Question

Interpreting directional derivatives A function \(f\) and a point \(P\) are given. Let \(\theta\) correspond to the direction of the directional derivative. a. Find the gradient and evaluate it at \(P\). b. Find the angles \(\theta\) (with respect to the positive \(x\) -axis) associated with the directions of maximum increase, maximum decrease, and zero change. c. Write the directional derivative at \(P\) as a function of \(\theta ;\) call this function \(g(\theta)\) d. Find the value of \(\theta\) that maximizes \(g(\theta)\) and find the maximum value. e. Verify that the value of \(\theta\) that maximizes \(g\) corresponds to the direction of the gradient. Verify that the maximum value of g equals the magnitude of the gradient. $$f(x, y)=e^{-x^{2}-2 y^{2}} ; P(-1,0)$$

Short answer

To summarize the solution, we find the gradient of the function \(f(x, y) = e^{-x^{2}-2y^{2}}\) and evaluate it at point \(P(-1, 0)\). We find that the angle of maximum increase is 0°, the angle of maximum decrease is 180°, and the angles of zero change are 90° and 270°. The directional derivative function at point P is \(g(\theta) = 2e^{-1}\cos\theta\). The maximum value of the directional derivative function is \(2e^{-1}\) and corresponds to an angle of 0°, which verifies our solution.

Step-by-step solution

Calculate the gradient of f

To calculate the gradient of f, we need to find the partial derivatives of f with respect to x and y, then combine them as a vector. $$\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$$ Calculate partial derivatives: $$\frac{\partial f}{\partial x} = \frac{d}{dx}\left(e^{-x^{2}-2y^{2}}\right) = -2x e^{-x^{2}-2y^{2}}$$ $$\frac{\partial f}{\partial y} = \frac{d}{dy}\left(e^{-x^{2}-2y^{2}}\right) = -4y e^{-x^{2}-2y^{2}}$$ Combine the partial derivatives into a vector: \(\nabla f(x, y) = (-2x e^{-x^{2}-2y^{2}}, -4y e^{-x^{2}-2y^{2}})\)

Evaluate the gradient at point P

To evaluate the gradient at point \(P(-1, 0)\), substitute the coordinates into the gradient vector. $$\nabla f(-1, 0) = (-2(-1) e^{-(-1)^{2}-2(0)^{2}}, -4(0) e^{-(-1)^{2}-2(0)^{2}}) = (2e^{-1}, 0)$$

Find the angles associated with direction properties

Since the gradient at point P is \((2e^{-1}, 0)\), it corresponds to the direction of maximum increase. Therefore, the angle is: $$\theta_\text{max increase} = \arctan\left(\frac{0}{2e^{-1}}\right) = 0°$$ As the direction of maximum decrease is the opposite direction of the gradient, the angle is: $$\theta_\text{max decrease} = 180°$$ Since the angle between the gradient and the direction of zero change is 90°, we have two angles for zero change: $$\theta_\text{zero change} = 90°, 270°$$

Write the function for the directional derivative at P as a function of θ

The directional derivative in the direction of a unit vector \(u = (\cos\theta, \sin\theta)\) is given by: $$g(\theta) = \nabla f(P) \cdot u = (2e^{-1}, 0) \cdot (\cos\theta, \sin\theta) = 2e^{-1}\cos\theta$$

Find the value of θ that maximizes g(θ) and find the maximum value

To find the maximum value, we can differentiate g(θ) w.r.t θ and equate it to zero: $$\frac{d}{d\theta}g(\theta) = -2e^{-1}\sin\theta$$ Setting the derivative equal to zero, and finding the values of theta which satisfy this condition: $$-2e^{-1}\sin\theta = 0$$ $$\sin\theta = 0$$ The solutions for θ lie in the range \([0, 2\pi)\): $$\theta = 0, \pi$$ The maximum value for \(g(\theta)\) occurs when \(\theta = 0\) since \(g(0) = 2e^{-1}\) and \(g(\pi) = -2e^{-1}\). Therefore, the maximum value is: $$\text{Max value} = g(0) = 2e^{-1}$$

Verify the maximum value and corresponding θ

The maximum value of \(g\) (\(2e^{-1}\)) is equal to the magnitude of the gradient at point P: $$||\nabla f(-1,0)|| = ||(2e^{-1},0)|| = \sqrt{(2e^{-1})^2+0^2} = 2e^{-1}$$ The value of θ that maximizes \(g(\theta)\) (0°) corresponds to the direction of the gradient. This verifies our solution.

