Problem 28 - Advanced Engineering Mathematics

Question

In Problems, find a vector that gives the direction in which the given function decreases most rapidly at the indicated point. Find the minimum rate. $$ f(x, y)=x^{3}-y^{3} ;(2,-2) $$

Step-by-Step Solution

Answer

The direction of most rapid decrease is \((-12, 12)\) with a rate of \(-12\sqrt{2}\).

Step 1: Compute the Gradient of f

The gradient of a function \( f(x, y) \) is a vector of its partial derivatives. Compute the partial derivative of \( f(x, y) = x^3 - y^3 \) with respect to \( x \) to get \( \frac{\partial f}{\partial x} = 3x^2 \). Compute the partial derivative with respect to \( y \) to get \( \frac{\partial f}{\partial y} = -3y^2 \). So, the gradient is \( abla f = (3x^2, -3y^2) \).

Step 2: Evaluate the Gradient at the Given Point

To find the direction in which the function decreases most rapidly at the point \((2, -2)\), substitute \( x = 2 \) and \( y = -2 \) into the gradient. Thus, \( abla f(2, -2) = (3(2)^2, -3(-2)^2) = (12, -12) \).

Step 3: Determine the Direction of Most Rapid Decrease

The direction of most rapid decrease is the negative of the gradient. So, the direction vector is \(-abla f(2, -2) = (-12, 12)\).

Step 4: Calculate the Minimum Rate of Decrease

The minimum rate of decrease is the magnitude of the gradient vector \( abla f \) evaluated at the point \((2, -2)\). Calculate the magnitude: \( \|abla f(2, -2)\| = \sqrt{12^2 + (-12)^2} = \sqrt{144 + 144} = \sqrt{288} = 12\sqrt{2} \). The minimum rate of decrease is \( -12\sqrt{2} \).

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, understanding partial derivatives is crucial. A partial derivative represents how a function changes as one specific variable changes, while all other variables are held constant. This concept is akin to taking a regular derivative, but it dissects the function's responsiveness to individual variables. For our function \(f(x, y) = x^3 - y^3\), the partial derivative with respect to \(x\) is calculated by treating \(y\) as a constant, resulting in \(\frac{\partial f}{\partial x} = 3x^2\). Similarly, the partial derivative with respect to \(y\) is obtained by treating \(x\) as constant, leading to \(\frac{\partial f}{\partial y} = -3y^2\). These derivatives provide insight into the slope or change rate of the function in each direction separately.

Gradient Vector

In multivariable calculus, the gradient vector, often denoted as \(abla f\), plays a pivotal role. It is a vector composed of the function's partial derivatives. For the function \(f(x, y) = x^3 - y^3\), the gradient is \(abla f = (3x^2, -3y^2)\). Think of the gradient as an arrow pointing in the direction of the steepest ascent from any given point. If you evaluate this vector at a specific point, for instance, \((2, -2)\), you substitute into the gradient and get \((12, -12)\). This vector not only tells us how steeply the function is rising but also can be used to find how steeply it might be decreasing by just taking its negative.

Rate of Decrease

The rate of decrease pertains to how fast a function is reducing its value. To find this, look at the direction opposite to the gradient, because the gradient indicates the direction of maximum increase. When you want to know how quickly it is decreasing, you find the magnitude of the gradient and consider its negative. For \(f(x, y)\) at \((2, -2)\), the gradient is \((12, -12)\), and thus its magnitude is determined by \(\|abla f(2, -2)\| = \sqrt{12^2 + (-12)^2} = 12\sqrt{2}\). Hence, the rate of decrease is \(-12\sqrt{2}\). This means the function decreases most rapidly in the direction \((-12, 12)\) with this specific rate.

Critical Points in Multivariable Calculus

Critical points are where the gradient vector is zero, meaning the function's slope is flat with respect to changes in all directions. To identify them for functions like \(f(x, y)\), you set \(abla f\) to zero and solve for the variables. This involves solving the equations \(3x^2 = 0\) and \(-3y^2 = 0\), leading to potential critical points at \(x = 0\) and \(y = 0\). These points are where local minima, maxima, or saddle points may occur. In optimization problems, finding critical points helps determine where the function could have significant behavior changes, crucial for understanding and solving equations in multivariable calculus contexts.