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

1. ## [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!

2. [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.

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

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

5. 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!!!