Problem 270 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The directional derivative of \( f(x, y) = y^{10} \) at \( P = (1, -1) \) in the direction of \( \mathbf{u} = \langle 0, -1 \rangle \) is 10.

Step 1: Understanding the Concept of Directional Derivative

The directional derivative of a function at a given point in a specific direction measures the rate of change of the function in that direction. It can be calculated using the gradient vector.

Step 2: Find the Gradient of the Function

Calculate the gradient, \( abla f \), of the function \( f(x, y) = y^{10} \). The gradient is given by:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]For the given function, \( \frac{\partial f}{\partial x} = 0 \) and \( \frac{\partial f}{\partial y} = 10y^9 \). So, \( abla f = (0, 10y^9) \).

Step 3: Evaluate the Gradient at Point P

Substitute the coordinates of point \( P = (1, -1) \) into the gradient:\[ abla f(1, -1) = (0, 10(-1)^9) = (0, -10) \]

Step 4: Normalize the Direction Vector

Normalize the direction vector \( \mathbf{u} = \langle 0, -1 \rangle \). Calculate its magnitude:\[ |\mathbf{u}| = \sqrt{0^2 + (-1)^2} = 1 \]So, the unit vector \( \mathbf{u} \) is already normalized: \( \mathbf{u} = \langle 0, -1 \rangle \).

Step 5: Compute the Directional Derivative

The directional derivative \( D_{\mathbf{u}} f \) is given by the dot product of the gradient and the unit vector \( \mathbf{u} \):\[ D_{\mathbf{u}} f = abla f \cdot \mathbf{u} = (0, -10) \cdot (0, -1) \]Calculate the dot product:\[ D_{\mathbf{u}} f = 0 \times 0 + (-10) \times (-1) = 10 \]

Key Concepts

Gradient Vector

To understand the concept of a directional derivative, one must first grasp the idea of the gradient vector. The gradient vector, often symbolized as \( abla f \), is a vector that contains all the partial derivatives of a function and points in the direction of the steepest ascent of that function. For a function \( f(x, y) \), the gradient vector is given by:

• \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \)

In our example, the function is \( f(x, y) = y^{10} \). Thus, the gradient is \( abla f = (0, 10y^9) \). Notice how the gradient tells us how much the function changes as we slightly vary \( x \) and \( y \). For this specific function, since \( x \) doesn't appear in \( y^{10} \), the gradient with respect to \( x \), \( \frac{\partial f}{\partial x} \), is zero. This means changes in \( x \) don't affect the function. Understanding these derivatives provide insights into how functions behave locally.

Partial Derivatives

Partial derivatives are a fundamental tool in calculus which allows us to study the change of multivariable functions. When a function has several variables, like \( f(x, y) \), a partial derivative represents how that function changes as one variable changes, keeping the others constant.

• The partial derivative \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) varies, locking \( y \) in place.
• Conversely, \( \frac{\partial f}{\partial y} \) measures change in \( f \) when only \( y \) changes.

In our context, for \( f(x, y) = y^{10} \), we calculate:

• \( \frac{\partial f}{\partial x} = 0 \)
• \( \frac{\partial f}{\partial y} = 10y^9 \)

This analysis highlights how one variable can dominate the behavior of a function. Here, change in \( y \) significantly influences \( f \), but change in \( x \) does not at all. Understanding partial derivatives is critical to using gradients effectively.

Dot Product

To find the directional derivative, we use a concept called the dot product. The dot product is a mathematical operation that combines two vectors to produce a scalar. For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)

This operation tells us how much one vector points in the direction of the other. It's especially useful in computing directional derivatives because it directly tells us the rate of change in a specific direction.

For our example, using the gradient vector \( abla f = (0, -10) \) and the direction vector \( \mathbf{u} = \langle 0, -1 \rangle \), the dot product is:

• \( D_{\mathbf{u}} f = (0 \times 0) + (-10) \times (-1) = 10 \)

This calculation implies that in the negative \( y \) direction, the function changes at a rate of 10. The dot product is thus a crucial step in finding how quickly functions change in specific directions.