Find the curve

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odyssey

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Find the curve whose curvature is 2, passes through the point \(\displaystyle (1,0)\) and whose tangent vector at \(\displaystyle (1,0)\) is \(\displaystyle [1/2 , \sqrt{3}/2 ].\)

I know I can apply the fundamental theorem of plane curves here. How will I be able to do that?

Also, I know there is a basic way other than using the fundamental theorem of plane curves since it gave us the curvature and tangent vector. Since the curvature is a constant I can conclude it is a circle thus the equation of a circle with radius r is \(\displaystyle (x-a)^2 + (y-b)^2 = r^2.\) But how is knowing the curvature, the value and tangent vector at (1,0) sufficient to construct the curve and how will I apply it?
 
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odyssey said:
Find the curve whose curvature is 2, passes through the point \(\displaystyle (1,0)\) and whose tangent vector at \(\displaystyle (1,0)\) is \(\displaystyle [1/2 , \sqrt{3}/2 ].\)

I know I can apply the fundamental theorem of plane curves here. How will I be able to do that?

Also, I know there is a basic way other than using the fundamental theorem of plane curves since it gave us the curvature and tangent vector. Since the curvature is a constant I can conclude it is a circle thus the equation of a circle with radius r is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2.\) But how is knowing the curvature, the value and tangent vector at (1,0) sufficient to construct the curve and how will I apply it?
Click to expand...
.

r = 1/k = 1/2
 
So I know the radius = 1/2 and the equation I use at (1,0) is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2 \implies (1-h)^2 + (0-k)^2 = (1/2)^2\) but where does my tangent vector come into use?
 
odyssey said:
So I know the radius = 1/2 and the equation I use at (1,0) is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2 \implies (1-h)^2 + (0-k)^2 = (1/2)^2\) but where does my tangent vector come into use?
Click to expand...

What is the slope of the tangent line?
 
I am not sure? But the tangent vector at the point (1,0) is \(\displaystyle [1/2, \sqrt{3}/2]\)
 
I was confused about his question, but the slope is the tangent vector which is given \(\displaystyle [\sqrt3, 2]\). But then how can I continue?

I get:

(x-a)^2 + (y-b)^2 = r^2, I get
(1-a)^2 + (0-b)^2 = (1/2)^2
(1-a)^2 + (b)^2 = 1/4
 
odyssey said:
I was confused about his question, but the slope is the tangent vector which is given \(\displaystyle [\sqrt3, 2]\). But then how can I continue?

I get:

(x-a)^2 + (y-b)^2 = r^2, I get
(1-a)^2 + (0-b)^2 = (1/2)^2
(1-a)^2 + (b)^2 = 1/4
Click to expand...

daon told you:

The tangent vector gives the direction (and also slope). Rise over run
Click to expand...

also:

slope of the tangent line (of a 2-D curve) is \(\displaystyle \dfrac{dy}{dx}{/tex] of the curve at that point.

Now equate those two to solve for 'a' and 'b' and you are done\)
 
But what do I take the derivative with? Do I take the derivative of \(\displaystyle (1-a)^2 + (b)^2 = 1/4\)?
 
Your objective is to find a and b such that the curve defined by the equation

\(\displaystyle
[*] (x-a)^2+(y-b)^2=\frac{1}{4}\) has the following properties:

1. The point \(\displaystyle (1,0)\) lies on the curve
2. \(\displaystyle \langle 1/2,\sqrt{3}/2\rangle\) is the tangent vector to the curve at (1,0)

The first bit of information tells you that

\(\displaystyle (1-a)^2+b^2=\frac{1}{4}\)

Implicitly differentiating \(\displaystyle
[*]\) and solving for \(\displaystyle \frac{dy}{dx}\) gives

\(\displaystyle \dfrac{dy}{dx}= \dfrac{a-x}{y-b}\)

The second piece of information tells you that \(\displaystyle \dfrac{dy}{dx} = (\sqrt{3}/2)/(1/2) = \sqrt{3}\) when \(\displaystyle x=1, y=0\).

Now you can obtain a second equation with a's and b's and solve for a and b!
 
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