Problem 17 - APEX Calculus

Question

A function \(z=f(x, y)\) and a point \(P\) are given. Find the directional derivative of \(f\) in the indicated directions. Note: these are the same functions as in Exercises 7 through 12. \(f(x, y)=x^{2}+2 y^{2}-x y-7 x, P=(4,1)\) (a) In the direction of \(\vec{v}=\langle-2,5\rangle\) (b) In the direction toward the point \(Q=(4,0)\).

Step-by-Step Solution

Answer

Both directional derivatives are 0.

Step 1: Find the Gradient of the Function

To find the directional derivative, we first need the gradient of the function. The gradient \( abla f(x, y) \) is a vector of the partial derivatives with respect to \( x \) and \( y \). Calculate: \[ \frac{\partial f}{\partial x} = 2x - y - 7 \]\[ \frac{\partial f}{\partial y} = 4y - x \]Thus, the gradient is:\[ abla f(x, y) = \langle 2x - y - 7, 4y - x \rangle \]

Step 2: Evaluate the Gradient at Point P

Evaluate the gradient of \( f \) at the point \( P = (4,1) \).\[ abla f(4,1) = \langle 2(4) - 1 - 7, 4(1) - 4 \rangle = \langle 0, 0 \rangle \]

Step 3: Normalize the Direction Vector \( \vec{v} \)

The direction vector given is \( \vec{v} = \langle -2, 5 \rangle \). First, we need to normalize this vector:Calculate the magnitude of \( \vec{v} \):\[ \| \vec{v} \| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \]The unit vector in the direction of \( \vec{v} \) is:\[ \hat{u} = \left\langle \frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle \]

Step 4: Calculate the Directional Derivative in Direction of \( \vec{v} \)

The directional derivative \( D_{\vec{u}}f \) of \( f \) at \( P \) in the direction of a vector \( \vec{u} \) is given by:\[ D_{\vec{u}}f = abla f(P) \cdot \hat{u} \]Since \( abla f(4,1) = \langle 0, 0 \rangle \), we have:\[ D_{\vec{u}}f = \langle 0, 0 \rangle \cdot \left\langle \frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle = 0 \]

Step 5: Determine the Direction Vector Toward Point Q

To find the directional derivative toward the point \( Q = (4, 0) \), find the direction vector from \( P \) to \( Q \):\[ \vec{PQ} = \langle 4 - 4, 0 - 1 \rangle = \langle 0, -1 \rangle \]The unit vector in this direction is:\[ \hat{u} = \langle 0, -1 \rangle \] (no need for normalization since it's already a unit vector)

Step 6: Calculate the Directional Derivative Toward Point Q

The directional derivative towards \( Q \) at \( P \) is:\[ D_{\vec{u}}f = abla f(P) \cdot \vec{PQ} \]Since \( abla f(4,1) = \langle 0, 0 \rangle \),\[ D_{\vec{u}}f = \langle 0, 0 \rangle \cdot \langle 0, -1 \rangle = 0 \]

Key Concepts

Gradient Vector

The gradient vector is a key concept in multivariable calculus, often symbolized as \( abla f \). It represents the rate at which a function increases or decreases at a specific point. For a function of two variables, such as \( f(x, y) \), the gradient is a vector consisting of the partial derivatives with respect to each variable. Therefore, it points in the direction of the steepest increase of the function.

For the function given in our exercise \( f(x, y) = x^2 + 2y^2 - xy - 7x \), the gradient is calculated as follows:

• Find the partial derivative with respect to \( x \): \( \frac{\partial f}{\partial x} = 2x - y - 7 \)
• Find the partial derivative with respect to \( y \): \( \frac{\partial f}{\partial y} = 4y - x \)

These components form the gradient vector: \( abla f(x, y) = \langle 2x - y - 7, 4y - x \rangle \). At any point \( (x, y) \), this vector points toward the direction where the function increases most rapidly.

Partial Derivatives

Partial derivatives are the derivatives of functions with several variables, taken with respect to one of those variables while keeping the others constant. They are denoted by \( \frac{\partial}{\partial x} \) and \( \frac{\partial}{\partial y} \), for a function \( f(x, y) \).

In our example, the partial derivative \( \frac{\partial f}{\partial x} \) is computed by differentiating \( f \) with respect to \( x \), treating \( y \) as a constant. Similarly, \( \frac{\partial f}{\partial y} \) involves differentiating with respect to \( y \), treating \( x \) as constant.

The process of finding these derivatives allows us to determine the rate of change of the function along each axis. This is crucial for constructing the gradient vector, which offers more insight into the behavior of the function around a given point.

Unit Vector

A unit vector is a vector with a length (or magnitude) of one. It serves as a means to indicate direction without influencing the magnitude of any operations performed with it. This is especially useful in directional derivatives, where only the direction matters.

When working with a direction vector \( \vec{v} \), such as \( \langle -2, 5 \rangle \), we first convert it into a unit vector. This is done by dividing each component by the vector's magnitude. The magnitude of \( \vec{v} \) is calculated as \( \| \vec{v} \| = \sqrt{(-2)^2 + 5^2} = \sqrt{29} \).

Hence, the unit vector \( \hat{u} \) in the direction of \( \vec{v} \) is \( \left\langle \frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle \). Using unit vectors, we ensure that only the direction is considered, not the magnitude, when calculating directional derivatives.

Normalization in Vectors

Normalization is the process of converting a vector into a unit vector. It is achieved by dividing each component of a vector by its magnitude. This is crucial when the goal is to consider direction without affecting magnitude, such as in the calculation of directional derivatives.

For a vector \( \vec{v} = \langle a, b \rangle \), the magnitude is given by \( \| \vec{v} \| = \sqrt{a^2 + b^2} \). Normalizing involves computing: \[ \hat{v} = \left\langle \frac{a}{\| \vec{v} \|}, \frac{b}{\| \vec{v} \|} \right\rangle \]

In our scenario, the vector \( \langle -2, 5 \rangle \) was normalized to become \( \left\langle \frac{-2}{\sqrt{29}}, \frac{5}{\sqrt{29}} \right\rangle \), ensuring that it only influences the direction of the change, not its magnitude. This technique is vital for precise calculations in vector-based operations, particularly in physics and engineering.