Problem 15 - Calculus with Concepts in Calculus

Question

Let \(f(x, y)=\sinh (y \ln x)\). Find the directional derivative of \(f\) at \((2,-1)\) in the direction away from the origin.

Step-by-Step Solution

Answer

The directional derivative at \((2, -1)\) is \(-\frac{\cosh(\ln 2)(1 + \ln 2)}{\sqrt{5}}\)."

Step 1: Find the Gradient of f

The gradient of a function \(f(x, y)\) is given by \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). We need to find the partial derivatives first. The function is \(f(x, y) = \sinh(y \ln x)\). Let's calculate the partial derivatives.

Step 1.1: Compute \(\frac{\partial f}{\partial x}\)

Using the chain rule, \(\frac{\partial f}{\partial x} = y \cosh(y \ln x) \cdot \frac{1}{x}\).

Step 1.2: Compute \(\frac{\partial f}{\partial y}\)

Using the chain rule, \(\frac{\partial f}{\partial y} = \ln x \cdot \cosh(y \ln x)\).

Step 2: Evaluate the Gradient at (2, -1)

Substitute \(x = 2\) and \(y = -1\) into the gradient components: \(\frac{\partial f}{\partial x} = -1 \cdot \cosh((-1) \ln 2) \cdot \frac{1}{2}\) and \(\frac{\partial f}{\partial y} = \ln 2 \cdot \cosh((-1) \ln 2)\).

Step 3: Simplify Gradient Components at (2, -1)

Calculate \(\ln 2\), then evaluate \(\cosh((-1) \ln 2)\), and simplify: \(abla f(2, -1) = \left(-\frac{\cosh(-\ln 2)}{2}, \ln 2 \cosh(-\ln 2)\right)\).

Step 4: Determine Direction Vector from Origin to (2, -1)

The direction vector is \((2, -1)\). Normalize it by dividing by its magnitude: \(\frac{(2, -1)}{\sqrt{2^2 + (-1)^2}} = \frac{(2, -1)}{\sqrt{5}}\).

Step 5: Compute Directional Derivative

The directional derivative in the direction of \( \vec{u} \) is \(abla f \cdot \vec{u}\). Compute: \(-\frac{\cosh(-\ln 2)}{2} \cdot \frac{2}{\sqrt{5}} + \ln 2 \cosh(-\ln 2) \cdot \frac{-1}{\sqrt{5}}\).

Step 6: Simplification

Simplify the expression from the previous step using properties of \(\cosh(-x)\). The directional derivative is \(-\frac{1}{\sqrt{5}} \cosh(\ln 2) - \frac{\ln 2}{\sqrt{5}} \cosh(\ln 2)\).

Key Concepts

Gradient

The gradient vector is an essential concept in multivariable calculus that helps us understand the direction and rate of change of a function. For a function of two variables, say \(f(x, y)\), the gradient is written as \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\). It essentially combines the partial derivatives of the function, dictating the direction in which the function increases the fastest.
For example, in our function \(f(x, y) = \sinh(y \ln x)\), we calculated the gradient by finding the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\). These components give us information about how the function \(f\) changes with respect to \(x\) and \(y\), respectively.
The crucial part about gradients is understanding how they point in the direction of steepest ascent. This is why, in our exercise, we evaluated the gradient at a specific point to determine how to proceed with calculating the directional derivative.

Partial Derivatives

Partial derivatives are the building blocks for understanding how multivariable functions change. They tell us how a function changes as we vary one variable while keeping others constant.

• For the function \(f(x, y) = \sinh(y \ln x)\), we calculate the partial derivative with respect to \(x\) and \(y\).
• Using the chain rule, we find that \(\frac{\partial f}{\partial x} = y \cosh(y \ln x) \cdot \frac{1}{x}\) and \(\frac{\partial f}{\partial y} = \ln x \cdot \cosh(y \ln x)\).

These equations tell us how \(f\) changes as \(x\) and \(y\) change, respectively. Partial derivatives are crucial because they form the components of the gradient, which we use to further compute directional derivatives.

Chain Rule

The chain rule is a fundamental concept in calculus used to compute the derivative of composite functions.
When dealing with functions like \(f(x, y) = \sinh(y \ln x)\), which include combinations of different functions, the chain rule helps us break down the derivatives into manageable parts.
For instance, when finding \(\frac{\partial f}{\partial x}\), the chain rule is applied as follows: we first consider the derivative of \(\sinh(y \ln x)\) with respect to \(y \ln x\), which is \(\cosh(y \ln x)\), and then multiply it by the derivative of \(y \ln x\) with respect to \(x\), which is \(y \cdot \frac{1}{x}\). This process allows us to effectively handle derivatives of complex expressions by systematically breaking them down.

Normalization

Normalization is the process of adjusting a vector so that it becomes a unit vector, which has a magnitude of 1. In the context of directional derivatives, normalization ensures that the directional vector points in the intended direction with a standard length.
For instance, in our exercise, we calculated the direction vector from the origin to the point (2, -1) to be \((2, -1)\). The magnitude of this vector is \(\sqrt{2^2 + (-1)^2} = \sqrt{5}\).
To normalize it, we divide each component by this magnitude, yielding \(\left(\frac{2}{\sqrt{5}}, \frac{-1}{\sqrt{5}}\right)\). Normalizing the vector is essential for calculating the directional derivative because it keeps the vector's length to 1, simplifying the calculation and interpretation of how the function changes in that direction. This concept of normalization is widely used across different areas of mathematics and physics to ensure consistency when working with vectors.