Equation of an Oscillating Circle

[Tricky557]

Homework Statement

Find an equation of the oscillating circle to y=ln(x) at the point (1,0)

Homework Equations

p will = the 2nd derivitive of y
u will = the 1st derivitive of y

i will = the 2nd derivitive of x
o will= the 1st derivitive of x

(po - ui)/(|V|^3) = k(curvature)

1/k = r(radius)

x=t
y=ln(t)

V=u ---> I think?

The Attempt at a Solution

o = 1
i = 0

u= 1/t
p= (-1/(2t^2))

I plugged those values into the curvature equation, (po - ui)/(|V|^3) = k

(1/t)/((1+(1/t^2))^(1/2))


The value I got for k was 1. So the radius also = 1. One of the issues I have with this problem is that I'm not sure I plugged in the correct velocity. I know that velocity = the 1st derivitive of the vector(so would it be u?).


So now with the radius, I need to find the tangent unit vector(right?).

T= V/|V|

And this is the point where I get stuck again. I'm really not sure if my value of the velocity is correct or not.

Once I get the Tangent unit vector, I think I'm supposed to add that vector to the point that I"m given, in this case (1,0) in order to find the center of the circle. And with the center, I can get the equation of the circle.




My apologies on the bad notation as far as the equations goes, I don't really know how to type out things like this.

 

[mighty2000]

I appreciate your attempt at solving this problem. However, there are a few things that need to be clarified in order to find the equation of the oscillating circle. Firstly, the curvature equation you have used is not applicable in this case as it is used for finding the curvature of a curve at a point, not for finding the equation of an oscillating circle.

To find the equation of an oscillating circle, we need to consider the general equation of a circle, which is (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center of the circle and r is the radius. In this case, since we are given the point (1,0), we can consider (1,0) as the center of the circle.

To find the radius, we can use the formula you have mentioned, 1/k = r. However, we need to find the value of k at the point (1,0). To do this, we can use the formula for curvature, k = |V|/r, where V is the velocity vector at the point (1,0). In this case, V = (1,1/t) since u = 1/t, as you have correctly calculated. Therefore, k = sqrt(1 + 1/t^2)/r. Plugging this into the formula 1/k = r, we get r = sqrt(1 + 1/t^2).

Now, to find the equation of the oscillating circle, we can substitute the values of a, b and r into the general equation of a circle, (x-a)^2 + (y-b)^2 = r^2. This gives us the equation (x-1)^2 + (y-0)^2 = (sqrt(1 + 1/t^2))^2, which simplifies to (x-1)^2 + y^2 = 1 + 1/t^2.

I hope this helps clarify the steps needed to find the equation of an oscillating circle. Please let me know if you have any further questions.