Problem 261 - Calculus 3

Question

For the following exercises, find the directional derivative using the limit definition only. \(f(x, y)=y^{2} \cos (2 x)\) at point \(P\left(\frac{\pi}{3}, 2\right)\) in the direction of \(\mathrm{u}=\left(\cos \frac{\pi}{4}\right) \mathrm{i}+\left(\sin \frac{\pi}{4}\right) \mathbf{j}\)

Step-by-Step Solution

Answer

The directional derivative is \( 2\sqrt{2} + \sqrt{6} \).

Step 1: Understand the Problem

We need to calculate the directional derivative of the function \( f(x, y) = y^2 \cos(2x) \) at point \( P\left( \frac{\pi}{3}, 2 \right) \) in the direction of a given unit vector \( \mathbf{u} = \left( \cos \frac{\pi}{4} \right) \mathbf{i} + \left( \sin \frac{\pi}{4} \right) \mathbf{j} \). The directional derivative is calculated using the limit definition.

Step 2: Define the Directional Derivative Using the Limit Definition

The directional derivative \( D_\mathbf{u} f(x, y) \) is defined by\[D_\mathbf{u} f(x, y) = \lim_{h \to 0} \frac{f(x + hu_1, y + hu_2) - f(x, y)}{h},\]where \( \mathbf{u} = (u_1, u_2) \) is a unit vector.

Step 3: Calculate the Unit Vector Components

Calculate the components of the unit vector \( \mathbf{u} \):\[ u_1 = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad u_2 = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}. \]

Step 4: Substitute Into the Limit Definition

Plug into the limit definition using the unit vector components and the point \( P \left( \frac{\pi}{3}, 2 \right) \):\[D_\mathbf{u} f\left( \frac{\pi}{3}, 2 \right) = \lim_{h \to 0} \frac{f\left( \frac{\pi}{3} + h\frac{\sqrt{2}}{2}, 2 + h\frac{\sqrt{2}}{2} \right) - f\left( \frac{\pi}{3}, 2 \right)}{h}.\]

Step 5: Compute \( f \left( x + hu_1, y + hu_2 \right) \)

Evaluate \( f \left( x + hu_1, y + hu_2 \right) \) for the given function:\[f\left( \frac{\pi}{3} + h\frac{\sqrt{2}}{2}, 2 + h\frac{\sqrt{2}}{2} \right) = \left( 2 + h\frac{\sqrt{2}}{2} \right)^2 \cos\left(2\left( \frac{\pi}{3} + h\frac{\sqrt{2}}{2} \right)\right).\]

Step 6: Evaluate \( f\left( \frac{\pi}{3}, 2 \right) \)

Substitute the values \( x = \frac{\pi}{3} \) and \( y = 2 \) into \( f(x, y) \):\[f\left( \frac{\pi}{3}, 2 \right) = (2)^2 \cos(2 \times \frac{\pi}{3}) = 4 \cos\left(\frac{2\pi}{3}\right) = 4 \left(-\frac{1}{2}\right) = -2.\]

Step 7: Simplify and Find the Limit

Substitute the expressions from Steps 5 and 6 into the limit expression and simplify:\[D_\mathbf{u} f\left( \frac{\pi}{3}, 2 \right) = \lim_{h \to 0} \frac{\left( 2 + h\frac{\sqrt{2}}{2} \right)^2 \cos\left( \frac{2\pi}{3} + h\sqrt{2} \right) + 2}{h}.\]Apply an expansion for \( \cos(2\pi/3 + h\sqrt{2}) \approx \cos(2\pi/3) - h\sqrt{2} \sin(2\pi/3) \) and simplify further to find that the limit evaluates to\[D_\mathbf{u} f\left( \frac{\pi}{3}, 2 \right) = 4\frac{\sqrt{2}}{2} + \sqrt{3}2\frac{\sqrt{2}}{2} = 2\sqrt{2} + \sqrt{6}.\]

Step 8: Conclusion

Thus, the directional derivative of \( f \) at point \( P \) in the direction \( \mathbf{u} \) is \( 2\sqrt{2} + \sqrt{6} \).

Key Concepts

Limit Definition

The concept of the limit definition of a directional derivative is foundational for understanding how a function changes as we move in a specific direction. Consider a multivariable function, say, \( f(x, y) \). The idea is to see how the function's value changes at a certain point when you slightly nudge it along a particular path. The directional derivative itself is defined as:

• \( D_{\mathbf{u}} f(x, y) = \lim_{h \to 0} \frac{f(x + hu_1, y + hu_2) - f(x, y)}{h} \)

This means:

• You take your function \( f(x, y) \) and look at the change when you shift TINY steps in the direction of a vector \( \mathbf{u} \).
• Here, \( h \) represents how small those steps are, gradually getting closer to zero.
• If the limit exists, it gives a concrete measure of the rate of change at that point in the chosen direction.

Break down the limit into clear steps:

• Calculate the initial function value \( f(x, y) \).
• Then find the function's value after moving a tiny bit in the direction, which involves adding \( hu_1 \) and \( hu_2 \) to \( x \) and \( y \) respectively.
• Lastly, assess the rate of that change by setting up your fraction: the difference in function values divided by \( h \).

This effectively captures how functions really feel about the particular path you're inquiring about.

Directional Vector

In mathematics, the directional vector plays a critical role, helping us determine how to measure change in a specific direction for functions of multiple variables. Let's consider the unit vector \( \mathbf{u} \), which represents this direction within the directional derivative formula.
The unit vector is usually comprised of components, here denoted as \( \mathbf{u} = (u_1, u_2) \). For our problem, we have:

• \( u_1 = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)
• \( u_2 = \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \)

This setup stems from basic trigonometric values. Each component coordinates with a part of our function's variables \( x \) and \( y \), making these crucial in determining how the function alters when we keep pace in \( \mathbf{u} \)’s direction.
When you express any direction as a unit vector, you ensure that the directional derivative represents pure direction-induced change, without the influence of vector length. This purity allows mathematicians and students alike to compare different paths' impacts directly. By normalizing the vector, i.e., making its length one, you verify that you're observing only the directional influences, not scaling effects.

Function Evaluation

Function evaluation is the part where we handle the substitution of values into our mathematical expression. This step transforms the theoretical definition of change into something real and measurable. Let's break it down:

• The original function given is \( f(x, y) = y^2 \cos(2x) \).
• At point \( P(\frac{\pi}{3}, 2) \), we evaluate \( f \) to find its immediate value: \( f(\frac{\pi}{3}, 2) = 4 \cos(\frac{2\pi}{3}) = -2 \).

To look at the change due to our directional vector:

• We substitute our shifted values \( x + hu_1 \) and \( y + hu_2 \) into the function.
• With the determined vector \( \mathbf{u} \), these become \( f(\frac{\pi}{3} + h\frac{\sqrt{2}}{2}, 2 + h\frac{\sqrt{2}}{2}) \), leading to complex expression transformations.

Quickly piecing together these values shows us what's changing, narrowing down to the real heart of the rate change delineated by the directional derivative. Each calculation involves intricate parts:

• First, evaluate how parameters interact in new variable places (\( x + hu_1, y + hu_2 \)).
• Second, compute resulting new function outcomes, cross-referenced with the original position state evaluations.

Going through these operations lets you fundamentally "talk" to the function by putting new values in old slots and sensing how it "feels" to go somewhere new at micro-levels.