Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find a unit vector in the direction in which \(f\) increases most rapidly at \(\mathbf{p} .\) What is the rate of change in this direction? \(f(x, y, z)=x e^{y z} ; \mathbf{p}=(2,0,-4)\)

Step-by-Step Solution

Answer

The unit vector is \(\langle 1, 0, 0 \rangle\) and the rate of change is 1.

Step 1: Find the Gradient of the Function

The gradient of a function \(f(x, y, z)\) is given by the vector of its partial derivatives. First, we need to find the partial derivatives of \(f(x, y, z) = x e^{yz}\). Compute the partial derivative with respect to \(x\):\[ f_x = \frac{\partial}{\partial x}(x e^{yz}) = e^{yz}. \]Compute the partial derivative with respect to \(y\):\[ f_y = \frac{\partial}{\partial y}(x e^{yz}) = xz e^{yz}. \]Compute the partial derivative with respect to \(z\):\[ f_z = \frac{\partial}{\partial z}(x e^{yz}) = xy e^{yz}. \] Thus, the gradient is given by:\[ abla f = \langle e^{yz}, xz e^{yz}, xy e^{yz} \rangle. \]

Step 2: Evaluate the Gradient at the Point

Now substitute the coordinates of point \(\mathbf{p} = (2,0,-4)\) into the gradient vector to evaluate it at this point. For \(x=2\), \(y=0\), and \(z=-4\):\[ abla f = \left\langle e^{0\cdot(-4)}, 2\cdot0\cdot e^{0\cdot(-4)}, 2\cdot0\cdot e^{0\cdot(-4)} \right\rangle = \langle 1, 0, 0 \rangle. \]

Step 3: Compute the Unit Vector in the Direction of the Gradient

The direction of maximum increase in a function is given by the gradient vector. To find the unit vector in this direction, divide the gradient vector by its magnitude.First, compute the magnitude of \(abla f\) at \(\mathbf{p}\):\[ \| abla f \| = \sqrt{1^2 + 0^2 + 0^2} = \sqrt{1} = 1. \]The unit vector in the direction of the gradient is:\[ \frac{abla f}{\| abla f \|} = \frac{\langle 1, 0, 0 \rangle}{1} = \langle 1, 0, 0 \rangle. \]

Step 4: Determine the Rate of Change in the Direction of the Gradient

The rate of change of \(f\) at the point \(\mathbf{p}\) in the direction of the gradient is given by the magnitude of the gradient vector. We've calculated the magnitude in the previous step as:\[ \| abla f \| = 1. \]Thus, the rate of change in this direction is 1.

Key Concepts

Unit Vector

A unit vector is a vector that has a magnitude of 1. Unit vectors are used to indicate direction without concern for length. In this context, they are helpful because they provide the direction of the most rapid increase of a function at a given point.

To create a unit vector from any vector, you divide the vector by its length, also known as its magnitude.

Here’s how to calculate the magnitude of a vector:

• First, square each component of the vector.
• Sum these squared values.
• Take the square root of this sum.

For our example, the gradient vector at point \( \mathbf{p} = (2, 0, -4) \) is \( \langle 1, 0, 0 \rangle \). Its magnitude is \( 1 \) because \( \sqrt{1^2 + 0^2 + 0^2} = 1 \).

Since its magnitude is already 1, it is its own unit vector, indicating that the direction of maximum increase in the function is in the positive x-direction.

Rate of Change

The rate of change in mathematics often measures how a quantity alters in relation to another. For instance, it can signify how quickly a function's value changes as we move in a specific direction.

In the case of a multivariable function, the rate of change in a direction is determined by the gradient at that point, as it provides the maximum rate of increase. This rate is computed as the magnitude of the gradient vector.

In our example with the function \( f(x, y, z) = x e^{yz} \), the gradient at \( \mathbf{p} = (2,0,-4) \) is \( \langle 1, 0, 0 \rangle \) with a magnitude of 1. Therefore, the rate of change in this direction, the direction of the gradient, is also 1. This means that if you move in the direction of this unit vector, the function increases at a rate of 1 per unit of distance traveled.

Partial Derivatives

Partial derivatives account for how a function changes as one of its variables changes, keeping the others constant. In functions of multiple variables, partial derivatives help understand how each variable independently affects the function's output.

To find a partial derivative, you differentiate with respect to one variable, treating all other variables as constants. This offers insights into the function’s behavior along one coordinate axis at a time.

For the function \( f(x, y, z) = x e^{yz} \), its partial derivatives are:

• With respect to \( x \): \( f_x = e^{yz} \)
• With respect to \( y \): \( f_y = xz e^{yz} \)
• With respect to \( z \): \( f_z = xy e^{yz} \)

These partial derivatives form the components of the gradient vector, \(abla f = \langle e^{yz}, xz e^{yz}, xy e^{yz} \rangle \), helping us determine the direction and rate of the most rapid increase of the function at a given point.