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If \(f(x, y, z)=2 z y^{2},\) then the gradient at the point (2,2,4) is \(\nabla f(2,2,4)=\) _____________.

Short Answer

Expert verified
The gradient of \(f(x, y, z) = 2zy^2\) at the point (2,2,4) is: \(\nabla f(2,2,4) = \begin{bmatrix} 0\\ 32\\ 8 \end{bmatrix}\).

Step by step solution

01

Compute the partial derivatives

To find the gradient of the function, we need to compute the partial derivatives with respect to all three variables (x, y, and z). The partial derivative of f with respect to x is: \(\frac{\partial f}{\partial x} = 0\) The partial derivative of f with respect to y is: \(\frac{\partial f}{\partial y} = 4zy\) The partial derivative of f with respect to z is: \(\frac{\partial f}{\partial z} = 2y^2\)
02

Evaluate partial derivatives at the given point (2,2,4)

Now, we'll evaluate the partial derivatives we computed in Step 1 at the point (2,2,4). \(\frac{\partial f}{\partial x}(2,2,4) = 0\) \(\frac{\partial f}{\partial y}(2,2,4) = 4(2)(4) = 32\) \(\frac{\partial f}{\partial z}(2,2,4) = 2(2^2) = 8\)
03

Construct the gradient vector

Finally, we can construct the gradient vector using the values of the partial derivatives at the point (2,2,4). \(\nabla f(2,2,4) = \begin{bmatrix} \frac{\partial f}{\partial x}(2,2,4)\\ \frac{\partial f}{\partial y}(2,2,4)\\ \frac{\partial f}{\partial z}(2,2,4) \end{bmatrix} = \begin{bmatrix} 0\\ 32\\ 8 \end{bmatrix}\) The gradient of \(f(x, y, z) = 2zy^2\) at the point (2,2,4) is: \(\nabla f(2,2,4) = \begin{bmatrix} 0\\ 32\\ 8 \end{bmatrix}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Partial Derivatives
Understanding partial derivatives is essential in multivariable calculus as they express how a function changes as only one of its variables varies, other variables being held constant. Imagine you have a function, like a temperature map of a mountain. A partial derivative tells you how the temperature changes if you move north, but not up or down.

For a function with more than one variable, such as our textbook example function, \( f(x, y, z)=2zy^2 \), the partial derivative with respect to each variable is taken separately. With \( \frac{\partial f}{\partial x} = 0 \), it represents that moving in the x-direction (holding y and z constant) does not change the function's value. Think of this as being on a flat plateau on our mountain where moving horizontally doesn't affect the temperature. On the contrary, \( \frac{\partial f}{\partial y} = 4zy \) and \( \frac{\partial f}{\partial z} = 2y^2 \) suggest that temperature does change with movement in y and z directions - akin to moving vertically or horizontally on a slope.

To improve comprehension, remember that partial derivatives can be visualized as the slopes of the function in each coordinate direction. They represent the immediate rate of change, like a snapshot of a hike's steepness at a moment in time.
Multivariable Calculus
Diving into multivariable calculus opens up the world of functions that have more than one input, like our example \( f(x, y, z)=2zy^2 \). It's like stepping up from looking at the profile of a single mountain trail to exploring a 3D map of an entire mountain range. Here, we study how functions change in more than one direction and learn to describe phenomena like varying temperatures at different points on the mountain.

In this broader context, understanding partial derivatives is just the beginning. Multivariable calculus also includes studying integrals over paths and surfaces, which could represent, for example, the total heat energy over an area. It brings powerful tools like the gradient, divergence, and curl that help us describe physical phenomena in three dimensions.

For a more solid grasp, students should visualize multivariable functions as surfaces or landscapes with various peaks, valleys, and curves. Each point on this landscape has unique slopes in different directions, all of which are captured by partial derivatives.
Gradient Vector
The gradient vector is a concept from multivariable calculus that is critical in fields like physics and engineering. It's like having a compass that not only tells you which way is north but also shows the steepest uphill direction from your current location. For our function \( f(x, y, z)=2zy^2 \), the gradient vector at any point tells us the direction of the steepest ascent.

In the context of the solved exercise, the gradient vector at the point (2,2,4), denoted as \( abla f(2,2,4) \), is found by taking the partial derivatives and evaluating them at the specific point. It results in \( \begin{bmatrix} 0\ 32\ 8 \end{bmatrix} \), meaning that at that point, moving in the direction of the y-axis leads to the steepest climb, while there is no change in the x-direction. For students, picturing the gradient as a 3D arrow pointing uphill can be tremendously helpful.

Intuitively, the gradient vector can be seen as the 'slope' of a multivariable function. In machine learning, finding the minimum or maximum of a function, which can be thought of as the lowest valley or the highest peak on a landscape, is often done by following the gradient (or opposite of it) - a process known as gradient descent or ascent.

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Most popular questions from this chapter

One mole of ammonia gas is contained in a vessel which is capable of changing its volume (a compartment sealed by a piston, for example). The total energy \(U\) (in Joules) of the ammonia is a function of the volume \(V\) (in cubic meters) of the container, and the temperature \(T\) (in degrees Kelvin) of the gas. The differential \(d U\) is given by \(d U=840 d V+27.32 d T\). (a) How does the energy change if the volume is held constant and the temperature is decreased slightly? \(\odot\) it increases slightly \(\odot\) it does not change \(\odot\) it decreases slightly (b) How does the energy change if the temperature is held constant and the volume is increased slightly? \(\odot\) it does not change \(\odot\) it increases slightly \(\odot\) it decreases slightly (c) Find the approximate change in energy if the gas is compressed by 150 cubic centimeters and heated by 3 degrees Kelvin. Change in energy = ___________. Please include units in your answer.

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If a continuous function \(f\) of a single variable has exactly one critical number with a relative maximum at that critical point, then the value of \(f\) at that critical point is an absolute maximum. In this exercise we see that the same is not always true for functions of two variables. Let \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) (from "' The Only Critical Point in Town"" Test by Ira Rosenholz and Lowell Smylie in the Mathematics Magazine, VOL 58 NO 3 May \(1985 .\) ). Show that \(f\) has exactly one critical point, has a relative maximum value at that critical point, but that \(f\) has no absolute maximum value. Use appropriate technology to draw the surface defined by \(f\) to see graphically how this happens.

Use the chain rule to find \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t},\) where $$z=e^{x y} \tan y, x=4 s+5 t, y=\frac{3 s}{3 t}$$ First the pieces: \(\frac{\partial z}{\partial x}=\) ________ \(\frac{\partial z}{\partial y}=\) \(\frac{\partial x}{\partial s}=\) ________ \(\frac{\partial x}{\partial t}=\) \(\frac{\partial y}{\partial s}=\) ________ \(\frac{\partial y}{\partial t}=\)

Calculate all four second-order partial derivatives of \(f(x, y)=(2 x+4 y) e^{y}\) \(f_{x x}(x, y)=\) ___________. \(f_{x y}(x, y)=\) ___________. \(f_{y x}(x, y)=\) ___________. \(f_{y y}(x, y)=\) ___________.

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