Problem 268 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=3 x+4 y+7, \mathbf{u}=\left\langle\frac{3}{5}, \frac{4}{5}\right\rangle, P\left(0, \frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

The directional derivative is 5.

Step 1: Calculate the Gradient of the Function

The gradient of a function \( f(x, y) \) is a vector composed of its partial derivatives. For the given function \( f(x, y) = 3x + 4y + 7 \), its gradient is denoted as \( abla f(x, y) \). The components of the gradient are the partial derivatives with respect to \( x \) and \( y \). Calculate them as follows:\[ \frac{\partial f}{\partial x} = 3 \quad \text{and} \quad \frac{\partial f}{\partial y} = 4 \]Thus, the gradient is:\[ abla f(x, y) = \langle 3, 4 \rangle \]

Step 2: Evaluate the Gradient at Point P

We need to evaluate the gradient vector at the given point \( P(0, \frac{\pi}{2}) \). Since the gradient \( abla f(x, y) = \langle 3, 4 \rangle \) is constant (as it doesn't depend on \( x \) or \( y \)), it remains the same at every point:\[ abla f(0, \frac{\pi}{2}) = \langle 3, 4 \rangle \]

Step 3: Find the Directional Derivative

The directional derivative of a function \( f \) at a point \( P \) in the direction of a vector \( \mathbf{u} \) is given by the dot product of the gradient and the direction vector:\[ D_{\mathbf{u}}f = abla f(x, y) \cdot \mathbf{u} \]Here, \( \mathbf{u} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \). Thus, the directional derivative is:\[ D_{\mathbf{u}}f = \langle 3, 4 \rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \]Compute this dot product to find:\[ D_{\mathbf{u}}f = 3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = \frac{9}{5} + \frac{16}{5} = \frac{25}{5} = 5 \]

Key Concepts

Gradient

The gradient is a fundamental concept in multivariable calculus and is particularly useful when dealing with functions of two variables, like in this exercise. For any function \( f(x, y) \), the gradient is written as \( abla f(x, y) \), which is a vector composed of the function's partial derivatives. Think of the gradient as providing the direction of the steepest ascent of the function.

• The gradient vector points in the direction where the function increases most rapidly.
• Its magnitude gives the rate of this increase.

For our function \( f(x, y) = 3x + 4y + 7 \), calculating the gradient involves finding its partial derivatives with respect to \( x \) and \( y \). Since the terms involving \( x \) and \( y \) are linear and independent, their partial derivatives are constant:

• \( \frac{\partial f}{\partial x} = 3 \)
• \( \frac{\partial f}{\partial y} = 4 \)

As a result, the gradient \( abla f(x, y) \) is \( \langle 3, 4 \rangle \). This vector is the same everywhere because our function is linear.

Partial Derivatives

Partial derivatives are derivatives of multivariable functions taken with respect to one variable, treating all other variables as constants. They are crucial because they provide insights into how a function changes as each specific variable changes independently.In the context of \( f(x, y) = 3x + 4y + 7 \), the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \) tell us how the function changes as \( x \) or \( y \) changes separately:

• \( \frac{\partial f}{\partial x} = 3 \): This is how much \( f \) changes with a small change in \( x \) while keeping \( y \) constant.
• \( \frac{\partial f}{\partial y} = 4 \): This is how much \( f \) changes with a small change in \( y \) while keeping \( x \) constant.

Understanding partial derivatives allows us to build the gradient vector. These derivatives are like the building blocks that construct the full picture of how a function behaves locally.

Dot Product

The dot product is a way to multiply two vectors to obtain a scalar. It plays a key role in finding directional derivatives, as seen in this exercise. The directional derivative gives the rate at which a function changes as you move in a particular direction. This direction is specified by a unit vector \( \mathbf{u} \).For our exercise, we calculate the directional derivative using the dot product of the gradient \( abla f(x, y) \) and the unit direction vector \( \mathbf{u} \):

• The gradient vector \( \langle 3, 4 \rangle \) represents the direction of steepest ascent.
• The unit vector \( \mathbf{u} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \) guides us along a particular direction.

By performing the dot product \( \langle 3, 4 \rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle \), we get the directional derivative, which is 5. This is a straightforward calculation that combines the gradient's effect with the chosen direction. The dot product brings these concepts together, showing how changes in the function scale along the given vector direction.