Problem 284 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(Q\). $$ f(x, y)=x^{2}+3 y^{2}, P(1,1), \quad Q(4,5)$$

Step-by-Step Solution

Answer

The directional derivative is 6.

Step 1: Determine the gradient of the function

To find the directional derivative, we first need to find the gradient of the function \( f(x, y) = x^2 + 3y^2 \). The gradient \( abla f(x, y) \) is the vector of partial derivatives. Compute each partial derivative: - The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x \). - The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 6y \). Thus, \( abla f(x, y) = (2x, 6y) \).

Step 2: Evaluate the gradient at point P

Substitute the coordinates of point \( P(1,1) \) into the gradient vector \( abla f(x, y) \). - \( abla f(1, 1) = (2 \times 1, 6 \times 1) = (2, 6) \). Thus, at point \( P \), the gradient is \( (2, 6) \).

Step 3: Determine the vector from P to Q

Find the vector from point \( P(1,1) \) to point \( Q(4,5) \). This is done by subtracting the coordinates of \( P \) from \( Q \).- Vector \( \overrightarrow{PQ} = (4-1, 5-1) = (3, 4) \). This is the direction vector.

Step 4: Normalize the direction vector

The directional derivative uses a unit direction vector. Normalize \( \overrightarrow{PQ} = (3,4) \) by dividing by its magnitude.- The magnitude of \( \overrightarrow{PQ} \) is \( \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \).- The unit vector in the direction of \( \overrightarrow{PQ} \) is \( \left(\frac{3}{5},\frac{4}{5}\right) \).

Step 5: Calculate the directional derivative

The directional derivative of \( f \) at \( P \) in the direction of \( \overrightarrow{PQ} \) is the dot product of the gradient \( abla f(1, 1) \) and the unit direction vector. - Compute \( abla f(1, 1) \cdot \left(\frac{3}{5}, \frac{4}{5}\right) = (2, 6) \cdot \left(\frac{3}{5}, \frac{4}{5}\right) = 2 \times \frac{3}{5} + 6 \times \frac{4}{5} = \frac{6}{5} + \frac{24}{5} = \frac{30}{5} = 6 \).

Step 6: Conclusion: Directional Derivative Result

The directional derivative of \( f(x, y) = x^2 + 3y^2 \) at point \( P(1,1) \) in the direction of point \( Q(4,5) \) is \( 6 \).

Key Concepts

Gradient Vector

Imagine the gradient vector as a tool that tells us how steep the 'slope' of a function is at any given point. It is essentially a vector that contains both the direction and the rate of the steepest ascent of the function. For a function of two variables, like in this exercise, it is made up of partial derivatives.Formally, the gradient \( abla f(x, y) \) is expressed as a vector where each component corresponds to the function's partial derivatives.

• The first component is the partial derivative with respect to the first variable (\( x \)).
• The second component is with respect to the second variable (\( y \)).

In our case, the partial derivative of \( f(x, y) = x^2 + 3y^2 \) with respect to \( x \) is \( 2x \), and with respect to \( y \) is \( 6y \). Therefore, the gradient is \( (2x, 6y) \). When evaluated at point \( P(1,1) \), this becomes \( (2, 6) \), providing crucial data about how the function behaves around this point.

Partial Derivatives

Partial derivatives are like the function's version of regular derivatives, but one variable at a time. They help us understand how a function changes as one variable changes, while keeping others constant.For a function \( f(x, y) = x^2 + 3y^2 \), the partial derivative with respect to \( x \) \( \left( \frac{\partial f}{\partial x} \right) \) is calculated by differentiating \( f \) as if \( y \) were a constant.

• Here, \( \frac{\partial f}{\partial x} = 2x \).

Similarly, for \( y \), treating \( x \) as constant:

• \( \frac{\partial f}{\partial y} = 6y \).

These derivatives give us the coefficients of the gradient vector, revealing the function's rate of change along the axes.

Unit Vector

A unit vector is a vector with a magnitude of one, pointing in a particular direction. It's like a compass direction, providing an "aim" without affecting size.To turn a vector into a unit vector, divide it by its magnitude. Let's take vector \( \overrightarrow{PQ} = (3, 4) \).

• First, find the magnitude: \( \sqrt{3^2 + 4^2} = 5 \).
• Then, normalize \( \overrightarrow{PQ} \) by dividing each component by 5: \( \left( \frac{3}{5}, \frac{4}{5} \right) \).

This unit vector points in the same direction as \( \overrightarrow{PQ} \), but with a length of 1, essential for the directional derivative.

Dot Product

The concept of a dot product is like a mathematical way to mix two vectors together. It gives a scalar (a single number), representing how much of one vector goes in the direction of another.To compute a dot product, multiply corresponding components of two vectors and add the results. For the gradient vector \( abla f(1, 1) = (2, 6) \) and the unit vector \( \left( \frac{3}{5}, \frac{4}{5} \right) \):

• Multiply: \( 2 \times \frac{3}{5} = \frac{6}{5} \)
• Multiply: \( 6 \times \frac{4}{5} = \frac{24}{5} \)
• Add them: \( \frac{6}{5} + \frac{24}{5} = \frac{30}{5} = 6 \)

The result is the directional derivative, providing how fast the function changes in that direction at \( P \).