Problem 22 - Calculus with Concepts in Calculus

Question

From (1) it follows that the directional derivative of a function \(f\) at a point is smallest in the direction opposite to the gradient of \(f\) at that point. Thus we say that a function decreases most rapidly in the direction opposite the gradient.Find the direction in which the function decreases most rapidly at the given point. $$ f(x, y)=\tan ^{-1}(x-y) ;(2,-2) $$

Step-by-Step Solution

Answer

The function decreases most rapidly in the direction \( \left( -\frac{1}{17}, \frac{1}{17} \right) \).

Step 1: Find the Gradient of the Function

The first step to finding the direction in which the function decreases most rapidly is to determine the gradient of the function, \( f(x,y) = \tan^{-1}(x-y) \). The gradient \( abla f(x, y) \) is a vector composed of the partial derivatives of \( f \) with respect to \( x \) and \( y \). Thus, calculate \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \):\[ \frac{\partial f}{\partial x} = \frac{1}{1+(x-y)^2} \] \[ \frac{\partial f}{\partial y} = - \frac{1}{1+(x-y)^2} \] So, the gradient is:\[ abla f(x, y) = \left( \frac{1}{1+(x-y)^2}, -\frac{1}{1+(x-y)^2} \right) \]

Step 2: Evaluate the Gradient at the Given Point

Substitute the point \((2, -2)\) into the gradient we found:\[ abla f(2, -2) = \left( \frac{1}{1+(2-(-2))^2}, -\frac{1}{1+(2-(-2))^2} \right) \] \[ = \left( \frac{1}{1+16}, -\frac{1}{1+16} \right) \] \[ = \left( \frac{1}{17}, -\frac{1}{17} \right) \]

Step 3: Determine the Opposite Direction to the Gradient

The function decreases most rapidly in the direction opposite to the gradient at the given point. Therefore, we need to find the negative of the gradient evaluated at the point \((2, -2)\):\[ -abla f(2, -2) = -\left( \frac{1}{17}, -\frac{1}{17} \right) = \left( -\frac{1}{17}, \frac{1}{17} \right) \]

Step 4: State the Direction of Most Rapid Decrease

The direction in which the function \( f(x, y) = \tan^{-1}(x-y) \) decreases most rapidly, at the point \((2, -2)\), is given by the vector \( \left( -\frac{1}{17}, \frac{1}{17} \right) \).

Key Concepts

Gradient Vector

In the field of multivariable calculus, the gradient vector is a crucial concept that helps us understand how a function changes as we move around in space. The gradient of a function, led by the symbol \(abla f\), is a vector that contains all the partial derivatives of the function. This makes it a concise way to show the rate and direction of change at any point in the function's domain.

When dealing with the function \(f(x, y) = \tan^{-1}(x-y)\), the gradient vector can be found by taking the partial derivatives of \(f\) with respect to \(x\) and \(y\). Here's what we have:

• \(\frac{\partial f}{\partial x} = \frac{1}{1+(x-y)^2}\)
• \(\frac{\partial f}{\partial y} = -\frac{1}{1+(x-y)^2}\)

The gradient vector, therefore, becomes:

• \(abla f(x, y) = \left( \frac{1}{1+(x-y)^2}, -\frac{1}{1+(x-y)^2} \right)\)

This tells us that the function's direction and rate of increase are determined by the gradient vector. So, by evaluating the gradient at a specific point, you can easily find the direction in which the function progresses most rapidly. The opposite of this vector will guide you in finding the direction of rapid decrease.

Partial Derivatives

Partial derivatives are central to understanding how functions behave in multidimensional spaces. They describe how a function changes as one variable is varied, while all other variables are held constant. For any function \(f(x, y)\), the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) provide insights into how the function varies along the \(x\) and \(y\) directions, respectively.

In the given function, \(f(x, y) = \tan^{-1}(x-y)\), we calculate the partial derivatives as follows:

• With respect to \(x\), \(\frac{\partial f}{\partial x} = \frac{1}{1+(x-y)^2}\)
• With respect to \(y\), \(\frac{\partial f}{\partial y} = -\frac{1}{1+(x-y)^2}\)

These derivatives tell us that, if we increase \(x\) slightly while keeping \(y\) constant, the function will increase because the derivative is positive. Conversely, if we increase \(y\) slightly, keeping \(x\) constant, the function decreases since the derivative is negative. Partial derivatives thus help us chart the path and slope of a change for each variable individually, forming the building blocks of the gradient vector.

Rapid Decrease Direction

Finding the rapid decrease direction involves identifying the direction opposite to the gradient vector. This is because the gradient vector points towards the steepest ascent, and the opposite direction leads to the steepest descent. This concept is vital when calculating how quickly a function value can be reduced at any point.

For the function \(f(x, y) = \tan^{-1}(x-y)\), we calculated the gradient vector at the point \((2, -2)\) as follows:

• \(abla f(2, -2) = \left( \frac{1}{17}, -\frac{1}{17} \right)\)

To find the direction of rapid decrease, take the opposite vector:

• \(-abla f(2, -2) = \left( -\frac{1}{17}, \frac{1}{17} \right)\)

Thus, the vector \(\left( -\frac{1}{17}, \frac{1}{17} \right)\) signifies the direction in which the function decreases the fastest at the point \((2, -2)\). Understanding this direction is especially valuable in optimization problems and in analyzing how systems behave under certain conditions.