Problem 3 - Calculus with Concepts in Calculus

Question

Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y)=\frac{x^{2}-y^{2}}{x^{2}+y^{2}} ; P=(3,4) ; \mathbf{a}=\frac{1}{2} \mathbf{i}-\frac{\sqrt{3}}{2} \mathbf{j} $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{192}{1250} + \frac{144\sqrt{3}}{1250} \).

Step 1: Find the Gradient of f

The gradient of a function \( f(x,y) \) is given by \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \), we compute the partial derivatives:\[ \frac{\partial f}{\partial x} = \frac{2x(x^2+y^2) - (x^2-y^2) 2x}{(x^2+y^2)^2} = \frac{4xy^2}{(x^2+y^2)^2} \] \[ \frac{\partial f}{\partial y} = \frac{-2y(x^2+y^2) + (x^2-y^2) 2y}{(x^2+y^2)^2} = \frac{-4yx^2}{(x^2+y^2)^2} \] Thus, \( abla f(x, y) = \left( \frac{4xy^2}{(x^2+y^2)^2}, \frac{-4yx^2}{(x^2+y^2)^2} \right) \).

Step 2: Evaluate the Gradient at Point P

Substitute \( P = (3, 4) \) into the gradient:\[abla f(3, 4) = \left( \frac{4 \times 3 \times 4^2}{(3^2 + 4^2)^2}, \frac{-4 \times 4 \times 3^2}{(3^2 + 4^2)^2} \right) \] Calculate step by step:- First, calculate each involving power: \( 3^2 = 9 \), \( 4^2 = 16 \), so \( 3^2 + 4^2 = 25 \).- \( 4xy^2 = 4 \times 3 \times 16 = 192 \), \(-4yx^2 = -4 \times 4 \times 9 = -144 \).- \((3^2 + 4^2)^2 = 625 \). Thus, \[ abla f(3, 4) = \left( \frac{192}{625}, \frac{-144}{625} \right) \].

Step 3: Find the Unit Vector in the Direction of a

The given vector \( \mathbf{a} = \frac{1}{2} \mathbf{i} - \frac{\sqrt{3}}{2} \mathbf{j} \) needs to be checked if it's already a unit vector. A unit vector \( \mathbf{u} \) has a magnitude of 1.Calculate:\[ ||\mathbf{a}|| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1 \]Since the magnitude is 1, \( \mathbf{a} \) is indeed a unit vector. Therefore, \( \mathbf{u} = \mathbf{a} \).

Step 4: Calculate the Directional Derivative

The directional derivative \( D_\mathbf{u}f \) of \( f \) in the direction of \( \mathbf{u} \) at point \( P \) is given by:\[ D_\mathbf{u}f = abla f(P) \cdot \mathbf{u} \]Substitute the values:\[ D_\mathbf{u}f = \left( \frac{192}{625}, \frac{-144}{625} \right) \cdot \left( \frac{1}{2}, -\frac{\sqrt{3}}{2} \right) \]Calculate the dot product:\[ D_\mathbf{u}f = \frac{192}{625} \cdot \frac{1}{2} + \frac{-144}{625} \cdot -\frac{\sqrt{3}}{2} \]\[ = \frac{192}{1250} + \frac{144\sqrt{3}}{1250} \]So, the directional derivative is \( \frac{192}{1250} + \frac{144\sqrt{3}}{1250} \).

Key Concepts

Gradient Vector

The gradient vector is a critical component when discussing directional derivatives. It is essentially a vector that contains the partial derivatives with respect to all variables involved in a multivariable function. For a function like \(f(x, y)\), the gradient is represented as \(abla f(x, y)\), which is a vector composed of \(\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)\).
This vector points in the direction of the greatest rate of increase of the function and its magnitude indicates how fast the function increase in that direction.

Partial Derivatives

Partial derivatives are derivatives where we only change one variable at a time while keeping all other variables constant. They form the components of the gradient vector. For example, when we have the function \( f(x, y) = \frac{x^2 - y^2}{x^2 + y^2} \), we find the partial derivative with respect to \(x\) and \(y\) independently to build our gradient vector.

• The partial derivative \(\frac{\partial f}{\partial x}\) helps understand how \(f\) changes as \(x\) changes, keeping \(y\) fixed.
• The partial derivative \(\frac{\partial f}{\partial y}\) shows the change in \(f\) when \(y\) changes, keeping \(x\) constant.

By integrating these derivatives, we find how the function behaves in multidimensional space.

Unit Vector

A unit vector is a vector with a magnitude of one. This is essential when calculating the directional derivative because we need to specify the direction precisely without altering the intensity. In our exercise, the vector \(\mathbf{a} = \frac{1}{2} \mathbf{i} - \frac{\sqrt{3}}{2} \mathbf{j}\) already has a magnitude of 1.
To check this, calculate the magnitude \(||\mathbf{a}|| = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{1} = 1 \). Since it holds true that \(||\mathbf{a}|| = 1\), \(\mathbf{a}\) is indeed a unit vector and can be used directly to determine the directional derivative.

Dot Product

The dot product is a method for multiplying two vectors, yielding a scalar. For the directional derivative, we utilize the dot product of the gradient vector and the direction unit vector. This operation combines their magnitudes and the cosine of the angle between them, providing the rate of change of the function in the specified direction.

• Given vectors \(\mathbf{A} = (a_1, a_2)\) and \(\mathbf{B} = (b_1, b_2)\), their dot product is \( \mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 \).
• The directional derivative formula: \( D_\mathbf{u}f = abla f(P) \cdot \mathbf{u} \), where \(abla f(P)\) is the gradient at point \(P\), and \(\mathbf{u}\) is the unit vector.

This calculation results in a scalar indicating how the function \(f\) changes as we move in the direction of \(\mathbf{u}\).