osculating circle

[logistic_guy]

here is the question

Find the osculating circle for the parabola \(\displaystyle x = y^2\) at \(\displaystyle t = 0\).

i need to parametrize the curve of this parabola

\(\displaystyle y = -\sqrt{x}\) and \(\displaystyle y = \sqrt{x}\)

can i say \(\displaystyle \bold{r}_1(t) = (t, -\sqrt{t})\) and \(\displaystyle \bold{r}_2(t) = (t, \sqrt{t})\)?

 

[BigBeachBanana]

here is the question

Find the osculating circle for the parabola \(\displaystyle x = y^2\) at \(\displaystyle t = 0\).

i need to parametrize the curve of this parabola

\(\displaystyle y = -\sqrt{x}\) and \(\displaystyle y = \sqrt{x}\)

can i say \(\displaystyle \bold{r}_1(t) = (t, -\sqrt{t})\) and \(\displaystyle \bold{r}_2(t) = (t, \sqrt{t})\)?

It's simplier to do [imath]x(t) = t^2 \text{ and } y(t) = t[/imath].

 

[logistic_guy]

It's simplier to do [imath]x(t) = t^2 \text{ and } y(t) = t[/imath].

thank

the curve is \(\displaystyle \bold{r}(t) = (t^2, t)\)

i'm still don't understand when to choose \(\displaystyle x\) or \(\displaystyle y\) as \(\displaystyle t\). is there a rule how to decide the parameterization?

 

[BigBeachBanana]

thank

the curve is \(\displaystyle \bold{r}(t) = (t^2, t)\)

i'm still don't understand when to choose \(\displaystyle x\) or \(\displaystyle y\) as \(\displaystyle t\). is there a rule how to decide the parameterization?

There's no strict rule as long as it follows the curve path. Often, you can pick a variable as t, i.e. x=t or y=t. In this case, it's simpler to choose y=t. What you have is not wrong, but a little more work.

 

[Dr.Peterson]

i'm still don't understand when to choose \(\displaystyle x\) or \(\displaystyle y\) as \(\displaystyle t\). is there a rule how to decide the parameterization?

It's like all problem-solving: You look for something to try, try it, and try something else if it doesn't work (or just gets too complicated). With experience, you get to recognize ahead of time what will likely be harder and not worth trying, so you'll find yourself usually choosing the easiest thing first.

In this case, having to split into two parts suggests trying something else first.

 

[mario99]

i'm still don't understand when to choose \(\displaystyle x\) or \(\displaystyle y\) as \(\displaystyle t\). is there a rule how to decide the parameterization?

I agree with BigBeachBanana that parametrization is a matter of taste. And I also I agree with professor Dave that with experience you will automatically use the easier parametrization.

But in this problem you will have to use the whole curve at once. You cannot have two separate parametrization curves because the point of interest is at the origin and it depends on the curve as a whole.

can i say \(\displaystyle \bold{r}_1(t) = (t, -\sqrt{t})\) and \(\displaystyle \bold{r}_2(t) = (t, \sqrt{t})\)?

It just happens this is a wrong parametrization in this particular problem. (But in general it is correct.)

 

[logistic_guy]

There's no strict rule as long as it follows the curve path. Often, you can pick a variable as t, i.e. x=t or y=t. In this case, it's simpler to choose y=t. What you have is not wrong, but a little more work.

thank

It's like all problem-solving: You look for something to try, try it, and try something else if it doesn't work (or just gets too complicated). With experience, you get to recognize ahead of time what will likely be harder and not worth trying, so you'll find yourself usually choosing the easiest thing first.

In this case, having to split into two parts suggests trying something else first.

thank

I agree with BigBeachBanana that parametrization is a matter of taste. And I also I agree with professor Dave that with experience you will automatically use the easier parametrization.

But in this problem you will have to use the whole curve at once. You cannot have two separate parametrization curves because the point of interest is at the origin and it depends on the curve as a whole.


It just happens this is a wrong parametrization in this particular problem. (But in general it is correct.)

thank

what i do now? as usual to find the curverutre?

 

[Dr.Peterson]

what i do now? as usual to find the curvature?

Apply the formula, and show us your work.

 

[BigBeachBanana]

what i do now?

Typically when solving problems, I keep the end goal in mind and devise a plan for how to get there.
The question asks for the "[equation of the] osculating circle", to write an equation for a circle you'd need 2 things: the radius and the center. Now, with the given information how can you find those components?
What is the definition of an osculating circle? That should help.

