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Chapter 10: Problem 7

The concentration of salt in a fluid at \((x, y, z)\) is given by \(F(x, y, z)=\) \(2 x^{2}+3 y^{4}+2 x^{2} z^{2} \mathrm{mg} / \mathrm{cm}^{3}\). You are at the point (-1,1,-1) . (a) In which direction should you move if you want the concentration to increase the fastest? direction: _________. (Give your answer as a vector.) (b) You start to move in the direction you found in part (a) at a speed of \(5 \mathrm{~cm} / \mathrm{sec} .\) How fast is the concentration changing? rate of change \(=\) ______________.

Short Answer

Expert verified
The direction in which you should move is \(-\frac{1}{4}\textbf{i} + \frac{3}{4}\textbf{j} - \frac{1}{4}\textbf{k}\), and the rate at which the concentration is changing is 55 mg/(cm鲁 sec).

Step by step solution

01

Compute the gradient vector of the function F

To compute the gradient vector, we need to find the partial derivatives with respect to x, y, and z. \[ \nabla F(x, y, z) = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = \left( 4x + 4xz^2, 12y^3, 4x^2z \right) \]
02

Evaluate the gradient vector at the given point

Now, we find the gradient vector at the point (-1, 1, -1) by plugging in the coordinates in the gradient vector formula. \[ \nabla F(-1, 1, -1) = \left( 4(-1) + 4(-1)(-1)^2, 12(1)^3, 4(-1)^2(-1) \right) = (-4, 12, -4) \]
03

Normalize the gradient vector to find the direction

To find the direction in which we should move to increase the concentration the fastest, we normalize the gradient vector. \[ \text{Direction} = \frac{(-4, 12, -4)}{\|(-4, 12, -4)\|} = \frac{(-4, 12, -4)}{\sqrt{(-4)^2 + (12)^2 + (-4)^2}} = \frac{(-4, 12, -4)}{16} = (-\frac{1}{4}, \frac{3}{4}, -\frac{1}{4}) \] So, the direction is \(-\frac{1}{4}\textbf{i} + \frac{3}{4}\textbf{j} - \frac{1}{4}\textbf{k}\).
04

Compute the rate of change in salt concentration

To find the rate of change in salt concentration while moving in the direction found in part (a) with a speed of 5 cm/s, we take the dot product of the gradient vector and the velocity vector. \[ \text{Velocity Vector} = 5 \times \text{Direction} = 5(-\frac{1}{4}\textbf{i} + \frac{3}{4}\textbf{j} - \frac{1}{4}\textbf{k}) = (-\frac{5}{4}\textbf{i} + \frac{15}{4}\textbf{j} - \frac{5}{4}\textbf{k}) \] \[ \text{Rate of Change} = \nabla F(-1,1,-1) \cdot \text{Velocity Vector} = (-4, 12, -4) \cdot (-\frac{5}{4}, \frac{15}{4}, -\frac{5}{4}) = 5 + 45 + 5 = 55 \text{ mg/(cm}^3\text{ sec)} \] The rate of change in salt concentration is 55 mg/(cm鲁 sec).

