Problem 7 - Active Calculus Multivariable

Question

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 \(=\) ______________.

Step-by-Step Solution

Answer

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/(cm³ sec).

Step 1: 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. \[ \nabla F(x, y, z) = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) = \left( 4x + 4xz^2, 12y^3, 4x^2z \right) \]

Step 2: 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. \[ \nabla F(-1, 1, -1) = \left( 4(-1) + 4(-1)(-1)^2, 12(1)^3, 4(-1)^2(-1) \right) = (-4, 12, -4) \]

Step 3: 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 \(-\frac{1}{4}\textbf{i} + \frac{3}{4}\textbf{j} - \frac{1}{4}\textbf{k}\).

Step 4: 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/(cm³ sec).

Key Concepts

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/(cm³ 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.