Question

Gradients in three dimensions 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. $$f(x, y, z)=1+4 x y z ; P(1,-1,-1) ;\left\langle\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right\rangle$$

Short answer

Based on the given exercise: a) Evaluate the gradient of the function \(f\) at point \(P(1, -1, -1)\). b) Find the unit vector in the maximum increase direction of \(f\) at point \(P\). c) Calculate the rate of change in the maximum increase direction at point \(P\). d) Compute the derivative in the direction of the given vector at point \(P\). Solution: a) The gradient of \(f\) at \(P(1, -1, -1)\) is \(\left\langle 4, -4, -4 \right\rangle\). b) The unit vector in the maximum increase direction at \(P\) is \(\left\langle\frac{2}{\sqrt{12}},-\frac{2}{\sqrt{12}},-\frac{2}{\sqrt{12}} \right\rangle\). c) The rate of change in the maximum increase direction at point \(P\) is \(4\sqrt{3}\). d) The directional derivative at point \(P\) in the direction of the given vector is \(\frac{4}{\sqrt{3}}\).

Step-by-step solution

Part a. Compute the gradient of \(f\) and evaluate it at \(P\).

To obtain the gradient of the function, you need to find the partial derivatives of \(f\) with respect to \(x\), \(y\), and \(z\): $$\frac{\partial f}{\partial x}=4 y z$$ $$\frac{\partial f}{\partial y}=4 x z$$ $$\frac{\partial f}{\partial z}=4 x y$$ Now, form the gradient vector by combining these partial derivatives: $$\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \left\langle 4yz, 4xz, 4xy \right\rangle$$ Next, evaluate the gradient vector at the point \(P(1, -1, -1)\): $$\nabla f (1, -1, -1) = \left\langle 4 (-1)(-1), 4 (1)(-1), 4 (1)(-1) \right\rangle = \left\langle 4, -4, -4 \right\rangle$$

Part b: Find the unit vector in the direction of maximum increase of \(f\) at \(P\).

The unit vector in the direction of maximum increase is simply the normalized gradient vector evaluated at the point P: $$\mathbf{u}_{max} = \frac{\nabla f (1, -1, -1)}{||\nabla f (1, -1, -1)||} = \frac{\left\langle 4, -4, -4 \right\rangle}{\sqrt{4^2+(-4)^2+(-4)^2}} = \frac{\left\langle 4, -4, -4\right\rangle}{\sqrt{48}} = \left\langle\frac{2}{\sqrt{12}},-\frac{2}{\sqrt{12}},-\frac{2}{\sqrt{12}} \right\rangle$$

Part c: Find the rate of change of the function in the direction of maximum increase at \(P\).

The rate of change in the maximum increase direction is simply the magnitude of the gradient vector evaluated at point P: $$R = ||\nabla f (1, -1, -1)|| = \sqrt{4^2+(-4)^2+(-4)^2} = \sqrt{48}=4\sqrt{3}$$

Part d: Find the directional derivative at \(P\) in the direction of the given vector.

To find the directional derivative in the given vector direction \(\mathbf{u}=\left\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\right\rangle\), you need to find the dot product of the gradient vector and the given unit vector at the point \(P\): $$D_\mathbf{u} f(P) = \nabla f(1, -1, -1) \cdot \mathbf{u} = \left\langle 4, -4, -4 \right\rangle \cdot \left\langle\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}} \right\rangle = \frac{4 - 4 + 4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$ The directional derivative at point P in the direction of the given vector is \(\frac{4}{\sqrt{3}}\).

Key concepts

Directional Derivative

The directional derivative is a measure of how a function changes as you move along a specific direction. Imagine you are hiking on a hill, and the hill's surface can be represented by a function. The directional derivative tells you how steep the hill is in the direction you choose to walk. Mathematically, to find the directional derivative of a function \( f \) at a point \( P \) in the direction of a unit vector \( \mathbf{u} \), you calculate the dot product between the gradient of \( f \) at \( P \) and \( \mathbf{u} \). This is written as: \[ D_\mathbf{u} f(P) = abla f(P) \cdot \mathbf{u} \] The gradient points in the direction of the steepest ascent, so the directional derivative can indicate whether you're climbing or descending, based on its sign and magnitude.

Partial Derivatives

Partial derivatives provide a way to understand how a multivariable function changes with respect to one variable while keeping others constant. They are the building blocks of the gradient. Consider a function \( f(x, y, z) \) that depends on three variables. Its partial derivative with respect to \( x \), denoted \( \frac{\partial f}{\partial x} \), measures how \( f \) changes when \( x \) changes, assuming \( y \) and \( z \) stay the same. The same goes for \( y \) and \( z \).

• \( \frac{\partial f}{\partial x} \) tells you the rate of change in the \( x \)-direction.
• \( \frac{\partial f}{\partial y} \) does the same for the \( y \)-direction.
• \( \frac{\partial f}{\partial z} \) shows it for the \( z \)-direction.

The gradient combines these into a vector, showing how the function changes in all directions.

Rate of Change

The rate of change in the context of functions of several variables can refer to various scenarios. It's essential to know this when examining gradients and directional derivatives. Often, when interested in how swiftly something changes at a point, you analyze the magnitude of the gradient \( ||abla f|| \). This represents how quickly the function \( f \) increases in the direction of greatest climb from that point. In practical terms:

• The gradient's magnitude gives you a handle on how severe the change is.
• In scenarios like temperature or elevation, a higher magnitude indicates a faster rate of temperature increase or steeper slope.

Understanding the rate of change helps in identifying features like peaks, valleys, or flat areas on surfaces represented by multivariable functions.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vectors and how they can be used to solve problems involving physical fields. This includes gradient, divergence, and curl. When we talk about the gradient in vector calculus, we are specifically referring to how a field changes at various points, and this is where gradients, partial derivatives, and vector operations come in.

• Gradient: Provides the direction and rate of the steepest ascent of a scalar function.
• Divergence and Curl: Deal with vector fields and have specific roles in understanding flow and rotational behavior.

With such tools, vector calculus allows us to solve complex physical problems, like fluid flow, electromagnetic fields, and forces in mechanical systems. It enables us to visualize and manipulate multidimensional functions effectively.