Finding a connection track to two railroads (circle and parabola)

In summary, the conversation discusses finding a straight line connection between two curves, represented by the equations y=(1/2)x2+7 and x2+y2=1. The conversation also includes a diagram and the derivatives of the curves, as well as a discussion about finding the correct points on each curve to create a tangent line. Ultimately, the goal is to find a line that joins the two curves and is tangent to both of them.
  • #1
dannyxyeah
11
0
There are two railroads, represented by the curves y=(1/2)x2+7, and x2+y2=1. I am supposed to find a straight line connection so that a atrain can get from one curve to the other.

I have drawn a simple diagram to help me picture things:
http://i50.tinypic.com/2saeccx.png
where (a,b) and (c,d) will be the points at which the line will connect with each curve.

I've found the derivatives to the curves to be y'=x and y'=-x/y respectively. I understand that these give me the slopes of the tangent lines for each curve, so would I be correct in reasoning that x=-x/y since the slopes should be the same?
 
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  • #2
dannyxyeah said:
There are two railroads, represented by the curves y=(1/2)x2+7, and x2+y2=1. I am supposed to find a straight line connection so that a atrain can get from one curve to the other.

I have drawn a simple diagram to help me picture things:
http://i50.tinypic.com/2saeccx.png
where (a,b) and (c,d) will be the points at which the line will connect with each curve.

I've found the derivatives to the curves to be y'=x and y=-x/y respectively.
Probably a typo, but the 2nd one should be y' = -x/y.
dannyxyeah said:
I understand that these give me the slopes of the tangent lines for each curve, so would I be correct in reasoning that x=-x/y since the slopes should be the same?
Yes.

Note that there is another path that works, from the left side of the circle to the right side of the parabola.
 
  • #3
Mark44 said:
Probably a typo, but the 2nd one should be y' = -x/y.
Yes, my mistake!

Mark44 said:
Yes.

Note that there is another path that works, from the left side of the circle to the right side of the parabola.
Right, that makes sense. Thanks! How do you reckon I should proceed from here? My mind is a total blank.
 
  • #4
Actually, I think this is a bit more complicated than setting the derivatives fo the two curves equal. You also want the line between (a, b) on the parabola and (c, d) on the circle to have the same slope as both derivatives you found.

The problem is that you could have two tangents on the two curves with the same slope, but they aren't necessarily connected, if you follow what I'm saying.
 
  • #5
That's exactly why I'm having trouble working out how to make sure that the tangent line passes through (a,b) and (c,d) :/
 
  • #6
dannyxyeah said:
That's exactly why I'm having trouble working out how to make sure that the tangent line passes through (a,b) and (c,d) :/

What's the equation of a line tangent to y = (1/2)x2 + 7 and passing through point (a, b) which is on the parabola?

What's the equation of a line tangent to x2 + y2 = 1 and passing through point (c, d) which is on the circle?
 
  • #7
SammyS said:
What's the equation of a line tangent to y = (1/2)x2 + 7 and passing through point (a, b) which is on the parabola?
y'=x
m=a
y-y1=m(x-x1)
y-b=(a)(x-a)

SammyS said:
What's the equation of a line tangent to x2 + y2 = 1 and passing through point (c, d) which is on the circle?
y'=-x/y
m=-a/b
y-y1=m(x-x1)
y-b=(-a/b)(x-a)

Is this what you mean?
 
  • #8
dannyxyeah said:
y'=x
m=a
y-y1=m(x-x1)
y-b=(a)(x-a)
That's good for the line tangent to the parabola at (a, b). But you can additionally express b in terms of a, since b = (1/2)a2 + 7 .

y'=-x/y
m=-a/b
y-y1=m(x-x1)
y-b=(-a/b)(x-a)

Is this what you mean?
For the circle, x + y2 = 1, you should use a different point, such as (c, d).

Then y-d = (-c/d)(x-c) .

Furthermore, as you already mentioned above, the slopes of the two lines must be equal, so now given a point, (c, d) on the unit circle, you can find the value of a, and plug that into the equation of the line tangent to the parabola.
 
  • #9
Furthermore, as you already mentioned above, the slopes of the two lines must be equal, so now given a point, (c, d) on the unit circle, you can find the value of a, and plug that into the equation of the line tangent to the parabola.
I don't really follow this last part; how do I know which point on the circle (or likewise, on the parabola) I am supposed to choose?
 
  • #10
dannyxyeah said:
I don't really follow this last part; how do I know which point on the circle (or likewise, on the parabola) I am supposed to choose?
What is the slope of the line y-b = (a)(x-a) ?

What is the slope of the line y-d = (-c/d)(x-c) ?

Equate those slopes.
 
  • #11
SammyS said:
What is the slope of the line y-b = (a)(x-a) ?

What is the slope of the line y-d = (-c/d)(x-c) ?

Equate those slopes.

a=(-c/d), but I've had that since the original post, no?
 
  • #12
dannyxyeah said:
a=(-c/d), but I've had that since the original post, no?
Beyond that, both lines must have the same y-intercept. Right?

There is some (maybe a lot of) algebra to do.
 
  • #13
There are actually four possible connections. You can have a and c both positive or both negative, and two possibilities where a is negative and c is positive or vice versa.
 
  • #14
I've mostly lost motivation in this problem. I don't understand the meaning of what I'm doing. :(
 
  • #15
dannyxyeah said:
I've mostly lost motivation in this problem. I don't understand the meaning of what I'm doing. :(
You're trying to find a line that:
1. Joins the parabola and the circle, and
2. Is tangent to both the parabola and the circle.
 
  • #16
dannyxyeah said:
I've mostly lost motivation in this problem. I don't understand the meaning of what I'm doing. :(
Unfortunately, in this new format for Physics Forums, there is no button to press which can provide the necessary motivation.

Solve the following for y, and try to simplify the right hand side.

[itex]\displaystyle y-d= \frac{-c}{d}(x-c)[/itex]

See if that leads you anywhere. Whether it does or not, please give your result.
 

1. How do I determine the coordinates of the connection track?

The coordinates of the connection track can be determined by finding the intersection point of the circle and the parabola. This can be done by setting the equations of the two curves equal to each other and solving for the x and y values.

2. Can there be more than one possible connection track?

Yes, there can be multiple connection tracks between a circle and a parabola. This will depend on the specific equations of the circle and parabola and the location of their intersection points.

3. Is there a specific method or formula for finding the connection track?

Yes, there are various methods and formulas that can be used to find the connection track. One common approach is to use the distance formula to find the shortest distance between the circle and the parabola, which will give the coordinates of the connection track.

4. What if the circle and parabola do not intersect?

If the circle and parabola do not intersect, then there is no connection track between them. In this case, you may need to adjust the equations of the curves or consider other methods for connecting them.

5. Can I use a computer program to find the connection track?

Yes, there are many mathematical software programs that can help you find the connection track between a circle and a parabola. These programs can save time and ensure accuracy in your calculations.

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