Problem 39
Question
Consider the following functions \(f\) and points \(P .\) Sketch the \(x y\) -plane showing \(P\) and the level curve through \(P\). Indicate (as in Figure 70 ) the directions of maximum increase, maximum decrease, and no change for \(f\). $$f(x, y)=8+4 x^{2}+2 y^{2} ; P(2,-4)$$
Step-by-Step Solution
Verified Answer
Question: Sketch the xy-plane showing the point P(2, -4) and the level curve through P for the given function f(x,y) = 8 + 4x^2 + 2y^2. Also, indicate the directions of maximum increase, maximum decrease, and no change for the function f at point P.
Answer: In the xy-plane, point P(2, -4) lies on the level curve f(x,y) = 8, which is an ellipse centered at the origin with the equation 4x^2 + 2y^2 = 8. The directions at point P for maximum increase, maximum decrease, and no change are given by the vectors:
- Maximum increase: \(\left< 16, -16 \right>\)
- Maximum decrease: \(\left< -16, 16 \right>\)
- No change: \(\left< 1, 1 \right>\)
1Step 1: Find the gradient of f
To find the gradient of the function f(x, y) = 8 + 4x^2 + 2y^2, we need to find the partial derivatives with respect to x and y. By applying the rules of differentiation, we find:
$$
\frac{\partial f}{\partial x} = 8x,
$$
$$
\frac{\partial f}{\partial y} = 4y.
$$
2Step 2: Find the gradient at the point P
Now, we have to find the gradient vector at the given point P(2, -4). To do this, we will evaluate the partial derivatives at point P:
$$
\frac{\partial f}{\partial x}(2, -4) = 8(2) = 16,
$$
$$
\frac{\partial f}{\partial y}(2, -4) = 4(-4) = -16.
$$
So, the gradient vector at point P is:
$$
\nabla f(2, -4) = \left< 16, -16 \right>.
$$
3Step 3: Find the level curve through P
To find the level curve through the point P, we need to plug the point P(2, -4) into the function f:
$$
f(2, -4) = 8 + 4(2^2) + 2(-4^2) = 8.
$$
The level curve through P is the equation f(x, y) = 8.
4Step 4: Sketch the \(x y\)-plane showing P and the level curve
On the xy-plane, plot the point P(2, -4) and the level curve f(x, y) = 8, which is an ellipse centered at the origin with semi-major axis along the x-axis, given by the equation:
$$
4x^2 + 2y^2 = 8.
$$
5Step 5: Indicate the directions of maximum increase, decrease, and no change
The direction of the maximum increase of f at the point P is given by the gradient vector \(\nabla f(2,-4)=\left< 16, -16 \right>\). The direction of maximum decrease is opposite to the gradient vector, i.e., \(\left< -16, 16 \right>\). The direction of no change is perpendicular to the gradient vector. In this case, you can pick any direction perpendicular to \(\left< 16, -16 \right>\). This could be, for example, the direction given by \(\left< 1, 1 \right>\).
In summary:
- Maximum increase: \(\left< 16, -16 \right>\)
- Maximum decrease: \(\left< -16, 16 \right>\)
- No change: \(\left< 1, 1 \right>\)
Now, indicate these directions on the xy-plane sketch.
Key Concepts
partial derivativeslevel curvemaximum increase directionmaximum decrease direction
partial derivatives
When working with functions of multiple variables, such as our exercise's function, partial derivatives serve as a means to understand how each variable individually affects the function. Let’s consider a function \(f(x, y)\). Partial derivatives help us grasp the rate at which the function changes as each variable, \(x\) or \(y\), changes.
- The partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), measures the change in the function as \(x\) changes, while keeping \(y\) constant. In this exercise, \(\frac{\partial f}{\partial x} = 8x\).
- Similarly, the partial derivative with respect to \(y\), denoted \(\frac{\partial f}{\partial y}\), measures changes in the function as \(y\) varies, keeping \(x\) constant. Here, it is \(\frac{\partial f}{\partial y} = 4y\).
By calculating these, we form the gradient vector, which offers deep insight into the function's behavior across its variables. This vector points in the direction of the steepest ascent from any given point in the function's domain.
