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# Find equations of the osculating circles of the parabola$y=\frac{1}{2} x^{2}$ at the points $(0,0)$ and $\left(1, \frac{1}{2}\right) .$ Graph both osculating circles and the parabola on the same screen.

## $$(x+1)^{2}+\left(y-\frac{5}{2}\right)^{2}=8$$

Vectors

Vector Functions

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

{'transcript': "So what we see here is that the Oscar waiting circle at a point touches the graph and has radius one over RK value. So this is the radius. Oscillating circle is always made on the concave down side of the graph. Um, so we note that the circle of the are the center of the escalating circle lies on the normal to the curve at that point. So the equation of the curve is given as y equals one half x squared. So we differentiate. So we take dy dx, and that's going to give us just X. Then we differentiate a second time, and that's going to give us one. So therefore, R radius of curvature is equal to magnitude of one over one plus X squared auto power of three halves. So this is gonna end up giving us that one over K or radius of convergence. One of our 86 emergency is going to be one plus x squared 23 halves. So if we find the escalating circle at 00 this is our circle right here. What we see is this right here will be here. Courage. Sure graph like that. Um, and the way we find. That is the fact that when we have this center at 00 we are curvature, which is one over. Okay, we're equal one, Um, and the normal to the proble 00 is the Y axis. So the center of the escalating circle will be zero was your plus one, which is 01 And then we know that the equation of the escalating circle will be X squared plus y minus one squared equals one. So it's gonna be a circle with a radius one, which is what we went for right about there, then finding the Oscar waiting circle at 11 half. What we end up getting is a one over the creature that's gonna give us to route to, um the slope at X equals one is one. So the slope of the normal would be negative one. So therefore, our center is going to be one minus two route two times, one over root, two and one half, plus two route to times one of our route to and that's going to give us a center of negative 1 5/2. And our equation, since that's the center, will be X plus one squared plus Y minus 5/2 squared equals eight because to route to was our radius. And so this is gonna look like now is the circle is going to be a bit more over here and then in terms of our wine, it'll keep coming down and still look like that."}

California Baptist University

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Vectors

Vector Functions

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

Lectures

Join Bootcamp