Problem 18 - Calculus Early Transcendentals

Question

Compute the directional derivative of the following functions at the given point P in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$f(x, y)=3 x^{2}+y^{3} ; P(3,2) ;\left\langle\frac{5}{13}, \frac{12}{13}\right\rangle$$

Step-by-Step Solution

Answer

The directional derivative of the function f(x, y) = 3x^2 + y^3 at the point P(3, 2) in the direction of the given vector <5/13, 12/13> is 234/13.

Step 1: Find the partial derivatives

We need to find the partial derivatives of the function f(x, y) with respect to x and y: $$f_x(x,y) = \frac{\partial f}{\partial x} = 6x \quad\text{and}\quad f_y(x,y) = \frac{\partial f}{\partial y} = 3y^2.$$ Now, evaluate these partial derivatives at the point P(3, 2): $$f_x(3, 2) = 6 \cdot 3 = 18 \quad\text{and}\quad f_y(3, 2) = 3 \cdot 2^2 = 12.$$

Step 2: Construct the gradient vector

Now that we have the partial derivatives at point P, we can form the gradient vector, which is $$\nabla f(P) = \left\langle f_x(3, 2), f_y(3, 2) \right\rangle = \left\langle 18, 12 \right\rangle.$$

Step 3: Find the dot product of the gradient vector with the given unit direction vector

Finally, we need to find the dot product of the gradient vector with the given unit direction vector: $$D_{\mathbf{u}} f(P) = \nabla f(P) \cdot \mathbf{u} = \left\langle 18, 12 \right\rangle \cdot \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle = 18 \cdot \frac{5}{13} + 12 \cdot \frac{12}{13}.$$ Evaluate the dot product: $$D_{\mathbf{u}} f(P) = 18 \cdot \frac{5}{13} + 12 \cdot \frac{12}{13} = \frac{90}{13} + \frac{144}{13} = \frac{234}{13}.$$ The directional derivative of the function f(x, y) = 3x^2 + y^3 at the point P(3, 2) in the direction of the given vector <5/13, 12/13> is $$D_{\mathbf{u}} f(P) = \frac{234}{13}.$$

Key Concepts

Partial Derivatives

Partial derivatives are a way to measure how a function changes as we slightly tweak one of its variables, while keeping the others fixed. When we have a multivariable function like \( f(x, y) = 3x^2 + y^3 \), we can consider how changes in \( x \) alone or \( y \) alone affect the value of the function.
For the function given, the partial derivative with respect to \( x \) is \( f_x(x,y) = \frac{\partial f}{\partial x} = 6x \), which considers the rate of change in the direction of the \( x \)-axis. Similarly, the partial derivative with respect to \( y \) is \( f_y(x,y) = \frac{\partial f}{\partial y} = 3y^2 \), considering changes along the \( y \)-axis.
To find the values of these partial derivatives at a particular point \( P(3, 2) \), we substitute \( x = 3 \) and \( y = 2 \) into each equation:

• \( f_x(3, 2) = 6 \cdot 3 = 18 \)
• \( f_y(3, 2) = 3 \cdot 2^2 = 12 \)

Gradient Vector

The gradient vector is a very useful tool that helps us see how a function changes in all directions at once.
For a function of two variables, the gradient vector \( abla f \) is formed by combining the partial derivatives with respect to each variable. In mathematical terms, it is expressed as:

• \( abla f = \left\langle f_x, f_y \right\rangle \)

For our function \( f(x, y) = 3x^2 + y^3 \) at the point \( P(3, 2) \), the gradient vector is

• \( abla f(P) = \left\langle 18, 12 \right\rangle \)

This vector gives us the direction in which the function increases most rapidly. Its magnitude indicates how fast the increase is.

Dot Product

The dot product is a crucial operation when working with vectors, especially when applying them to directional derivatives.
The dot product between two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) is defined as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \)

In the context of our problem, we calculate the dot product of the gradient vector \( \left\langle 18, 12 \right\rangle \) with the unit direction vector \( \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle \).

• \( D_{\mathbf{u}} f(P) = 18 \cdot \frac{5}{13} + 12 \cdot \frac{12}{13} = \frac{90}{13} + \frac{144}{13} = \frac{234}{13} \)

This calculation provides us the directional derivative, demonstrating how much \( f \) changes going directly along the specified direction vector.

Unit Vector

A unit vector has a magnitude of one and is often used to indicate direction. This is essential for calculating directional derivatives because we want to understand change exclusively based on direction, not magnitude.
In our problem, the given direction vector is \( \left\langle \frac{5}{13}, \frac{12}{13} \right\rangle \). This vector is already a unit vector since it satisfies the condition:

• The magnitude \( \sqrt{\left(\frac{5}{13}\right)^2 + \left(\frac{12}{13}\right)^2} = 1 \)

By using a unit vector, we ensure that the derivative we compute truly reflects the rate of change along the direction, without stretching or compressing it due to the vector's length. This consistency is vital for accurate interpretation of the directional derivative.