So I had to graph a function y=(-1/r)x+r with about 100 values for r. The result is the graph below. Now I need to find the exact equation for the turned parabola the first equation describes. Since equation y=(-1/r)x+r is the tangent to this parabola I tried to find it through calculating antiderivative of y=(-1/r)x+r, but my result y=(r^2/2)-xlog(r), when graphed is not the desired turned parabola. So could you help me to find this equation? Thanks in advance.
Interchange the usual roles of x & y:
The equation for the parabola is:The point lies on the parabola. The slope of the tangent line at this point is:
The x-intercept, b, for this line may be obtained from:
The equation of the tangent line is:
The family of lines that were graphed is given by:
This should give you a. (Also, r in terms of y_0, if you want that.)
There's a typo in my next to last line of my previous post:
The expression should be:Originally Posted by SammyS
(There actually was an "En dash" in place of the negative sign. That's a character that LaTeX doesn't understand.)
I haven't come up with an easy way.
Find the point of intersection (x_1, y_1) for two lines, one with r=R, the other with r=R+ε . Take the limit as ε → 0. That should be the point at which the line with r=R is tangent to the unknown curve.That's a start. The result should be consistent with the previous analysis which assumed a parabolic shape.
To be thorough, do the same with r=R-ε.
BTW: I found another mistake in my post #5 above.
The equation for the tangent line should be: because the x-intercept is