# [SOLVED] Differential Equation Geometry/Word Problem Help!

• Sep 22nd 2007, 08:43 PM
horan
[SOLVED] Differential Equation Geometry/Word Problem Help!
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,

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

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

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

Any insight would be greatly appreciated!
• Sep 23rd 2007, 02:38 AM
Opalg
[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 $\displaystyle (x_0,y_0)$ be a point on the curve. The tangent at this point has equation $\displaystyle y-y_0=f'(x_0)(x-x_0)$. It meets the y-axis at the point $\displaystyle (0,y_1)$, where y_1 is given by $\displaystyle y_1-y_0 = -f'(x_0)x_0$.

If we take the triangle to have its base along the y-axis then its base is $\displaystyle \pm y_1$ ( 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 $\displaystyle \textstyle A=\pm\frac12x_0y_1 = \pm\frac12x_0(y_0-x_0f'(x_0))$.

At this stage we need to change the notation. Up until now, $\displaystyle (x_0,y_0)$ has been fixed. But in order to get a differential equation for f(x), we need to see what happens as $\displaystyle (x_0,y_0)$ 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 $\displaystyle x(y-xy') = \pm2A$. 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.
• Sep 23rd 2007, 08:02 AM
Soroban
Hello, bherila!

Quote:

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 $\displaystyle A$.

Code:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

• Sep 23rd 2007, 12:22 PM
Opalg
The other solution is $\displaystyle \boxed{y =\frac{A}{x} + (3-A)x}$.
• Sep 23rd 2007, 02:43 PM
Jhevon
Quote:

Originally Posted by Soroban
Hello, bherila!

Code:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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