Key concepts

Gradient Vector

The concept of a gradient vector is crucial in understanding directional derivatives. In mathematical terms, the gradient vector, often represented as \( abla f \), gives the direction and rate of fastest increase of a function. This is crucial when trying to move through a landscape of a graph to find maximum or minimum points.
The gradient vector is derived by taking the partial derivatives of a function with respect to its variables and arranging them in a vector format. For example, for a function \( f(x, y) \), the gradient \( abla f \) is \( \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \).
This allows us to see at a glance how changes in the variables affect the value of the function, providing a map of change over the domain of \( f \).
In the original problem, we see this with the function \( f(x, y) = e^{-x^2 - 2y^2} \), where the gradient gives us a direction \( (2e^{-1}, 0) \) at the point \( (-1, 0) \). This vector points to where \( f \) increases most rapidly, emphasizing the powerful visualization that gradients offer in calculus.

Partial Derivatives

Partial derivatives are the building blocks of gradients and directional derivatives. When a function has multiple variables, each partial derivative measures the rate of change of the function with respect to one of those variables, with all others held constant.
Imagine a hilly landscape (a graph of the function), where each point is known by a pair of coordinates. By taking a partial derivative with respect to \( x \), you know how steep the hill is in the \( x \)-direction at each point. Similarly, by taking it with respect to \( y \), you understand the slope along the \( y \)-axis.
In the provided example, the function \( f(x, y) = e^{-x^2 - 2y^2} \) involves two variables. Its partial derivatives are:

• \( \frac{\partial f}{\partial x} = -2x e^{-x^2 - 2y^2} \)
• \( \frac{\partial f}{\partial y} = -4y e^{-x^2 - 2y^2} \)

These derivatives help in composing the gradient vector, marrying the idea of individual variable change back together into a coherent whole for analysis. Thus, partial derivatives provide insight into specific directions of a function's rate of change.

Angle of Maximum Increase

The angle of maximum increase is an intriguing concept that embodies the power of the gradient vector. In essence, this angle shows us the exact direction in which a function will grow at its steepest possible rate.
Mathematically, it is represented by the orientation of the gradient vector itself. Since the gradient points to the steepest ascent, the angle it forms with a reference axis (typically the positive \( x \)-axis) denotes its direction.
For the example function \( f \) in the original problem, the gradient \( (2e^{-1}, 0) \) means that at point \( (-1, 0) \), the path of sharpest increase aligns perfectly with the positive \( x \)-axis. Thus, the angle of maximum increase \( \theta_{\text{max increase}} \) is \( 0^\circ \).
Understanding this angle allows us to predict how a function behaves at any given point, simulating a sort of compass that points the way to the biggest heights in the graph.

Magnitude of the Gradient

The magnitude of the gradient vector provides valuable information about how steeply a function is climbing or falling at a given point. It's a measure of intensity, depicting just how much the function's value is changing.
Mathematically, the magnitude of a vector \( \mathbf{v} = (a, b) \) is given by \( ||\mathbf{v}|| = \sqrt{a^2 + b^2} \). For a gradient vector, this magnitude tells us the rate of maximum increase of the function at that point.
In the case of the function \( f(x, y) = e^{-x^2 - 2y^2} \), when the gradient at position \( (-1, 0) \) is \((2e^{-1}, 0)\), the magnitude is calculated as \( 2e^{-1} \).
This number, \( 2e^{-1} \), not only signifies how steep the function is at that spot, but it also turns up as the maximum value of the directional derivative at that point, showing deep connections between these concepts. This interplay ensures that we have a full understanding of movement on the function's surface.