 

[logistic_guy]

Apply the formula, and show us your work.

the process is along. before i calculate i just want to make sure finding the curveture will help me

Typically when solving problems, I keep the end goal in mind and devise a plan for how to get there.
The question asks for the "[equation of the] osculating circle", to write an equation for a circle you'd need 2 things: the radius and the center. Now, with the given information how can you find those components?
What is the definition of an osculating circle? That should help.

it say \(\displaystyle t = 0\) it mean we know the center of the circle. it is on the origin. is the curveture is the radius?

 

[BigBeachBanana]

it say \(\displaystyle t = 0\) it mean we know the center of the circle. it is on the origin. is the curveture is the radius?

No. See the definition of an osculating circle. You're looking for the blue circle.

 

[logistic_guy]

No. See the definition of an osculating circle. You're looking for the blue circle.
View attachment 38298

how can you visiulize the curve like this? this make the problem easier. i'm stupid when i visiualize the circle at the origin it won't be tangent to the curve. also i thought parabola mean curve go up or down, not to the right

with this information i'm still not sure if should find the curvture or not?

 

[Dr.Peterson]

with this information i'm still not sure if should find the curvature or not?

The definition you were given involves the curvature; so, YES, find it. That's what we've implied.

Then use that to find the radius of curvature, and then find the center that distance along the normal to the curve.

how can you visiulize the curve like this? this make the problem easier. i'm stupid when i visiualize the circle at the origin it won't be tangent to the curve. also i thought parabola mean curve go up or down, not to the right

What were you taught about graphing curves?

This equation expresses x as a quadratic function of y, so the axis of the parabola is horizontal.

Or, you could just plot a few points to get a sense of the curve. That should take you back to the very beginning of algebra.

 

[Otis]

i thought parabola mean curve go up or down

The word 'parabola' refers to certain curve shapes (see: conic sections). On a plane, those parabolic curves may be oriented in any direction (i.e., the axis of symmetry may be horizontal, vertical or oblique).
[imath]\;[/imath]

 

[logistic_guy]

The definition you were given involves the curvature; so, YES, find it. That's what we've implied.

Then use that to find the radius of curvature, and then find the center that distance along the normal to the curve.

i'll apply the formula

\(\displaystyle \bold{r}(t) = (t^2, t)\)

\(\displaystyle \bold{r}'(t) = (2t, 1)\)

\(\displaystyle \bold{T}(t) = \frac{\bold{r}(t)}{|\bold{r}(t)|} = \frac{(t^2,t)}{\sqrt{4t^2 + 1^2}} = \frac{(t^2,t)}{\sqrt{4t^2 + 1}}\)

\(\displaystyle \kappa(t) = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\)

i'm stuck. i can't find \(\displaystyle \bold{T}'(t)\)

What were you taught about graphing curves?

This equation expresses x as a quadratic function of y, so the axis of the parabola is horizontal.

Or, you could just plot a few points to get a sense of the curve. That should take you back to the very beginning of algebra.

i learn about graph before. i still don't understand them well. i studied curves and i think there is a relation ship between graph of a function and graph of a curve. if i treat the graph as curve then it isn't injective. this also mean the graph isn't represent a function. why you say function when one \(\displaystyle x\) give two \(\displaystyle y\)?

Or, you could just plot a few points to get a sense of the curve. That should take you back to the very beginning of algebra.

i'm uncapable of that. i'm still in a low level of graphing

The word 'parabola' refers to certain curve shapes (see: conic sections). On a plane, those parabolic curves may be oriented in any direction (i.e., the axis of symmetry may be horizontal, vertical or oblique).
[imath]\;[/imath]

this conic i see in some of algebra lessons. the only thing i know is that a circle can be strechted to become like an egg

i'll spend more time in algebra to learn graphs because i found when i know the graphe the question is easier

 

[blamocur]

i'm stuck. i can't find \(\bold{T}'(t)\)

Can you find [imath]\frac{d}{dt}\frac{t}{\sqrt{4t^2+1}}[/imath] ?

 

[BigBeachBanana]

i'll apply the formula

\(\displaystyle \bold{r}(t) = (t^2, t)\)

\(\displaystyle \bold{r}'(t) = (2t, 1)\)

\(\displaystyle \bold{T}(t) = \frac{\bold{r}(t)}{|\bold{r}(t)|} = \frac{(t^2,t)}{\sqrt{4t^2 + 1^2}} = \frac{(t^2,t)}{\sqrt{4t^2 + 1}}\)

\(\displaystyle \kappa(t) = \frac{|\bold{T}'(t)|}{|\bold{r}'(t)|}\)

i'm stuck. i can't find \(\displaystyle \bold{T}'(t)\)

This approach is OK but it's more tedious than the other version.