Problem 287 - Calculus Volume 3

Question

For the following exercises, find the derivative of the function at \(P\) in the direction of \(\mathbf{u}\). $$ f(x, y)=\ln (5 x+4 y), P(3,9), \quad \mathbf{u}=6 \mathbf{i}+8 \mathbf{j}$$

Step-by-Step Solution

Answer

The derivative in the direction of \( \mathbf{u} \) at \( P(3,9) \) is approximately 0.1216.

Step 1: Understand the Derivative in a Given Direction

To find the derivative of the function at point \( P \) in the direction of vector \( \mathbf{u} \), we need to calculate the directional derivative. The directional derivative \( D_\mathbf{u} f \) of a function \( f(x,y) \) at a point \( P(x_0, y_0) \) in the direction of a vector \( \mathbf{u} = \langle a, b \rangle \) is given by:\[D_\mathbf{u} f(x_0, y_0) = abla f(x_0, y_0) \cdot \frac{\mathbf{u}}{\|\mathbf{u}\|}\]where \( abla f(x_0, y_0) \) is the gradient of \( f \) at the point \( P \).

Step 2: Find the Gradient of the Function

The first step in calculating the directional derivative is to find the gradient \( abla f(x,y) \) of the function \( f(x,y) = \ln(5x + 4y) \). The gradient is a vector of partial derivatives:\[abla f(x,y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\]Calculating these, we get:\[\frac{\partial f}{\partial x} = \frac{5}{5x + 4y}, \quad \frac{\partial f}{\partial y} = \frac{4}{5x + 4y}\]

Step 3: Evaluate the Gradient at the Point

Now, substitute the point \( P(3, 9) \) into the gradient found in the previous step:\[\frac{\partial f}{\partial x} \bigg|_{(3,9)} = \frac{5}{5(3) + 4(9)} = \frac{5}{ 15 + 36 } = \frac{5}{51} \]\[\frac{\partial f}{\partial y} \bigg|_{(3,9)} = \frac{4}{5(3) + 4(9)} = \frac{4}{51}\]Thus, \( abla f(3,9) = \left( \frac{5}{51}, \frac{4}{51} \right) \).

Step 4: Normalize the Direction Vector

The next step is to normalize the direction vector \( \mathbf{u} = \langle 6, 8 \rangle \). The length (magnitude) of \( \mathbf{u} \) is given by:\[\|\mathbf{u}\| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\]Therefore, the unit vector in the direction of \( \mathbf{u} \) is:\[\frac{\mathbf{u}}{\|\mathbf{u}\|} = \left(\frac{6}{10}, \frac{8}{10}\right) = \left(0.6, 0.8\right) \]

Step 5: Calculate the Dot Product

Now calculate the directional derivative using the dot product:\[D_\mathbf{u} f(3, 9) = abla f(3, 9) \cdot \frac{\mathbf{u}}{\|\mathbf{u}\|}\]\[D_\mathbf{u} f(3, 9) = \left( \frac{5}{51}, \frac{4}{51} \right) \cdot \left(0.6, 0.8\right)\]\[= \frac{5}{51} \cdot 0.6 + \frac{4}{51} \cdot 0.8\]\[= \frac{3}{51} + \frac{3.2}{51}\]\[= \frac{6.2}{51} \approx 0.1216\]

Step 6: Conclusion

The derivative of the function \( f(x, y) = \ln(5x + 4y) \) at the point \( P(3, 9) \) in the direction of the vector \( \mathbf{u} = 6\mathbf{i} + 8\mathbf{j} \) is approximately 0.1216.

Key Concepts

Gradient

In calculus, a gradient is a vector that indicates how a function changes at any given point. It consists of partial derivatives with respect to each variable of the function. For a function \( f(x, y) \), the gradient \( abla f \) is written as:

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

This means the gradient points in the direction of the steepest increase of the function and its magnitude gives the rate of increase in that direction.
The gradient is a fundamental tool because it helps determine the behavior of multivariable functions, making it vital in optimization and in finding tangent planes at points.

Partial Derivatives

Partial derivatives are used to find the rate of change of multivariable functions with respect to one variable, keeping the others constant. For a function like \( f(x, y) = \ln(5x + 4y) \), you find the partial derivative with respect to \( x \) by differentiating while treating \( y \) as a constant:

• \( \frac{\partial f}{\partial x} = \frac{5}{5x + 4y} \)

Similarly, the partial derivative with respect to \( y \):

• \( \frac{\partial f}{\partial y} = \frac{4}{5x + 4y} \)

These allow us to understand how changes in each variable impact the function, which is crucial for building the gradient vector.

Normalizing Vectors

Normalizing a vector involves scaling it to have a length of one, which is often needed when working with directional vectors. To normalize, divide each component of the vector by its magnitude. For the vector \( \mathbf{u} = \langle 6, 8 \rangle \), calculate the magnitude:

• \( \|\mathbf{u}\| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \)

Then, divide each component by the magnitude:

• Normalized vector: \( \left( \frac{6}{10}, \frac{8}{10} \right) = (0.6, 0.8) \)

This unit vector keeps the direction of \( \mathbf{u} \) while standardizing its length, making it easier to work with in calculations like finding the directional derivative.

Dot Product

The dot product is a fundamental operation that combines two vectors to produce a scalar. It's calculated by multiplying corresponding components of the vectors and summing the results. In the context of the directional derivative, the dot product is used to find how much of the gradient is going in the direction of the vector \( \mathbf{u} \):

• If \( abla f(3,9) = \left( \frac{5}{51}, \frac{4}{51} \right) \) and the normalized vector is \( (0.6, 0.8) \), the dot product is calculated as:
• \( D_\mathbf{u} f(3, 9) = \left( \frac{5}{51}\right) \cdot 0.6 + \left( \frac{4}{51} \right) \cdot 0.8 \)
• The result gives the rate of change of the function in that specific direction.

Understanding the dot product is essential because it helps consolidate the influence of multiple variables into a single derivative value when considering directions.