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Chapter 12: Problem 51

Estimate the closest point on the paraboloid \(z=x^{2}+y^{2}\) to the point (1,0,0)

Short Answer

Expert verified
Upon solving, we will end up with four potential points that are closest to (1,0,0) on the paraboloid, along with four distances. Each of these points needs to be checked to verify which actually gives the shortest distance. Finally, the point(s) resulting in the smallest such value corresponds to the closest point(s) on the paraboloid.

Step by step solution

01

Define the Objective Function

The distance between a point (x,y,z) on the paraboloid and the reference point (1,0,0) can be represented using the Euclidean distance formula like this: d^2 = (x-1)^2+(y-0)^2+(z-0)^2. However, knowing that z = x^2 + y^2 (from the given paraboloid's equation), our objective function can be expressed in terms of x and y alone. Therefore, we can replace z in the previous function, obtaining an objective function: f(x,y) = (x-1)^2 + y^2 + (x^2 + y^2)^2.
02

Find the Gradient of f(x,y)

Now find the two first partial derivatives of our function f(x,y), df/dx and df/dy, by applying the standard rules of differentiation. These can be computed as: \[\frac{{df}}{{dx}} = 2(x-1) + 2x(1 + 2x^2 + 2y^2)\] and \[\frac{{df}}{{dy}} = 2y(1 + 2x^2 + 2y^2).\]
03

Solve for the Optimal Points

Next, we can set both df/dx and df/dy to zero and solve the two equations to find the optimal points of x and y respectively. We have thus two following equations: 2(x-1) + 2x(1 + 2x^2 + 2y^2) = 0 and 2y(1 + 2x^2 + 2y^2) = 0. Solving these equations yields the points (x,y) that will result in the minimum value for f(x,y).
04

Validate the Results

By substituting obtained x and y values into our original paraboloid equation z = x^2 + y^2, we can find the z-coordinate(s) associated with these points. These points, then, are the points on the paraboloid that are closest to our given point of (1,0,0). We have to compute d^2 for each of these points to ensure that they do indeed yield a minimum value.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Paraboloid
A paraboloid is a surface in three-dimensional space that can take the form of an upward or downward "bowl" shape. In the context of this problem, the paraboloid is described by the equation \(z = x^2 + y^2\). This specific type is known as an elliptic paraboloid.

Key characteristics of an elliptic paraboloid include:
  • It is symmetric around the z-axis.
  • At any cross-section parallel to the xy-plane, it forms a circle or an ellipse.
  • Each point on the surface satisfies the equation \(z = x^2 + y^2\).

Understanding the shape and equation of a paraboloid can help in visualizing optimization problems, where the objective is often to find a minimum or maximum point on the surface.
Euclidean Distance
The Euclidean distance is a measure used to calculate the "straight-line" distance between two points in space, much like measuring it with a ruler. For points in three-dimensional space, such as (x, y, z) and a reference point like (1, 0, 0), the Euclidean distance formula is: \(d = \sqrt{(x-1)^2 + (y-0)^2 + (z-0)^2}\).

For the optimization problem, we're interested in minimizing this distance to find the closest point on the paraboloid. However, instead of working with \(d\) directly, we use \(d^2\) because it simplifies calculations and does not affect where the minimum occurs. Minimizing \(d^2\) vs. \(d\) yields the same optimal point.
Partial Derivatives
In calculus, partial derivatives represent how a function changes as each individual variable changes, keeping the other variables constant. In this problem, we're dealing with a function \(f(x, y)\) that describes the squared distance in terms of \(x\) and \(y\).

To find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\), we apply rules of differentiation. For this problem they are:
  • \(\frac{\partial f}{\partial x} = 2(x-1) + 2x(1 + 2x^2 + 2y^2)\)
  • \(\frac{\partial f}{\partial y} = 2y(1 + 2x^2 + 2y^2)\)

These derivatives help us understand the rate of change of the distance function in each direction, which is crucial for finding minimum points through optimization.
Gradient
The gradient of a function in multi-dimensional calculus is a vector that points in the direction of the greatest rate of increase of the function. For a two-variable function \(f(x, y)\), like our squared distance function, the gradient \(abla f\) is given by:
  • \(abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\)

In this problem, the gradient helps determine the direction in which the function increases most rapidly, and thus, by going against this gradient (i.e., following the negative gradient), we can find where the function achieves its minimum. For optimization, we set both components of the gradient equal to zero to find critical points, which are potential minimum points on the paraboloid's surface.

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Most popular questions from this chapter

Graph \(z=\sin (x+y) .\) Compute \(\nabla \sin (x+y)\) and explain why the gradient gives you the direction that the sine wave travels. In which direction would the sine wave travel for \(z=\sin (2 x-y) ?\)

In this exercise, we visualize the linear approximation of example 4.3. Start with a contour plot of \(f(x, y)=2 x+e^{x^{2}-y}\) with \(-1 \leq x \leq 1\) and \(-1 \leq y \leq 1 .\) Then zoom in on the point (0,0) of the contour plot until the level curves appear straight and equally spaced. (Level curves for \(z\) -values between 0.9 and 1.1 with a graphing window of \(-0.1 \leq x \leq 0.1\) and \(-0.1 \leq y \leq 0.1 \text { should work. })\) You will need the \(z\) -values for the level curves. Notice that to move from the \(z=1\) level curve to the \(z=1.05\) level curve you move 0.025 unit to the right. Then \(\frac{\partial f}{\partial x}(0,0) \approx \frac{\Delta z}{\Delta x}=\frac{0.05}{0.025}=2 .\) Verify graphically that \(\frac{\partial f}{\partial y}(0,0) \approx-1 .\) Explain how to use the contour plot to reproduce the linear approximation \(1+2 x-y.\)

In exercise \(3,\) there is a saddle point at \((0,0) .\) This means that there is (at least) one trace of \(z=x^{3}-3 x y+y^{3}\) with a local minimum at (0,0) and (at least) one trace with a local maximum at \((0,0) .\) To analyze traces in the planes \(y=k x\) (for some constant \(k\) ), substitute \(y=k x\) and show that \(z=\left(1+k^{3}\right) x^{3}-3 k x^{2} .\) Show that \(f(x)=\left(1+k^{3}\right) x^{3}-3 k x^{2}\) has a local minimum at \(x=0\) if \(k<0\) and a local maximum at \(x=0\) if \(k>0 .\) (Hint: Use the Second Derivative Test from section \(3.5 .)\)

Sketch a contour plot. $$f(x, y)=x^{2}+4 y^{2}$$

The accompanying data show the average number of points professional football teams score when starting different distances from the opponents' goal line. (For more information, see Hal Stern's "A Statistician Reads the Sports Pages" in Chance, Summer \(1998 .\) The number of points is determined by the next score, so that if the opponent scores next, the number of points is negative.) Use the linear model to predict the average number of points starting (a) 60 yards from the goal line and (b) 40 yards from the goal line. $$\begin{array}{|c|c|c|c|c|c|}\hline \text { Yards from goal } & 15 & 35 & 55 & 75 & 95 \\\\\hline \text { Average points } & 4.57 & 3.17 & 1.54 & 0.24 & -1.25 \\\\\hline\end{array}$$
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