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Find the point on the parabola $x=t, y=t^{2},-\infty<t<\infty$closest to the point $(2,1 / 2) .$ (Hint: Minimize the square of thedistance as a function of $t .$ )

$(1,1)$ is the point on parabola which is closest to $\left(2, \frac{1}{2}\right)$

Calculus 2 / BC

Chapter 11

Parametric Equations and Polar Coordinates

Section 1

Parametrizations of Plane Curves

Parametric Equations

Polar Coordinates

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in this question, we are given a brochure ization off the standards So x this is given by T and the corresponding white Gordon. It is C squared. So let's make it speak, drawing what it looks like we were. And why axes? And then it just looks like this guy. That's terrible, drawing more like that and it goes through zero. Then we're also giving the points, too. 1/2 which is roughly here, let's say and we're justifying the shortest distance to this point. So, um, well, this is an optimization question so that that's try to figure out which question we need to open eyes. Let's first figure out the distance as a function off time. Not any distance in this case is just getting by. Um, decorous. So suppose we are over here when we get this nice writing triangle. So the distance is the distance in the extraction, which is no to minus except e weird, um, plus the distance in the wind direction, which is, uh, 1/2 minus. Why FT here? Um noted, technically, the distance in the ex direction is the absolute value off to minus X. Um, so just two minus X, But since we're squaring it anyway, keeping track of those upbeat values, um, isn't gonna do anything for us. So now let's plug in. Uh, tea are in the lift. Logan are known expressions for X and y. So we have this equals two minus T squared, plus one minus. He's weird square. Now we can work out these squares. Well, see, this will become for my for tea plus t squared. And this because a order minus steets were less You 2 to 4. So you can just check that this works out using foil, or like the normal rules from a publication. Um, now, let's addle like terms together and organized. It's a bit nicer, so that's starting to deter the fourth. There's no t cute. Um, we have plus t squared minus d squared. So there's no teeth where it's left over. We do have this minus 40 and then finally, we have the four plus in order. So we get, uh, this for wasn't order and I can simplify this to 17 over four. Yeah, but we'll leave it like this for now. So now we want to figure out, um When, uh, it is minimal. So we want to minimize the distance function. The distance is a function of time. So to do that, we computed derivative prime off T. Um, so we get for she cubes minus for, uh, zero. So 40 cute, minus fourth derivative. And since you're optimizing the tea for which the distance takes some extreme value, so a local minimum or maximum is as a time for which interrupted zero So great d prime off t the road. Isn't it easier if the plane we're after is zero four times t zero pubes minus four equals zero. Well, we can divide by four and we find a tease era pubes equals one removed four to the other side before. And that tells us that in fact T zero, there's one. So be it. Distance is minimized at time, equal to p zero. So what is this? What is the corresponding points next of T zero equals one and why off zero equals one? So this point here we were after wait 11

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