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Chapter 10: Problem 5

Find the minimum distance from the point (1,1,11) to the paraboloid given by the equation \(z=x^{2}+y^{2}\). Minimum distance = ___________.

Short Answer

Expert verified
The minimum distance between the point (1, 1, 11) and the paraboloid \(z = x^2 + y^2\) is approximately \(10.598\).

Step by step solution

01

Write down the given information

We are given a point \(P(1, 1, 11)\) and a paraboloid defined by the equation \(z = x^2 + y^2\).
02

Write down the distance formula in 3D

The distance formula in 3D, between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), is given by \[ d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2} \]
03

Plug in the given point and the paraboloid equation into the distance formula

Since we are finding the distance from point \(P(1, 1, 11)\) to the paraboloid \(z = x^2 + y^2\), we can set \((x_1, y_1, z_1) = (1, 1, 11)\) and \((x_2, y_2, z_2) = (x, y, x^2 + y^2)\) in the distance formula: \[ d = \sqrt{(1 - x)^2 + (1 - y)^2 + (11 - (x^2 + y^2))^2} \]
04

Minimize the square of the distance function

To simplify the problem, we can minimize the square of the distance function, because minimizing the square of a function will also minimize the function itself: \[ d^2 = (1 - x)^2 + (1 - y)^2 + (11 - (x^2 + y^2))^2 \]
05

Find the partial derivatives of the distance squared function

Calculate the partial derivatives of the distance squared function with respect to x and y: \[ \frac{\partial d^2}{\partial x} = -2(1-x) + 4x(11-(x^2+y^2)) \] \[ \frac{\partial d^2}{\partial y} = -2(1-y) + 4y(11-(x^2+y^2)) \]
06

Set the partial derivatives to zero and solve the system of equations

Set the partial derivatives to zero to find the critical points of the distance squared function: \[ \begin{cases} -2(1-x) + 4x(11-(x^2+y^2)) = 0 \\ -2(1-y) + 4y(11-(x^2+y^2)) = 0 \end{cases} \] Solving this system of equations, we get \(x = \frac{1}{3}\) and \(y = \frac{1}{3}\).
07

Find the minimum distance by plugging the critical points into the distance formula

Plug the critical points (\(x = \frac{1}{3}\) and \(y = \frac{1}{3}\)) back into the distance formula: \[ d = \sqrt{(1 - \frac{1}{3})^2 + (1 - \frac{1}{3})^2 + (11 - ((\frac{1}{3})^2 + (\frac{1}{3})^2))^2} \] Evaluating this expression, we get the minimum distance \(d \approx 10.598 \).
08

Final Answer

The minimum distance between the point (1, 1, 11) and the paraboloid \(z = x^2 + y^2\) is approximately \(10.598\).

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Key Concepts

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

Distance Formula in 3D
When tackling geometric problems in three dimensions, an essential tool is the distance formula in 3D. This formula allows us to compute the distance between two points in a three-dimensional space. Specifically, given two points, \(P_1(x_1, y_1, z_1)\) and \(P_2(x_2, y_2, z_2)\), the distance \(d\) between them is defined by the following equation:
\[d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}\]
This formula is a direct extension of Pythagoras' theorem into three dimensions. When measuring the distance to a surface, such as a paraboloid, you consider a point on the surface and the given point, computing the distance between them by plugging their coordinates into the distance formula.
To optimize this calculation in the context of finding a minimum distance, we may only need to consider certain aspects or simplify the problem, such as by minimizing the square of the distance instead for easier computation, which leads us into optimization techniques in multivariable calculus.
Partial Derivatives
In multivariable calculus, we often deal with functions of several variables, and partial derivatives are the tools we use to find how a function changes with respect to each variable independently. The partial derivative of a function \(f(x, y)\) with respect to \(x\) is denoted as \(\frac{\partial f}{\partial x}\) and represents the rate of change of \(f\) with respect to \(x\) while keeping \(y\) constant.
In practical applications, such as finding the minimum distance to a surface, partial derivatives help us understand the behavior of the distance function with respect to each individual coordinate. By taking the partial derivative of the squared distance function with respect to \(x\) and \(y\), we set the stage for finding the critical points where the distance might be at a minimum or maximum.
Optimization in Multivariable Calculus
The process of finding either the maximum or minimum values of a function falls under the umbrella of optimization in multivariable calculus. When the function involves multiple variables, the solution often involves finding points where all partial derivatives of the function are zero; these points are called critical points. Once the partial derivatives of the function are set to zero, we can solve the resulting system of equations to find these critical points.
In the context of finding the minimum distance between a point and a paraboloid, the distance function involves variables \(x\) and \(y\) of the point on the paraboloid. By setting the partial derivatives with respect to \(x\) and \(y\) equal to zero, we locate the critical points which are potential candidates for the minimum distance. Additional checks, like the second derivative test, can confirm the nature of these critical points, ensuring they correspond to a minimum rather than a maximum.

