# Solving Antiderivative of y=(-1/r)x+r with Graph

• Jan 22nd 2011, 12:04 PM
bsev
Solving Antiderivative of y=(-1/r)x+r with Graph
Hello Guys,
So I had to graph a function y=(-1/r)x+r with about 100 values for r. The result is the graph below. Attachment 20556 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.
• Jan 22nd 2011, 12:11 PM
dwsmith
Quote:

Originally Posted by bsev
Hello Guys,
So I had to graph a function y=(-1/r)x+r with about 100 values for r. The result is the graph below. Attachment 20556 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.

Should your derivative be respect to x, y, or r? Is r a constant or variable? If you want a parabola symmetric to the x-axis, it needs to be in y^2.
• Jan 22nd 2011, 01:32 PM
bsev
The derevative should be in respect to x. Where r is a variable.
• Jan 22nd 2011, 01:34 PM
dwsmith
Quote:

Originally Posted by bsev
The derevative should be in respect to x. Where r is a variable.

If it is with respect to x, you parabola will open up or down not to the left.
• Jan 22nd 2011, 03:36 PM
SammyS
Quote:

Originally Posted by bsev
Hello Guys,
So I had to graph a function y=(-1/r)x+r with about 100 values for r. The result is the graph below. http://www.mathhelpforum.com/math-he...raph-1.low.jpg
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 anti-derivative 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.

Knowing that it's a parabola helps. From your graph, the parabola opens to the left, and as dwsmith points out, should have y². It looks as if the vertex of the parabola is at the origin. Indeed, solving y=(-1/r)x+r for x gives x=–r·y+r². x→0 as r→0. Therefore, the equation of the parabola may be written as x = –a·y².

Interchange the usual roles of x & y:

The equation for the parabola is: $x = -a\,y^2 \,$
$\displaystyle \to\quad{{dx}\over{dy}}=-2a\,y\,.$
The point $(-a\,{y_0}^2,\,y_0)$ lies on the parabola. The slope of the tangent line at this point is: $m=-2a\,y_0\,.$

The x-intercept, b, for this line may be obtained from: $x_0=m\,y_0+b$
$\to b=-m\,y_0+x_0=(2a\,y_0)(y_0)-a\,{y_0}^2=a\,{y_0}^2$

The equation of the tangent line is: $x=(-2a\,y_0)y+\,{y_0}^2$

The family of lines that were graphed is given by: $x=–r\,y+r^2\,.$

This should give you a. (Also, r in terms of y_0, if you want that.)
• Jan 23rd 2011, 06:12 AM
bsev
Thanks SammyS! So eventually I got $a={{-1\,}\over{2}}$ and my parabola equation is then $x={{-1\,}\over{2}}y^2 \,$.

I'm just wondering, what if this shape is only parabola-like, but in fact is not. Is there a way to see that?
• Jan 24th 2011, 09:18 PM
SammyS
Quote:

Originally Posted by bsev
Thanks SammyS! So eventually I got $a={{-1\,}\over{2}}$ and my parabola equation is then $x={{-1\,}\over{2}}y^2 \,$.

I'm just wondering, what if this shape is only parabola-like, but in fact is not. Is there a way to see that?

The original post stated that these were tangent lines for a turned parabola.

There's a typo in my next to last line of my previous post:
Quote:

Originally Posted by SammyS
The family of lines that were graphed is given by: $x=–r\,y+r^2\,.$

The expression should be: $x=-r\,y+r^2\,.$

(There actually was an "En dash" in place of the negative sign. That's a character that LaTeX doesn't understand.)
• Jan 30th 2011, 06:53 AM
bsev
Thanks SammyS!
I know I wrote it is a parabola, but I did so only because it looked like one.
Still, do you think there is a way to examine whether or not it is a parabola?
• Jan 30th 2011, 11:04 AM
SammyS
Quote:

Originally Posted by bsev
Thanks SammyS!
I know I wrote it is a parabola, but I did so only because it looked like one.
Still, do you think there is a way to examine whether or not it is a parabola?

I'm convinced there is a way to show that this is a parabola.

I haven't come up with an easy way.

One idea:
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.
To be thorough, do the same with r=R-ε.
That's a start. The result should be consistent with the previous analysis which assumed a parabolic shape.

BTW: I found another mistake in my post #5 above.

The equation for the tangent line should be: $x=(-2a\,y_0)y+\,a{y_0}^2\,,$ because the x-intercept is $a{y_0}^2\,.$