Problem 296 - Calculus 3

Question

For the following exercises, find the derivative of the function. \(f(x, y)=\arctan \left(\frac{y}{x}\right)\) at point \((-9,9) \quad\) in the direction the function increases most rapidly

Step-by-Step Solution

Answer

The direction of most rapid increase is along the vector \((-1/18, -1/18)\).

Step 1: Understand the Context

To find the direction in which the function increases most rapidly, we need to compute the gradient of the function. The gradient is a vector that points in the direction of the greatest rate of increase.

Step 2: Setting Up for Derivatives

The function given is \(f(x, y) = \arctan \left(\frac{y}{x}\right)\). This is a composition of the \(\arctan\) function and the division \(\frac{y}{x}\). We need to calculate the partial derivatives with respect to \(x\) and \(y\).

Step 3: Compute Partial Derivative with Respect to x

Using the chain rule, the partial derivative of \(f(x,y)\) with respect to \(x\) is \(f_x = \frac{d}{dx} \left(\arctan \left(\frac{y}{x}\right)\right) = \frac{-y}{x^2 + y^2}\). This is because the derivative of \(\arctan(u)\) is \(\frac{1}{1+u^2}\), and the derivative of \(\frac{y}{x}\) with respect to \(x\) is \(\frac{-y}{x^2}\).

Step 4: Compute Partial Derivative with Respect to y

Similarly, using the chain rule, the partial derivative with respect to \(y\) is \(f_y = \frac{x}{x^2 + y^2}\). The partial derivative of \(\frac{y}{x}\) with respect to \(y\) is \(\frac{1}{x}\).

Step 5: Evaluate Gradient at Point

Now evaluate these partial derivatives at the point \((-9, 9)\). For \(f_x(-9, 9)\): \[f_x(-9, 9) = \frac{-9}{(-9)^2 + 9^2} = \frac{-9}{162} = -\frac{1}{18}\]For \(f_y(-9, 9)\): \[f_y(-9, 9) = \frac{-9}{(-9)^2 + 9^2} = \frac{-9}{162} = -\frac{1}{18}\].

Step 6: Form Gradient Vector

The gradient vector at \((-9, 9)\) is\[ abla f(-9, 9) = \left(f_x(-9, 9), f_y(-9, 9)\right) = \left(-\frac{1}{18}, \frac{-1}{18}\right).\]

Step 7: Direction of Rapid Increase

The direction of greatest increase of the function is the same as the direction of the gradient vector. Thus, the function increases most rapidly in the direction of the vector \(-\frac{1}{18}, \frac{-1}{18}\).

Key Concepts

Partial Derivative

In calculus, partial derivatives are used to take the derivative of a function with more than one variable. When you have a function of two variables, such as \(f(x, y) = \arctan \left(\frac{y}{x}\right)\), you can differentiate with respect to each variable independently.
The partial derivative with respect to \(x\) is denoted as \(f_x\) and with respect to \(y\), it’s \(f_y\).

• To find \(f_x\) in this example, we treat \(y\) as a constant and differentiate \(\frac{y}{x}\) as though \(x\) is the only variable.
• Similarly, for \(f_y\), we treat \(x\) as constant.

Partial derivatives help us understand how a function changes with respect to each variable independently, essential in multi-variable calculus and various fields such as physics and engineering.

Chain Rule

The chain rule is a fundamental tool in calculus, used to differentiate composite functions — functions made up by combining two or more functions.
In the context of our example, the function \( \arctan \left(\frac{y}{x}\right)\) can be seen as the composition of the function \( \arctan(u)\) with \(u = \frac{y}{x}\).

• To differentiate this using the chain rule, we first differentiate \( \arctan(u)\) about \(u\), which gives \( \frac{1}{1+u^2}\).
• Then, we multiply this by the derivative of \(u\) with respect to the chosen variable (either \(x\) or \(y\)).

This rule makes it possible to compute derivatives of complex functions efficiently and accurately.

Directional Derivative

The directional derivative provides the rate of change of a function in any specified direction.
It gives more insight than just knowing how a function changes with one variable at a time.
To calculate the directional derivative of a function \(f\) at a point along a vector \(\mathbf{v}\), the following formula is used:
\[ D_\mathbf{v}f = abla f \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|} \]

• Here, \(abla f\) is the gradient of \(f\), and it points in the direction of the steepest ascent.
• The vector \(\mathbf{v}\) is made into a unit vector by dividing by its magnitude \(\|\mathbf{v}\|\).
• The dot product \(abla f \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|}\) gives the rate of increase in the direction of \(\mathbf{v}\).

Understanding directional derivatives is crucial for fields like gradient descent optimization in machine learning.

Rate of Increase

The rate of increase of a function informs us about how quickly the function's value is changing, and in what direction.
In multivariable calculus, this is linked closely with the gradient. At a given point, the gradient vector not only gives the direction of the quickest increase but also its magnitude.

• The magnitude of the gradient vector represents the maximum rate of increase at that point.
• For a function \(f(x, y)\), the gradient \(abla f = (f_x, f_y)\)
• If \(abla f = \left(-\frac{1}{18}, -\frac{1}{18}\right)\), then the magnitude is found using \( \sqrt{f_x^2 + f_y^2} \).

This concept can be applied to various real-world problems, such as determining the steepest path down a hill, or maximum profit increment in a business model.