Analyzing the Setup
Imagine you are standing on a coordinate plane, looking at a collection of circles. They are all the same size—a radius of 5—but they are sliding horizontally, their centers locked onto the line y=2.
In mathematics, a family is defined by a parameter that allows the curve to move or change shape. Here, that parameter is the x-coordinate of the center, which we call h.
The equation of any circle in this family is:
(x−h)2+(y−2)2=25
Our goal is to find the differential equation that governs this entire family. We want an equation that describes the slope y′ at any point (x,y) on any of these circles, without needing to know the specific value of h.
The Calculus of Elimination
To find the differential equation, we must exorcise the 'ghost' parameter h. Since there is only one arbitrary constant, we differentiate the equation exactly once with respect to x.
Applying the power rule and the chain rule:
dxd[(x−h)2]+dxd[(y−2)2]=dxd[25]
This yields:
2(x−h)+2(y−2)y′=0
Simplifying this by dividing by
2, we get:
(x−h)+(y−2)y′=0
This is a beautiful, simple relationship. It tells us that the horizontal distance from the center (x−h) is directly related to the slope y′ and the vertical distance from the line y=2.
The Final Substitution
Now, we perform the final algebraic dance. We isolate
(x−h) to get:
(x−h)=−(y−2)y′
We then take this expression and plug it back into our original circle equation. Substituting this into
(x−h)2+(y−2)2=25, we get:
[−(y−2)y′]2+(y−2)2=25
Expanding the squared term, the negative sign vanishes, leaving us with:
(y−2)2(y′)2+(y−2)2=25
Rearranging to match the standard form, we arrive at the final differential equation:
(y−2)2(y′)2=25−(y−2)2
This is the differential equation of our family of circles. It is elegant, precise, and captures the essence of the entire family in one stroke. You have successfully navigated the transition from geometry to calculus!