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Most popular questions from this chapter

The temperature on an unevenly heated metal plate positioned in the first quadrant of the \(x y\) -plane is given by $$C(x, y)=\frac{25 x y+25}{(x-1)^{2}+(y-1)^{2}+1}$$ Assume that temperature is measured in degrees Celsius and that \(x\) and \(y\) are each measured in inches. (Note: At no point in the following questions should you expand the denominator of \(C(x, y) .)\) a. Determine \(\left.\frac{\partial C}{\partial x}\right|_{(x, y)}\) and \(\left.\frac{\partial C}{\partial y}\right|_{(x, y)}\). b. If an ant is on the metal plate, standing at the point (2,3) , and starts walking in the direction parallel to the positive \(y\) axis, at what rate will the temperature the ant is experiencing change? Explain, and include appropriate units. c. If an ant is walking along the line \(y=3\) in the positive \(x\) direction, at what instantaneous rate will the temperature the ant is experiencing change when the ant passes the point (1,3)\(?\) d. Now suppose the ant is stationed at the point (6,3) and walks in a straight line towards the point (2,0) . Determine the average rate of change in temperature (per unit distance traveled) the ant encounters in moving between these two points. Explain your reasoning carefully. What are the units on your answer?

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

Are the following statements true or false? (a) The gradient vector \(\nabla f(a, b)\) is tangent to the contour of \(f\) at \((a, b)\). (b) \(f_{\vec{u}}(a, b)=\|\nabla f(a, b)\| .\) (c) \(f_{\vec{u}}(a, b)\) is parallel to \(\vec{u}\). (d) If \(\vec{u}\) is perpendicular to \(\nabla f(a, b),\) then \(f_{\vec{u}}(a, b)=\langle 0,0\rangle\). (e) If \(\vec{u}\) is a unit vector, then \(f_{\vec{u}}(a, b)\) is a vector. (f) Suppose \(f_{x}(a, b)\) and \(f_{y}(a, b)\) both exist. Then there is always a direction in which the rate of change of \(f\) at \((a, b)\) is zero. (g) If \(f(x, y)\) has \(f_{x}(a, b)=0\) and \(f_{y}(a, b)=0\) at the point \((a, b)\), then \(f\) is constant everywhere. (h) \(\nabla f(a, b)\) is a vector in 3 -dimensional space.

Suppose that you are climbing a hill whose shape is given by \(z=902-\) \(0.07 x^{2}-0.1 y^{2},\) and that you are at the point (40,70,300) In which direction (unit vector) should you proceed initially in order to reach the top of the hill fastest? (_________________) If you climb in that direction, at what angle above the horizontal will you be climbing initially (radian measure)?

In a simple electric circuit, Ohm's law states that \(V=I R,\) where \(\mathrm{V}\) is the voltage in volts, I is the current in amperes, and \(\mathrm{R}\) is the resistance in ohms. Assume that, as the battery wears out, the voltage decreases at 0.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.02 ohms per second. When the resistance is 100 ohms and the current is 0.02 amperes, at what rate is the current changing? ___________ amperes per second .
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