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Key Concepts

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

Partial Derivatives
Partial derivatives help us understand how a function changes as we tweak one of its variables, while keeping the rest constant. Imagine a multivariable function like the concentration of salt in a fluid, expressed in terms of coordinates \( x, y, \) and \( z \). To grasp how the concentration changes at each coordinate, we look at the partial derivatives. In our exercise, the function is given by:
\[ F(x, y, z) = 2x^2 + 3y^4 + 2x^2z^2 \]
The partial derivative with respect to \( x \) means varying \( x \) while \( y \) and \( z \) remain fixed. For this function, it results in:\[ \frac{\partial F}{\partial x} = 4x + 4xz^2 \]Similarly, for \( y \):\[ \frac{\partial F}{\partial y} = 12y^3 \]And for \( z \):\[ \frac{\partial F}{\partial z} = 4x^2z \]
These partial derivatives form a vector, known as the gradient, which gives us significant insight into the behavior of the function in various dimensions. They tell us the rate at which the concentration changes in each direction.
Rate of Change
The rate of change is a concept that tells us how fast something is changing over time. When discussing functions, the rate of change in a given direction can be determined using the gradient vector.
In our problem, once we find the gradient vector at the point \((-1, 1, -1)\), we compute:\[ abla F(x, y, z) = (-4, 12, -4) \]This vector tells us how the concentration of salt changes with respect to each spatial coordinate.
To find the actual rate of change as you move in a specific direction, it's essential to calculate the dot product of the gradient vector with the velocity vector. This step effectively multiplies the directional influence of each component by how fast we're moving in that direction.
The velocity vector is found by scaling the direction found earlier with a speed factor, here being 5 cm/s:\[ \text{Velocity Vector} = (-\frac{5}{4}, \frac{15}{4}, -\frac{5}{4}) \]The rate of change calculation goes as follows:\[ (-4, 12, -4) \cdot (-\frac{5}{4}, \frac{15}{4}, -\frac{5}{4}) = 55 \text{ mg/(cm}^3\text{ sec)} \]This value, 55 mg/(cm鲁 sec), tells us how rapidly the salt concentration is increasing as we move in our specified direction.
Vector Normalization
Vector normalization plays a critical role in finding a direction vector of unit length. This means making the vector's magnitude equal to 1, which allows us to focus solely on its direction.
In the exercise, after determining the gradient vector at point \((-1, 1, -1)\), we had:\[ (-4, 12, -4) \]The next step was to normalize it, which involves dividing the vector by its magnitude. The magnitude can be calculated using:\[ \|(-4, 12, -4)\| = \sqrt{(-4)^2 + (12)^2 + (-4)^2} = 16 \]Once the magnitude is found, the normalized vector becomes:\[ (-\frac{1}{4}, \frac{3}{4}, -\frac{1}{4}) \]
This new vector directs us at a pace that would increase the salt concentration the fastest at that particular point. It's crucial in applications requiring precision in directional decisions, as it disregards the scale of the vector, focusing instead entirely on its direction.

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Most popular questions from this chapter

For each of the following prompts, provide an example of a function of two variables with the desired properties (with justification), or explain why such a function does not exist. a. A function \(p\) that is defined at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} p(x, y)\) does not exist. b. A function \(q\) that does not have a limit at \((0,0),\) but that has the same limiting value along any line \(y=m x\) as \(x \rightarrow 0\). c. A function \(r\) that is continuous at \((0,0),\) but \(\lim _{(x, y) \rightarrow(0,0)} r(x, y)\) does not exist. d. A function \(s\) such that \(\lim _{(x, x) \rightarrow(0,0)} s(x, x)=3\) and \(\lim _{(x, 2 x) \rightarrow(0,0)} s(x, 2 x)=6\) for which \(\lim _{(x, y) \rightarrow(0,0)} s(x, y)\) exists. e. A function \(t\) that is not defined at (1,1) but \(\lim _{(x, y) \rightarrow(1,1)} t(x, y)\) does exist.

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 \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(d U\) is given by \(d U=840 d V+27.32 d T\). (a) How does the energy change if the volume is held constant and the temperature is decreased slightly? \(\odot\) it increases slightly \(\odot\) it does not change \(\odot\) it decreases slightly (b) How does the energy change if the temperature is held constant and the volume is increased slightly? \(\odot\) it does not change \(\odot\) it increases slightly \(\odot\) it decreases slightly (c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin. Change in energy = ___________. Please include units in your answer.

Find the absolute maximum and minimum of the function \(f(x, y)=x^{2}+\) \(y^{2}\) subject to the constraint \(x^{4}+y^{4}=6561\). As usual, ignore unneeded answer blanks, and list points in lexicographic order. Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, ) Absolute minimum value: ____________________. attained at ( ____________, ______________) ( ____________, )
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