Problem 4 - Calculus Early Transcendentals

Question

\(4-6\) Find the directional derivative of \(f\) at the given point in the direction indicated by the angle \(\theta .\) $$f(x, y)=x^{2} y^{3}-y^{4}, \quad(2,1), \quad \theta=\pi / 4$$

Step-by-Step Solution

Answer

The directional derivative is \(6\sqrt{2}\).

Step 1: Find the Gradient of the Function

The gradient of a function \(f(x, y)\) is the vector of its partial derivatives. For \(f(x, y) = x^2y^3 - y^4\), calculate the partial derivatives:- Partial derivative with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2y^3 - y^4) = 2xy^3 \]- Partial derivative with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2y^3 - y^4) = 3x^2y^2 - 4y^3 \]Thus, the gradient is:\[ abla f(x, y) = \langle 2xy^3, 3x^2y^2 - 4y^3 \rangle \]

Step 2: Evaluate the Gradient at the Given Point

Substitute the point \((2, 1)\) into the gradient. - Evaluate \(f_x(2, 1): 2 \times 2 \times 1^3 = 4\)- Evaluate \(f_y(2, 1): 3 \times 2^2 \times 1^2 - 4 \times 1^3 = 12 - 4 = 8\)The gradient at \((2, 1)\) is:\[ abla f(2, 1) = \langle 4, 8 \rangle \]

Step 3: Determine the Direction Vector Using Angle \(\theta\)

The direction is given by the angle \(\theta = \frac{\pi}{4}\). The direction vector based on this angle is \(\mathbf{u} = \langle \cos(\pi/4), \sin(\pi/4) \rangle\).- \(\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}\), so\[ \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \]

Step 4: Compute the Dot Product for the Directional Derivative

The directional derivative \(D_\mathbf{u}f(2,1)\) in the direction \(\mathbf{u}\) is given by the dot product of the gradient and the direction vector: \[ D_\mathbf{u}f(2, 1) = abla f(2, 1) \cdot \mathbf{u} \]Calculate the dot product:\[ D_\mathbf{u}f(2, 1) = \langle 4, 8 \rangle \cdot \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \]\[ = 4 \cdot \frac{\sqrt{2}}{2} + 8 \cdot \frac{\sqrt{2}}{2} \]\[ = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2} \]

Step 5: Conclusion

The directional derivative of the function at the point \((2, 1)\) in the direction given by the angle \(\theta = \frac{\pi}{4}\) is \(6\sqrt{2}\).

Key Concepts

Gradient

The concept of the Gradient is pivotal when exploring functions of multiple variables in calculus. Essentially, the gradient symbolizes the direction of the steepest ascent of a function. It is represented as a vector consisting of all the partial derivatives with respect to each variable.

To illustrate, consider a function of two variables, such as \( f(x, y) = x^2y^3 - y^4 \). The gradient, denoted as \( abla f(x, y) \), is formed by:

• The partial derivative with respect to \( x \), \( f_x(x, y) = 2xy^3\).
• The partial derivative with respect to \( y \), \( f_y(x, y) = 3x^2y^2 - 4y^3 \).

Thus, the gradient vector here is \( abla f(x, y) = \langle 2xy^3, 3x^2y^2 - 4y^3 \rangle \).

This vector not only directs the rate of change of the function but also gives the magnitude of change. It's particularly helpful for optimization problems, as it shows how to adjust the input variables for maximum benefit.

Partial Derivative

When working on functions with multiple variables, Partial Derivatives help us understand how the function changes with respect to one variable, while keeping others constant. It's akin to holding other measurements static while seeing how one alone influences the outcome.

For the function \( f(x, y) = x^2y^3 - y^4 \), let's determine its partial derivatives:

• Regarding \( x \), the partial derivative is \( f_x(x, y) = 2xy^3 \). Notice how \( y \) is treated as a constant here.
• For \( y \), the partial derivative is \( f_y(x, y) = 3x^2y^2 - 4y^3 \), where \( x \) is held constant.

These derivatives are crucial, especially when constructing gradient vectors. They tell us how sensitive the function is with changes in each respective variable. In calculus and multidimensional analysis, this becomes essential for identifying critical points and analyzing behavior within functions.

Dot Product

Dot Product is an essential operation in vector mathematics, connecting vectors to quantities. In the context of directional derivatives, it helps establish the rate of change of the function in a specified direction.

To find the dot product, we multiply corresponding components of two vectors and sum the results. Let's apply this principle:

• Considering the gradient vector \( abla f(2, 1) = \langle 4, 8 \rangle \).
• And, the direction vector derived from angle \( \theta \), \( \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \).

The dot product, thus, is calculated as:
\[ D_\mathbf{u}f(2, 1) = \langle 4, 8 \rangle \cdot \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2} \]

This result signifies the directional derivative, indicating the function's rate of change in that particular direction. It's a fundamental technique used broadly in physics and engineering fields for analyzing vector quantities.

Angle in Radians

Angles in radians provide a natural way to express angles in calculus due to their relation to the radius of a circle. When dealing with directional derivatives, specifying the direction often involves angles like \( \theta = \frac{\pi}{4} \).

Understanding radians is crucial:

• One radian corresponds to the angle formed when the arc length is equal to the radius of a circle.
• In situations involving trigonometry, such as calculating direction vectors, using radians simplifies the computation.

For instance, to find a direction vector using \( \theta = \frac{\pi}{4} \), use:

• \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
• \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)

Thus, the vector is \( \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \). Using radians keeps the math rigorous and precise, allowing for a seamless connection between angle measures and distance units.