Solving the Differential Equation for Curve Passing Through (1,1) with Equidistant Normal

In summary: If you are trying to find the equation of a line that passes through the point (1,1), you would use the equation y=mx+c.
  • #1
I need help. I am not getting the solution to this:

A curve passes through the point (1,1) and has the property that the normal at any point on the curve is equidistant from the x-axis and the origin. Form the differential equation for the curve, find the curve and state which quadrant(s) the curve lies in.

Here is what I did,
The slope of the normal at any point would be -1/(dy/dx) and the equation of the normal at the point would be y=mx+c where m= -1/(dy/dx). If this is the normal at (1,1), then 1=m+c, => c=1-m

Therefore, the equation of the line comes out to be y=mx+1-m. Rearranging this, y-mx-1+m=0. The distance of this line from the origin is :
|-1+m| / ((1+m*m)^1/2) which is equal to 1 (distance from the x-axis as this is the normal at (1,1) ).

After this, you substitute m= -1/(dy/dx) and you get your differential equation... only you dont... what am I doing wrong?
 
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  • #2
I find the statement "the normal at any point on the curve is equidistant from the x-axis and the origin" confusing. The normal at that point is a line. I know what is meant by the distance from a line to a point but what is the "distance" between two lines in a plane? At the end, you appear to be using the distance from the point (1,1) to the x-axis as the "distance between the normal and the x-axis". That certainly isn't the way I would interpret the phrase.
 
  • #3
If that isn't it, then how would you look at the problem? What could a possible solution be?
 
  • #4
[tex] y - mx - 1 + m = 0 \Rightarrow y -(-\frac{1}{\frac{dy}{dx}})x-1-\frac{1}{\frac{dy}{dx}} = 0 [/tex]

[tex] y +\frac{1}{\frac{dy}{dx}}(x-1) - 1 = 0 [/tex]

[tex] \frac{1}{\frac{dy}{dx}}(x-1) = 1- y [/tex][tex] \frac{1}{\frac{dy}{dx}} = \frac{1-y}{x-1} [/tex]

[tex] \frac{dy}{dx} = \frac{x-1}{1-y} [/tex]
 
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  • #5
Where do you use the condition that the perpendicular distance of the normal at a point is equal to its y co-ordinate? The answer given is d(x*x-y*y)=(2xy)dy. Is there something I am missing?
 
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1. What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It is used to model the relationship between a quantity and its rate of change.

2. How do you solve a differential equation?

There are various methods for solving a differential equation, depending on its type and complexity. Some common methods include separation of variables, using integrating factors, and using power series. In this specific case, we can use the method of equating coefficients to find a solution.

3. What does it mean for a curve to pass through (1,1) with equidistant normal?

When a curve passes through a point (1,1) with equidistant normal, it means that the distance between the point (1,1) and any point on the curve is equal to the distance between the normal line at that point and the point (1,1). In other words, the curve is perpendicular to the normal line at all points.

4. Why is it important to solve this type of differential equation?

Differential equations are used in various fields of science and engineering to model real-world phenomena. Solving this type of differential equation allows us to find a specific curve that satisfies certain conditions, which can help us understand and predict the behavior of a system in a given scenario.

5. Can this type of differential equation have multiple solutions?

Yes, this type of differential equation can have multiple solutions. In fact, for this specific problem, we can find an infinite number of curves that pass through (1,1) with equidistant normal. This is because the equation only provides one condition and does not uniquely determine a single curve.

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