Find the point on the ellipse x=2 cost, y=sint, 0 ≤t ≤2 πclosest to the point (3 / 4,0) . (Hint:

Find the point on the ellipse x=2 cost, y=sint, 0 ≤t ≤2...

Key Concepts

Optimization

Optimization in calculus involves finding the maximum or minimum values of a function. In problems involving distance, one often minimizes the square of the distance instead of the distance itself to simplify the differentiation since the square function is monotonic for non-negative values.

Parametric Equations

A parametric equation represents a curve by expressing the coordinates as functions of a parameter. In the context of ellipses, using trigonometric functions (such as cosine and sine) allows the representation of the ellipse in a way that can simplify the computation of distances to a fixed point.

Distance Function

The distance function measures the separation between two points. By squaring the distance, one obtains a function that is often easier to manipulate and differentiate because it removes the square root, making it more straightforward to find critical points using calculus.

Derivative and Critical Point Analysis

Finding the critical points of a function by taking its derivative and setting it equal to zero is a fundamental technique in calculus. This method is used to identify potential points of minima or maxima, and in the context of distance minimization, it helps determine the point on a curve that is closest to a given fixed point.

Transcript

00:01 So for this question, we are given a parameterization of an ellipse.

00:06 X is 2 cosine of t and y is just sign of t, where t is between 0 and 2 pi.

00:22 So drawing this out, we get something like this nice, what on earth happened there.

00:34 So we get a nice or not so nice ellipse.

00:45 We're also given a points.

00:47 3 quarters of 0.

00:49 So let's call that p.

00:53 So x is three quarters, y is zero.

00:56 So that's going to be hereish.

01:01 And we're asked to find the points on the ellipse.

01:04 So they're going to be roughly here and here.

01:06 So we're only known from the picture that there's going to be two solutions.

01:10 We're asked to find a point or points for which the distance between p and the ellipse is minimal.

01:18 So let's figure out what the distance is as we traverse through the ellipse.

01:26 So we want to find the distance between a point on the ellipse and p and the function of time.

01:34 Well, in x, y coordinates, so we have the...

01:43 X of t y of t so the dotted line is the difference in the x distance different in the x's squared plus the difference in the y squared um so the distance is or rather um the distance squared let's say it like that so d squared of t is minus three quarters squared plus sine of t minus zero squared.

02:35 So i'm going to use a trick here that is useful in that it makes the analysis a lot simpler.

02:44 Squaring a distance, because distances are always positive, doesn't change the ordering.

02:50 What i mean by that is if you have two numbers, a less than b positive, then their squares will be in the same order.

03:12 So squaring something doesn't change ordering.

03:14 You can see this because the positive branch of the parabola is strictly increasing.

03:21 So if you move further in that direction, that direction, you also move further in that direction.

03:27 Direction.

03:29 The derivative of x squared is to x, so if x is positive, the derivative is positive.

03:34 That's essentially what i'm saying.

03:38 So what this means is that finding the minimum, so the minimum of d, or sort of the point at which that let me let me let me let me phrase this correctly, because this is a very useful trick.

04:03 So the point where d of t, so the actual distance, which is the square root of the expression we have up there, is minimal, is the same as the points where d squared, so the square of the distance is minimal.

04:46 And this is because taking a square doesn't change the size of your number.

04:51 Now mind you, you do need that these numbers are positive because otherwise it makes no sense, right? so you could take like minus three and zero and then square both and then the orring gets reversed.

05:04 So you can only do this with distances.

05:06 But for distances, it's very useful.

05:08 Because it allows us to avoid square roots.

05:14 And square roots are major sources of error.

05:21 Ok, so back to the problem at hand and away from this general...

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