Question

Consider the following functions \(f,\) points \(P,\) and unit vectors \(\mathbf{u}\). a. Compute the gradient of \(f\) and evaluate it at \(P\). b. Find the unit vector in the direction of maximum increase of \(f\) at \(P\). c. Find the rate of change of the function in the direction of maximum increase at \(P\) d. Find the directional derivative at \(P\) in the direction of the given vector. $$f(x, y, z)=\frac{x-z}{y-z} ; P(3,2,-1) ;\left\langle\frac{1}{3}, \frac{2}{3},-\frac{2}{3}\right\rangle$$

Short answer

Based on the given information, calculate the directional derivative at point P(3, 2, -1) in the direction of the given vector <1/3, 2/3, -2/3>. Answer: The directional derivative at point P(3, 2, -1) in the direction of the given vector <1/3, 2/3, -2/3> is 19/27.

Step-by-step solution

1. Compute the gradient of the function f(x, y, z)

We need to find the gradient of the function, which is a vector consisting of the partial derivatives of the function: \(\nabla f = \langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\rangle\). $$ \nabla f(x, y, z) = \left\langle\frac{\partial}{\partial x}\left(\frac{x - z}{y - z}\right), \frac{\partial}{\partial y}\left(\frac{x - z}{y - z}\right), \frac{\partial}{\partial z}\left(\frac{x - z}{y - z}\right)\right\rangle $$ Now, we compute the partial derivatives: $$ \frac{\partial f}{\partial x} = \frac{1}{y - z},\ \frac{\partial f}{\partial y} = \frac{z - x}{(y - z)^2},\ \frac{\partial f}{\partial z} = \frac{x - y}{(y - z)^2} $$ So, the gradient is: $$ \nabla f(x, y, z) = \left\langle\frac{1}{y - z}, \frac{z - x}{(y - z)^2}, \frac{x - y}{(y - z)^2}\right\rangle $$

2. Evaluate the gradient of f at point P(3, 2, -1)

Now, we need to evaluate the gradient at the point \(P(3, 2, -1)\). We plug in the coordinates of \(P\) into the gradient: $$ \nabla f(3, 2, -1) = \left\langle\frac{1}{2 - (-1)}, \frac{-1 - 3}{(2 - (-1))^2}, \frac{3 - 2}{(2 - (-1))^2}\right\rangle = \left\langle3, -\frac{4}{9}, \frac{1}{9}\right\rangle $$

3. Find the unit vector in the d direction of maximum increase of the function f at point P

The direction of maximum increase of a function at a point is given by the gradient at that point, which we have already computed: \(\left\langle3, -\frac{4}{9}, \frac{1}{9}\right\rangle\). To find the unit vector, we need to find the magnitude and then normalize it: $$ \|\nabla f(3, 2, -1)\| = \sqrt{3^2 + \left(-\frac{4}{9}\right)^2 + \left(\frac{1}{9}\right)^2} = \sqrt{9 + \frac{16}{81}+\frac{1}{81}} = \sqrt{\frac{729+16+1}{81}}=\sqrt{\frac{746}{81}} $$ Now, we normalize the gradient vector: $$ \frac{\nabla f(3, 2, -1)}{\|\nabla f(3, 2, -1)\|} = \frac{\left\langle3, -\frac{4}{9}, \frac{1}{9}\right\rangle}{\sqrt{\frac{746}{81}}} = \left\langle\frac{9\sqrt{746}}{82}, -\frac{4\sqrt{746}}{82}, \frac{\sqrt{746}}{82}\right\rangle $$

4. Find the rate of change of the function in the direction of maximum increase at point P

The rate of change of a function in the direction of the gradient at a point is given by the magnitude of the gradient at that point: $$ \text{Rate of change} = \|\nabla f(3, 2, -1)\| = \sqrt{\frac{746}{81}} $$

5. Find the directional derivative at point P in the direction of the given vector

Lastly, we need to find the directional derivative at point \(P\) in the direction of the given vector \(\left\langle\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\right\rangle\): $$ \nabla f(3, 2, -1) \cdot \left\langle\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\right\rangle = \left\langle3, -\frac{4}{9}, \frac{1}{9}\right\rangle \cdot \left\langle\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\right\rangle = 3\cdot\frac{1}{3} +-\frac{4}{9}\cdot\frac{2}{3} +\frac{1}{9}\cdot-\frac{2}{3} $$ The directional derivative is: $$ = 1 - \frac{8}{27} - \frac{2}{27} = \frac{19}{27} $$

Key concepts

Partial Derivatives

Partial derivatives are essential in the realm of multivariable calculus. They measure how a function changes as only one input changes while keeping the rest constant. Imagine you're walking on a hilly trail, and at any point, you can measure the slope in the direction you're facing (north, east, etc.)—that's what a partial derivative tells us.

For a function like our example, \( f(x, y, z) = \frac{x-z}{y-z} \), we consider how \( f \) changes with respect to each variable individually. For instance, \( \frac{\partial f}{\partial x} = \frac{1}{y - z} \) means that, if you move east (along the x-axis), the rate at which the elevation changes is \( \frac{1}{y - z} \), considering that your north and vertical positions (y and z) don't change. This concept is fundamental to solving problems in physics, engineering, and economics, where multivariable functions are common, and understanding their individual influences is crucial.

Directional Derivative

Moving beyond the fixed compass directions, the directional derivative is about finding the rate of change of a function in any direction given by a vector. Think of flying a drone over the same hilly terrain—this time, you're not limited to the cardinal directions and can move in a diagonal direction, which is a combination of north, east, and upward movements.

The directional derivative along a vector \( \mathbf{u} \) is symbolically written as \( abla f \cdot \mathbf{u} \). It tells us the slope of the hill, or the rate of change of our function \( f \), in the particular direction of \( \mathbf{u} \). So, if \( \mathbf{u} \) was the direction of your drone's flight (like the given vector in the exercise), the directional derivative would reveal how rapidly the function's value would increase or decrease along that specific flight path.

Rate of Change

The rate of change is a universal concept in mathematics and science, describing how a quantity varies with respect to some change, such as time, distance, or any other variable. It is analogous to velocity, which is the rate of change of position with time.

In the context of our function \( f(x, y, z) \), the rate of change in the direction of the gradient vector is the steepest. Imagine you're at the bottom of a valley and looking for the fastest way up to the top of the hill; following the gradient's direction is like hiking the steepest path directly to the summit. Mathematically, this is simply the magnitude of the gradient. In our example, we calculated that the rate of change at point P in the direction of the gradient, basically the 'steepness' of our hill at P, is \( \sqrt{\frac{746}{81}} \). It's an important measure that appears in a myriad of fields, from describing how fast a population grows to how quickly heat spreads across a surface.