Problem 270 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=y^{10}, \quad \mathbf{u}=\langle 0,-1\rangle, \quad P=(1,-1)$$

Step-by-Step Solution

Answer

The directional derivative is 10.

Step 1: Understand the Given Problem

We are tasked with finding the directional derivative of the function \( f(x, y) = y^{10} \) at the point \( P = (1, -1) \) in the direction of the vector \( \mathbf{v} = \langle 0, -1 \rangle \). The directional derivative represents the rate of change of the function in the specified direction.

Step 2: Normalize the Direction Vector

To find the directional derivative, the direction vector must be a unit vector. Our given direction vector is \( \mathbf{v} = \langle 0, -1 \rangle \). The magnitude of this vector is \( \| \mathbf{v} \| = \sqrt{0^2 + (-1)^2} = 1 \), hence the unit vector \( \mathbf{u} \) is \( \langle 0, -1 \rangle \). This vector is already a unit vector.

Step 3: Calculate the Gradient of the Function

The gradient of \( f(x, y) = y^{10} \) is calculated by finding the partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is 0 because the function does not depend on \( x \). The partial derivative with respect to \( y \) is \( \frac{\partial}{\partial y}(y^{10}) = 10y^9 \). Thus, the gradient \( abla f \) is \( \langle 0, 10y^9 \rangle \).

Step 4: Evaluate the Gradient at Point P

Substitute the coordinates of point \( P = (1, -1) \) into the gradient \( abla f = \langle 0, 10y^9 \rangle \) to get \( abla f(1, -1) = \langle 0, 10(-1)^9 \rangle = \langle 0, -10 \rangle \).

Step 5: Compute the Directional Derivative

The directional derivative of \( f \) at point \( P \) in the direction of the unit vector \( \mathbf{u} = \langle 0, -1 \rangle \) is the dot product of \( abla f(1, -1) \) and \( \mathbf{u} \). Thus, \( D_{\mathbf{u}}f = \langle 0, -10 \rangle \cdot \langle 0, -1 \rangle = 0 \times 0 + (-10) \times (-1) = 10 \).

Key Concepts

Gradient Vector

The gradient vector is like a compass for functions. It tells you which direction to go if you want to increase the function's value the fastest. For a function of two variables, like our function \( f(x, y) = y^{10} \), the gradient is a vector. This vector contains the partial derivatives with respect to each variable. These partial derivatives indicate how the function changes as you vary one of the variables, keeping the others constant.

• For any function \( f(x, y) \), the gradient is represented as \( abla f = \langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle \).
• In our problem, \( f(x, y) = y^{10} \), so it doesn’t change with \( x \), making \( \frac{\partial f}{\partial x} = 0 \).
• For \( y \), the function changes significantly because \( \frac{\partial f}{\partial y} = 10y^9 \).

By evaluating the gradient at a specific point, you can precisely measure the function's rate of increase in every direction around that point.

Partial Derivative

Partial derivatives are fundamental in multivariable calculus. They allow us to understand how a function changes as we tweak each variable independently. For our function \( f(x, y) = y^{10} \), we compute partial derivatives with respect to both \( x \) and \( y \).

• To find the partial derivative with respect to \( x \), treat \( y \) as a constant. This gives us \( \frac{\partial}{\partial x}(y^{10}) = 0 \).
• To find the partial derivative with respect to \( y \), treat \( x \) as constant. This results in \( \frac{\partial}{\partial y}(y^{10}) = 10y^9 \).

Partial derivatives capture how a small change in one variable influences the function, keeping all other variables constant. They are building blocks for the gradient vector, guiding us on how the function behaves locally.

Unit Vector

A unit vector provides direction without changing the magnitude. It's like a focused arrow pointing in a particular direction, consisting of a length of 1. When finding directional derivatives, the direction must be expressed using a unit vector.

• For our example, we have \( \mathbf{v} = \langle 0, -1 \rangle \). This vector is already a unit vector because its length, calculated as \( \sqrt{0^2 + (-1)^2} = 1 \), confirms this.

Using a unit vector simplifies calculations of directional derivatives and allows us to precisely understand how functions change in specific directions.

Dot Product

The dot product is a key operation in vector mathematics. It helps calculate the directional derivative by combining the gradient vector and the direction of interest into a single number. The dot product measures how much one vector extends in the direction of another vector.

• For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is \( \mathbf{a} \cdot \mathbf{b} = a_1 \cdot b_1 + a_2 \cdot b_2 \).
• In our problem, the dot product \( abla f(1, -1) \cdot \mathbf{u} = \langle 0, -10 \rangle \cdot \langle 0, -1 \rangle = 0 \times 0 + (-10) \times (-1) = 10 \).

The dot product is crucial for determining how much of the gradient's direction aligns with our specified direction, ultimately giving us the rate of change in that direction.