed{y =\frac{A}{x} + (3-A)x}$ .

**Jhevon** · September 23rd 2007, 02:43 PM

Originally Posted by Soroban

Hello, bherila!

Code:

        |
        *                 * y = f(x)
        |    *
        |        *       *
        |             *  P(p,q)
        |           o /
        |       o   /
        |   o     /
      B o       /
        |     /
        |   /
        | /
    - - * - - - - - - - - - - -
        |O
        |

We have a function: $y \,=\,f(x)$

Let $P(p,q)$ be a point on $y \,=\,f(x).$

$OP$ is the segment from the origin to point $P.$

$BP$ is the tangent at $P$ .
. . The slope of this tangent is $y'$ and it contains $P(p,q).$
. . Its equation is: . $y - q \:=\:y'(x - p)\quad\Rightarrow\quad y \:=\:y'x - y'p + q$
. . Its y-intercept is: . $B(0,\,q - py')$

The base of $\Delta OBP$ is: . $OB \:=\:q - py'$
The height of the triangle is: . $p$

The area of $\Delta OBP\!:\;\;\frac{1}{2}(q - py')p \;=\;A$

\. . which simplifies to: . $p^2y' - pq \:=\:-2A$

Divide by $p^2\!:\;\;y' - \frac{1}{p}\,q \:=\:-\frac{2A}{p^2}$

Hence, we have: . $\frac{dy}{dx} - \frac{1}{x}\,y \;=\;-\frac{2A}{x^2}$

Integrating factor: . $I \:=\:e^{\int\left(-\frac{1}{x}\right)dx} \:=\:e^{-\ln x} \:=\:e^{\ln\frac{1}{x}} \:=\:\frac{1}{x}$

So we have: . $\frac{1}{x}\,\frac{dy}{dx} - \frac{1}{x^2}\,y \;=\;-\frac{2A}{x^3}$

Then: . $\frac{d}{dx}\left(\frac{1}{x}\!\cdot\!y\right) \;=\;-2Ax^{-3}$

Integrate: . $\frac{1}{x}\!\cdot\!y \;=\;-Ax^{-2} + C$

Multiply by $x\!:\;\;y \;=\;-\frac{A}{x} + Cx$

Since $(1,3)$ is on the curve,
. . $3 \;=\;-\frac{A}{1} + C\quad\Rightarrow\quad C \:=\:A + 3$

Therefore: . $\boxed{y \;=\;-\frac{A}{x} + (A + 3)x}$

very nice. that was a lot easier than i thought it would be. Good stuff as always, Soroban!!!

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