Problem 267 - Calculus Volume 3

Question

For the following exercises, find the directional derivative of the function at point \(P\) in the direction of \(\mathbf{v}\). $$f(x, y)=x^{2}-y^{2}, \mathbf{u}=\left\langle\frac{\sqrt{3}}{2}, \frac{1}{2}\right\rangle, P(1,0)$$

Step-by-Step Solution

Answer

The directional derivative is \(\sqrt{3}\).

Step 1: Compute the Gradient of f

The gradient of a function in two variables, \(f(x, y)\), is given by \(abla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle\). For \(f(x, y) = x^2 - y^2\), compute the partial derivatives: \(\frac{\partial f}{\partial x} = 2x\) and \(\frac{\partial f}{\partial y} = -2y\). Thus, \(abla f = \langle 2x, -2y \rangle\).

Step 2: Evaluate the Gradient at Point P

Substitute \(P(1,0)\) into the gradient to find \(abla f(1, 0)\). This gives \(abla f(1, 0) = \langle 2(1), -2(0) \rangle = \langle 2, 0 \rangle\).

Step 3: Normalize Vector v

The directional vector \(\mathbf{v} = \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle\) is already given as a unit vector, so \(\mathbf{u} = \mathbf{v}\). No further normalization is required.

Step 4: Compute the Directional Derivative

The directional derivative of \(f\) at point \(P\) in the direction of the unit vector \(\mathbf{u}\) is given by the dot product \(abla f(P) \cdot \mathbf{u}\). So, calculate \(\langle 2, 0 \rangle \cdot \left \langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right \rangle = 2 \cdot \frac{\sqrt{3}}{2} + 0 \cdot \frac{1}{2} = \sqrt{3}\).

Key Concepts

Gradient

The gradient is a powerful concept from vector calculus. It is a vector that points in the direction of the greatest increase of a function. For a function in two variables, like the given example \(f(x, y) = x^2 - y^2\), the gradient \(abla f\) is calculated using partial derivatives. The formula is:
\[abla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle\]
Partial derivatives measure how a function changes as each variable is varied independently. In our example:

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

Thus, the gradient becomes \(abla f = \langle 2x, -2y \rangle\).
The gradient evaluated at point \(P(1, 0)\) gives \(abla f(1,0) = \langle 2, 0 \rangle\). This vector shows the direction where the function increases fastest and its magnitude tells how fast the increase happens.

Unit Vector

A unit vector is a vector that has a length or magnitude of 1. It is primarily used to indicate direction. In the context of the directional derivative problem, we need a direction to obtain the rate of change of our function.
If you have a vector \(\mathbf{v} = \langle a, b \rangle\), its magnitude is \(\sqrt{a^2 + b^2}\). To convert it into a unit vector, divide each component by the magnitude:
\[\mathbf{u} = \left\langle \frac{a}{\sqrt{a^2 + b^2}}, \frac{b}{\sqrt{a^2 + b^2}} \right\rangle\]
In our case, the given vector \(\mathbf{u} = \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle\) is already a unit vector, meaning its length is already 1, so we can directly use it without further normalization.

Dot Product

The dot product is an operation that takes two equal-length sequences of numbers and returns a single number. This operation is essential when calculating the directional derivative of a function.
Given two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), their dot product is
\[\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\]
For our example, we find the dot product of the gradient at point \(P\), \(abla f(1,0) = \langle 2, 0 \rangle\), with the unit vector \(\mathbf{u} = \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle\):
\[\langle 2, 0 \rangle \cdot \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle = 2 \cdot \frac{\sqrt{3}}{2} + 0 \cdot \frac{1}{2} = \sqrt{3}\]
The result, \(\sqrt{3}\), represents the rate of change of the function \(f\) at point \(P\) in the direction of \(\mathbf{u}\).

Partial Derivative

Partial derivatives are fundamental in multivariable calculus. They allow us to understand how a function changes when altering just one of its variables while keeping others constant.
For a function \(f(x, y)\), the partial derivative with respect to \(x\) is written as \(\frac{\partial f}{\partial x}\). It tells us the rate of change of the function as \(x\) changes, fixing \(y\). Similarly, \(\frac{\partial f}{\partial y}\) is the rate of change with respect to \(y\).
In our specific function \(f(x, y) = x^2 - y^2\), the partial derivatives are:

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

These derivatives tell us how the function will increase or decrease as \(x\) or \(y\) is varied, giving insight into the function's behavior in each direction. They are the building blocks for calculating gradients, and hence, directional derivatives.