Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-8, find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=x^{2} y ; \mathbf{p}=(1,2) ; \mathbf{a}=3 \mathbf{i}-4 \mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative is \(\frac{8}{5}\).

Step 1: Understand the Concept

The directional derivative of a function \(f(x, y)\) in the direction of a vector \(\mathbf{a}\) at point \(\mathbf{p}\) measures the rate at which the function changes at \(\mathbf{p}\) in that direction. It is formulated as \( D_\mathbf{u}f(p) = abla f(p) \cdot \mathbf{u} \), where \(\mathbf{u}\) is the unit vector in the direction of \(\mathbf{a}\).

Step 2: Compute the Gradient \(\nabla f(x,y)\)

The gradient \(abla f(x, y)\) is a vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = x^2 y \), calculate the partial derivatives:- \( \frac{\partial f}{\partial x} = 2xy \)- \( \frac{\partial f}{\partial y} = x^2 \)Thus, \( abla f(x, y) = (2xy, x^2) \).

Step 3: Evaluate the Gradient at \(\mathbf{p}\)

Substitute \(\mathbf{p} = (1, 2)\) into the gradient:\( abla f(1, 2) = (2 \cdot 1 \cdot 2, 1^2) = (4, 1) \).

Step 4: Determine Unit Vector \(\mathbf{u}\)

Normalize the direction vector \( \mathbf{a} = 3\mathbf{i} - 4\mathbf{j} \) to get the unit vector \(\mathbf{u}\). The magnitude of \(\mathbf{a}\) is \(\| \mathbf{a} \| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\).Therefore, \(\mathbf{u} = \left( \frac{3}{5}, \frac{-4}{5} \right) \).

Step 5: Compute the Directional Derivative

The directional derivative is computed as the dot product: \(D_\mathbf{u}f(1, 2) = abla f(1, 2) \cdot \mathbf{u} = (4, 1) \cdot \left( \frac{3}{5}, \frac{-4}{5} \right) \).Calculate the dot product:\(4 \cdot \frac{3}{5} + 1 \cdot \frac{-4}{5} = \frac{12}{5} - \frac{4}{5} = \frac{8}{5}\).

Step 6: Conclusion

The directional derivative of \( f(x, y) = x^2 y \) at the point \( (1, 2) \) in the direction of the vector \( 3\mathbf{i} - 4\mathbf{j} \) is \( \frac{8}{5} \).

Key Concepts

Gradient

The gradient of a function is a vector that points in the direction of the greatest rate of increase of the function. If you have a function of two variables like \(f(x, y)\), the gradient \(abla f(x, y)\) is made up of the partial derivatives of \(f\) with respect to each variable. This gives you a cool tool to find out how a function changes in multi-dimensional space.
The full formula for the gradient of a function \(f(x, y)\) is:

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

In this exercise, you calculated the partials for the function:

• \(\frac{\partial f}{\partial x} = 2xy\)
• \(\frac{\partial f}{\partial y} = x^2\)

So, the gradient \(abla f(x, y)\) became \((2xy, x^2)\). At the point \(\mathbf{p} = (1, 2)\), this evaluated to \((4, 1)\).
Understanding the gradient helps you see how your function behaves at a point and guides you toward the optimal direction for various applications like optimization and robotics.

Partial Derivative

Partial derivatives are a foundational concept in calculus, especially when dealing with functions of several variables. They show how a function changes as one of its input variables is varied, keeping the others constant.
For example, if your function is \(f(x, y) = x^2 y\), then the partial derivative with respect to \(x\) is \(2xy\) while keeping \(y\) unchanged. Similarly, the partial with respect to \(y\) is \(x^2\), keeping \(x\) constant.
This concept gets tricky sometimes because you're only focusing on one variable at a time. But it's useful for creating the gradient or finding tangent planes to a surface.
Think of partial derivatives as the ingredients you need to fully analyze how a multivariable function behaves. They are the building blocks for gradients and other vector calculus operations.

Unit Vector

A unit vector is a vector that has a magnitude of 1. It helps to signify direction without influencing magnitude with size. In directional derivatives, we use unit vectors to determine the direction for measuring change in the function.
To find a unit vector from any vector like \(\mathbf{a} = 3\mathbf{i} - 4\mathbf{j}\), you must normalize it. This means you divide each component of the vector by its magnitude.
The magnitude \(\| \mathbf{a} \|\) is calculated as:

• \(\sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5\)

Then, normalize \(\mathbf{a}\) to get the unit vector \(\mathbf{u} = \left( \frac{3}{5}, \frac{-4}{5} \right)\).
This process is vital because it ensures that the direction is preserved, but measures like angles and rates remain accurate — an essential aspect when calculating directional derivatives.