- The partial derivative with respect to \(x\), denoted \(\frac{\partial f}{\partial x}\), measures the change in the function as \(x\) changes, while keeping \(y\) constant. In this exercise, \(\frac{\partial f}{\partial x} = 8x\).
- Similarly, the partial derivative with respect to \(y\), denoted \(\frac{\partial f}{\partial y}\), measures changes in the function as \(y\) varies, keeping \(x\) constant. Here, it is \(\frac{\partial f}{\partial y} = 4y\).
By calculating these, we form the gradient vector, which offers deep insight into the function's behavior across its variables. This vector points in the direction of the steepest ascent from any given point in the function's domain.
level curve
Level curves, also known as contour lines, are incredibly useful in visualizing functions in two dimensions. They represent paths of constant function values in the plane. When examining a function like \(f(x, y) = 8 + 4x^2 + 2y^2\), level curves help us identify points that share the same function value.
To find the level curve through a specific point, such as \(P(2, -4)\), we substitute this point into the function to ascertain its value. In this case, \(f(2, -4) = 8\). Thus, the level curve through \(P\) is represented by the equation \(f(x, y) = 8\), which simplifies to a geometric shape, in this instance, an ellipse given by \(4x^2 + 2y^2 = 8\), centered at the origin.
Level curves are parallel to areas of equal elevation and play a crucial role in understanding landscapes of mathematical functions without needing three-dimensional graphs.
To find the level curve through a specific point, such as \(P(2, -4)\), we substitute this point into the function to ascertain its value. In this case, \(f(2, -4) = 8\). Thus, the level curve through \(P\) is represented by the equation \(f(x, y) = 8\), which simplifies to a geometric shape, in this instance, an ellipse given by \(4x^2 + 2y^2 = 8\), centered at the origin.
Level curves are parallel to areas of equal elevation and play a crucial role in understanding landscapes of mathematical functions without needing three-dimensional graphs.
maximum increase direction
The direction of maximum increase for a function at a particular point is crucial to understand as it indicates where the function's value is growing the fastest. For any differentiable function like our given \(f(x, y)\), the gradient vector \(abla f(x, y)\) shows the steepest ascent direction.
- At point \(P(2, -4)\), the gradient vector is \(\langle 16, -16 \rangle\).
- This vector points towards the direction where an increase in \(x\) and a decrease in \(y\) yields the quickest elevation in \(f(x, y)\).
In essence, wherever you encounter a multidimensional function, the gradient vector will not only indicate how quickly you’ll rise, but also guide you through the path of steepest ascent in the landscape of the function.
- At point \(P(2, -4)\), the gradient vector is \(\langle 16, -16 \rangle\).
- This vector points towards the direction where an increase in \(x\) and a decrease in \(y\) yields the quickest elevation in \(f(x, y)\).
In essence, wherever you encounter a multidimensional function, the gradient vector will not only indicate how quickly you’ll rise, but also guide you through the path of steepest ascent in the landscape of the function.
maximum decrease direction
Conversely, the maximum decrease direction tells us where the function decreases fastest. This is closely related to the gradient concept. While the gradient vector tells us the path of steepest ascent, the steepest descent is just on the opposite path.
- To find this direction at the point \(P(2, -4)\), we take the negative of the gradient vector.
- Hence, the maximum decrease direction is in the direction \(\langle -16, 16 \rangle\).
Think of it as walking down a hill: if upwards (gradient direction) is \(\langle 16, -16 \rangle\), the downward path will be the exact opposite, \(\langle -16, 16 \rangle\). Understanding these directions helps one navigate and interpret multidimensional function "terrains" effectively.
- To find this direction at the point \(P(2, -4)\), we take the negative of the gradient vector.
- Hence, the maximum decrease direction is in the direction \(\langle -16, 16 \rangle\).
Think of it as walking down a hill: if upwards (gradient direction) is \(\langle 16, -16 \rangle\), the downward path will be the exact opposite, \(\langle -16, 16 \rangle\). Understanding these directions helps one navigate and interpret multidimensional function "terrains" effectively.
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