Problem 20 - APEX Calculus

Question

A function \(z=f(x, y)\) and a point \(P\) are given. (a) Find the direction of maximal increase of \(f\) at \(P\). (b) What is the maximal value of \(D_{\vec{u}} f\) at \(P ?\) (c) Find the direction of minimal increase of \(f\) at \(P\). (d) Give a direction \(\vec{u}\) such that \(D_{\vec{u}} f=0\) at \(P\). Note: these are the same functions and points as in Exercises 13 through 18 . $$ f(x, y)=\sin x \cos y, P=\left(\frac{\pi}{4}, \frac{\pi}{3}\right) $$

Step-by-Step Solution

Answer

Maximal increase is in the direction \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{6}}{2} \right) \) with a value \( \frac{1}{2} \). Minimal increase is opposite, and any orthogonal vector gives zero directional derivative.

Step 1: Find the Gradient of f

The gradient of the function \( f(x, y) = \sin x \cos y \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Compute these partial derivatives:\[ \frac{\partial f}{\partial x} = \cos x \cos y \]\[ \frac{\partial f}{\partial y} = -\sin x \sin y \]So the gradient is \( abla f = (\cos x \cos y, -\sin x \sin y) \).

Step 2: Evaluate the Gradient at Point P

Substitute \( P = \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \) into the gradient to find \( abla f(P) \).\[ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \]\[ abla f(P) = \left( \frac{\sqrt{2}}{2} \cdot \frac{1}{2}, - \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} \right) = \left( \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \right) \]

Step 3: Determine Direction of Maximal Increase

The direction of maximal increase of \( f \) at \( P \) is given by the direction of the gradient \( abla f(P) \). This vector is \( \left( \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \right) \). Normalize this vector by finding its magnitude:\[ \|abla f(P)\| = \sqrt{\left(\frac{\sqrt{2}}{4}\right)^2 + \left(-\frac{\sqrt{6}}{4}\right)^2} = \sqrt{\frac{2}{16} + \frac{6}{16}} = \frac{1}{2} \]The unit vector in the direction of maximal increase is \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{6}}{2} \right) \).

Step 4: Compute Maximal Value of Directional Derivative

The maximal value of \( D_{\vec{u}}f \) at \( P \) is \( \|abla f(P)\| \), which we computed as \( \frac{1}{2} \).

Step 5: Determine Direction of Minimal Increase

The direction corresponding to minimal increase is opposite to the direction of maximal increase. This direction is given by \( -abla f(P) \), i.e., \( \left( -\frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{2} \right) \).

Step 6: Direction for Zero Directional Derivative

A direction vector \( \vec{u} \) such that \( D_{\vec{u}} f = 0 \) is orthogonal to \( abla f(P) \). As \( abla f(P) = \left( \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \right) \), a vector orthogonal to this can be \( \left( -\frac{\sqrt{6}}{4}, -\frac{\sqrt{2}}{4} \right) \), ensuring the dot product is zero.

Key Concepts

Gradient Vector

The gradient vector of a function is a powerful tool in multi-variable calculus. For a function of two variables, say \( f(x, y) \), the gradient is denoted \( abla f \) and is a vector that contains all the partial derivatives of the function.

In mathematical terms, this means \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This vector essentially acts like an indicator for the direction in which the function increases most quickly.

Imagine standing on a hill, and you're trying to figure out which direction will lead you up the steepest path. The gradient vector at any point on a surface or graph serves that purpose; it points to the direction of the steepest ascent.

In our exercise example, for the function \( f(x, y) = \sin x \cos y \), the partial derivatives were calculated to find the gradient vector. Once these derivatives are computed, substituting the values for the specific point \( P \) helps in finding the gradient at that point.

Partial Derivatives

Partial derivatives are a foundational concept in calculus, particularly when dealing with functions of several variables. These derivatives give us the rate at which the function changes as each variable is altered, while all other variables are held constant.

For a function \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( \frac{\partial f}{\partial x} \), measures how the function changes as \( x \) changes while \( y \) remains fixed. Similarly, \( \frac{\partial f}{\partial y} \) observes the change with respect to \( y \).

In our exercise, we found these partial derivatives directly as:

• \( \frac{\partial f}{\partial x} = \cos x \cos y \)
• \( \frac{\partial f}{\partial y} = -\sin x \sin y \)

These calculations are crucial for forming the gradient vector, as they represent the components of this vector.

Maximal Increase Direction

The direction of maximal increase for a function at a point is determined by the gradient vector. As mentioned earlier, the gradient vector points in the direction where the function increases most rapidly.

To make this direction a unit vector, you normalize the gradient. This means you scale the vector so its length is one. The length or magnitude of the gradient vector is calculated using the formula \( \sqrt{\left(\frac{\sqrt{2}}{4}\right)^2 + \left(-\frac{\sqrt{6}}{4}\right)^2} \).

In our exercise, this magnitude turned out to be \( \frac{1}{2} \).

Then, the direction of maximal increase at the point \( P \) \( \left( \frac{\pi}{4}, \frac{\pi}{3} \right) \) is given by the unit vector \( \left( \frac{\sqrt{2}}{2}, -\frac{\sqrt{6}}{2} \right) \). This direction reflects the steepest uphill path on our function's surface starting from point \( P \).

Orthogonal Vectors

Orthogonal vectors occur frequently in calculus and linear algebra, representing vectors that are perpendicular to each other. When dealing with the directional derivative, if you want the derivative to be zero in a certain direction, you choose a direction that is orthogonal to the gradient vector.

In our context, if \( D_{\vec{u}} f = 0 \), this implies no increase or decrease is observed in that direction. To find such a vector, you make sure the dot product with the gradient vector is zero, which confirms they are perpendicular.

For our exercise, the gradient vector at \( P \) is \( \left( \frac{\sqrt{2}}{4}, -\frac{\sqrt{6}}{4} \right) \), and an example of an orthogonal vector to this one would be \( \left( -\frac{\sqrt{6}}{4}, -\frac{\sqrt{2}}{4} \right) \).

This orthogonality ensures the function neither increases nor decreases in the direction of vector \( \vec{u} \), making \( D_{\vec{u}} f \) zero at the specified point.