Find the point on the parabola x=t, y=t^2,-∞<t<∞closest to the point (2,1 / 2) . (Hint: Minimize

Find the point on the parabola x=t, y=t^2,-∞<t<∞closest to...

Key Concepts

Parametric Equations

Parametric equations represent curves using a parameter to express the coordinates, allowing the description of paths that might not be functions in the traditional sense. This method is useful in problems where points are defined as functions of a single variable, making it easier to perform differentiation and substitution during analysis.

Distance Formula

The distance formula is used to compute the distance between two points in the Cartesian plane based on their coordinates. In many optimization problems, particularly in minimizing distances, the square of the distance is used to simplify calculations by removing the square root, while still preserving the location of the minimum.

Optimization with Derivatives

Optimization with derivatives involves finding the critical points of a function by setting its derivative equal to zero and then testing these points to determine if they correspond to minima or maxima. This technique is fundamental in calculus for solving problems that seek to minimize or maximize particular quantities.

Squared Distance Minimization

Squared distance minimization is an approach where the function to be minimized is the square of the distance between points, rather than the distance itself. This simplifies the differentiation process because it avoids the complications introduced by the square root, while still identifying the same point of minimum distance due to the monotonicity of the squaring function.

Transcript

00:01 In this question, we are given a parameterization of the standard or ebola.

00:07 So x is just given by t and the corresponding y coordinate is t squared.

00:15 So let's make a big drawing of what this looks like.

00:18 So we have our x and then it just looks like this guy.

00:27 That's a terrible drawing.

00:33 More like that and it goes through zero.

00:36 Then we're also given the point to one half, which is roughly here, let's say.

00:45 And we're asked to find the shortest distance to this point.

00:52 So, well, this is an optimization question.

00:58 So let's try to figure out which question we need to optimize.

01:04 Let's first figure out the distance as a function of time.

01:09 Now the distance in this case is just given by pythagoras.

01:15 So suppose we're over here, then we get this nice right angle of triangle.

01:24 So the distance is the distance in the x direction, which is 2 minus x of t squared, plus the distance in the y direction, which is 1 1 half, minus y of t squared.

01:48 Note that technically the distance in the x direction is the absolute value of 2 minus x, instead of just 2 minus x.

01:56 But since we're squaring it anyway, keeping track of those absolute values isn't going to do anything for us.

02:04 So now let's plug in t, or let's plug in our known expressions for x and y...

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