Problem 11 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14, a function \(f\), a point \(P\), and a unit vector \(\mathbf{U}\) are given. Find (a) the gradient of \(f\) at \(P\), and (b) the rate of change of the function value in the direction of \(\mathrm{U}\) at \(P\). \(f(x, y)=x^{2}-4 y ; P=(-2,2) ; \mathbf{U}=\cos \frac{1}{3} \pi \mathbf{i}+\sin \frac{1}{3} \pi \mathbf{j}\)

Step-by-Step Solution

Answer

The gradient at \(P\) is \((-4, -4)\) and the rate of change is \(2 - 2\sqrt{3}\).

Step 1 - Calculate Partial Derivatives

Find the partial derivatives of the function. We have \[ f(x, y) = x^2 - 4y \] The partial derivative with respect to \(x\) is \[ f_x = \frac{\partial}{\partial x}(x^2 - 4y) = 2x \] The partial derivative with respect to \(y\) is \[ f_y = \frac{\partial}{\partial y}(x^2 - 4y) = -4 \]

Step 2 - Evaluate the Gradient at P

Evaluate the gradient of the function at the given point \(P = (-2, 2)\). The gradient of \(f\) is given by the vector \(abla f = (f_x, f_y)\).\[ abla f(-2, 2) = (f_x(-2, 2), f_y(-2, 2)) = (2(-2), -4) = (-4, -4) \]

Step 3 - Find the Unit Vector

Simplify the given unit vector. The unit vector \(\mathbf{U}\) is given by \[ \mathbf{U} = \cos \frac{1}{3} \pi \mathbf{i} + \sin \frac{1}{3} \pi \mathbf{j} \]Using trigonometric values, this simplifies to \[ \mathbf{U} = -\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \]

Step 4 - Compute the Rate of Change in the Direction of U

Compute the dot product of the gradient vector and the unit vector \(\mathbf{U}\). The rate of change is given by \[ D_\mathbf{U}f = abla f \cdot \mathbf{U} \]Substitute the values: \[ D_\mathbf{U}f = (-4, -4) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = (-4)(-\frac{1}{2}) + (-4)(\frac{\sqrt{3}}{2}) \]This simplifies to \[ 2 - 2\sqrt{3} \]

Key Concepts

Partial Derivatives

Partial derivatives are essential in multivariable calculus. Unlike ordinary derivatives, which deal with functions of a single variable and rate of change in one direction, partial derivatives deal with functions of multiple variables.
To find a partial derivative concerning a specific variable, treat all other variables as constants and differentiate concerning that variable.
For example, given the function \( f(x, y) = x^2 - 4y \), the partial derivative with respect to \( x \) is \( f_x = \frac{\partial}{\partial x}(x^2 - 4y) = 2x \). Here, \( y \) is treated as a constant. Similarly, the partial derivative with respect to \( y \) is \( f_y = \frac{\partial}{\partial y}(x^2 - 4y) = -4 \). Here, \( x \) is treated as a constant.
To summarize:

• Partial derivatives help us understand how a function changes as one variable changes while others remain constant.
• They are fundamental in finding gradients for multivariable functions.

Rate of Change

The rate of change of a function in a particular direction tells us how fast the function's value changes as we move in that direction. For multivariable functions, we often compute this change using the gradient vector and a unit vector in the direction of interest.
The gradient \( abla f \) of a function \( f \) at a point gives us the direction of the steepest ascent and its rate. It combines all partial derivatives into one vector.
For instance, for the function \( f(x, y) = x^2 - 4y \) at the point \( P = (-2, 2) \), the gradient is \( abla f(-2, 2) = (-4, -4) \).
To find the rate of change in a given direction, we take the dot product of the gradient vector and the unit vector in that direction. For example, with the unit vector \( \mathbf{U} = -\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \), the rate of change is \( (-4, -4) \cdot \left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = 2 - 2\sqrt{3} \).
Key points to remember:

• The gradient vector provides the direction and rate of steepest ascent.
• The dot product of the gradient vector and a unit vector gives the rate of change in the direction of the unit vector.

Unit Vector

A unit vector is a vector of length 1. It is often used to indicate a specific direction without scaling the magnitude. In the context of gradient vector calculus, the unit vector helps determine the rate of change of the function in that specified direction.
A unit vector \( \mathbf{U} \) is typically represented in terms of its components along the coordinate axes. For example, \( \mathbf{U} = \cos \left( \frac{1}{3} \pi \right) \mathbf{i} + \sin \left( \frac{1}{3} \pi \right) \mathbf{j} \). Using trigonometric values, this simplifies to \( \mathbf{U} = -\frac{1}{2} \mathbf{i} + \frac{\sqrt{3}}{2} \mathbf{j} \).
Key properties of unit vectors include:

• The length or magnitude of a unit vector is always 1.
• They are used to specify directions without affecting magnitudes.
• Unit vectors are crucial in dot products to find directional rates of change.

Understanding unit vectors:

• The direction they specify can be determined using trigonometric identities.
• They are helpful to normalize vectors to a standard length of 1.