/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Find the partial derivatives wit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 10

Find the partial derivatives with respect to \(x, y,\) and \(z\) of the following functions: (a) \(f(x, y, z)=\) \(a x^{2}+b x y+c y^{2},(\mathbf{b}) g(x, y, z)=\sin \left(a x y z^{2}\right),(\mathbf{c}) h(x, y, z)=a e^{x y / z^{2}},\) where \(a, b,\) and \(c\) are constants. Remember that to evaluate \(\partial f / \partial x\) you differentiate with respect to \(x\) treating \(y\) and \(z\) as constants.

Short Answer

Expert verified
For function (a), partial derivatives: \( \frac{\partial f}{\partial x} = 2ax + by \), \( \frac{\partial f}{\partial y} = bx + 2cy \), \( \frac{\partial f}{\partial z} = 0 \). For function (b): \( \frac{\partial g}{\partial x} = ayz^2 \cos(ax yz^2) \), \( \frac{\partial g}{\partial y} = axz^2 \cos(ax yz^2) \), \( \frac{\partial g}{\partial z} = 2axyz \cos(ax yz^2) \). For function (c): \( \frac{\partial h}{\partial x} = a\frac{y}{z^2} e^{xy/z^2} \), \( \frac{\partial h}{\partial y} = a\frac{x}{z^2} e^{xy/z^2} \), \( \frac{\partial h}{\partial z} = -a\frac{2xy}{z^3} e^{xy/z^2} \).

Step by step solution

01

Find \( \frac{\partial f}{\partial x} \)

To find the partial derivative of the function \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. The derivative of \(ax^2\) with respect to \(x\) is \(2ax\).3. The derivative of \(bxy\) with respect to \(x\) is \(by\).4. The term \(cy^2\) is constant with respect to \(x\), so its derivative is \(0\).Thus, \( \frac{\partial f}{\partial x} = 2ax + by \).
02

Find \( \frac{\partial f}{\partial y} \)

To find the partial derivative of \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. The derivative of \(ax^2\) with respect to \(y\) is \(0\).3. The derivative of \(bxy\) with respect to \(y\) is \(bx\).4. The derivative of \(cy^2\) with respect to \(y\) is \(2cy\).Thus, \( \frac{\partial f}{\partial y} = bx + 2cy \).
03

Find \( \frac{\partial f}{\partial z} \)

To find the partial derivative of \(f(x, y, z) = ax^2 + bxy + cy^2 \) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. As there are no terms in the function involving \(z\), all derivatives with respect to \(z\) are \(0\).Thus, \( \frac{\partial f}{\partial z} = 0 \).
04

Find \( \frac{\partial g}{\partial x} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. Using the chain rule, derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(x\) is \(ayz^2\).Thus, \( \frac{\partial g}{\partial x} = \cos(ax yz^2) \cdot ayz^2 \).
05

Find \( \frac{\partial g}{\partial y} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. Use the chain rule: derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(y\) is \(axz^2\).Thus, \( \frac{\partial g}{\partial y} = \cos(ax yz^2) \cdot axz^2 \).
06

Find \( \frac{\partial g}{\partial z} \)

To find the partial derivative of \(g(x, y, z) = \sin(ax yz^2)\) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. Use the chain rule: derivative of \(\sin(u)\) is \(\cos(u)\) multiplied by the derivative of \(u\), for \(u = ax yz^2\).3. The derivative of \(u = ax yz^2\) with respect to \(z\) is \(2ax yz\).Thus, \( \frac{\partial g}{\partial z} = \cos(ax yz^2) \cdot 2ax yz \).
07

Find \( \frac{\partial h}{\partial x} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(x\):1. Treat \(y\) and \(z\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(x\) is \(y/z^2\).Thus, \( \frac{\partial h}{\partial x} = ae^{xy/z^2} \cdot \frac{y}{z^2} \).
08

Find \( \frac{\partial h}{\partial y} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(y\):1. Treat \(x\) and \(z\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(y\) is \(x/z^2\).Thus, \( \frac{\partial h}{\partial y} = ae^{xy/z^2} \cdot \frac{x}{z^2} \).
09

Find \( \frac{\partial h}{\partial z} \)

To find the partial derivative of \(h(x, y, z) = ae^{xy/z^2}\) with respect to \(z\):1. Treat \(x\) and \(y\) as constants.2. Using the chain rule: The derivative of \(e^u\) is \(e^u\) times the derivative of \(u\), for \(u = xy/z^2\).3. The derivative of \(u = xy/z^2\) with respect to \(z\) is \(-2xy/z^3\).Thus, \( \frac{\partial h}{\partial z} = ae^{xy/z^2} \cdot \left(-\frac{2xy}{z^3}\right) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Understanding the Chain Rule in Multivariable Calculus
When working with multivariable functions, the chain rule becomes an invaluable tool. It helps us differentiate functions composed of multiple sub-functions. Imagine you have a function like \(g(x, y, z) = \sin(ax yz^2)\). Here, the dependent variable depends on two or more variables. The chain rule can simplify the differentiation process by breaking the function into parts.
For a function \(f(u)\), where \(u\) is another function like \(u = ax yz^2\), the chain rule states that the derivative of \(f\) is the derivative of \(f\) with respect to \(u\) multiplied by the derivative of \(u\) with respect to the variable of interest.
In practice:
  • Differentiate the outer function. For \(\sin(u)\), it becomes \(\cos(u)\).
  • Next, find the derivative of the inner function \(u\) as it relates to your variable of interest, like \(x\), \(y\), or \(z\).
  • Multiply these derivatives to get the final result.
The power of the chain rule lies in unraveling complex expressions into manageable components.
Exploring Multivariable Calculus
Multivariable calculus extends single-variable calculus to functions of two or more variables. In contrast to single-variable calculus, where you only deal with one independent variable, multivariable calculus involves several independent variables.
This allows for a more sophisticated description of the world, as most systems and processes are inherently multivariable.
With functions like \(f(x, y, z) = ax^2 + bxy + cy^2\), you can calculate how one variable uniquely influences the overall function while the others remain constant. This is done using partial derivatives, a central concept in multivariable calculus.
Applications of multivariable calculus are vast and include physics, engineering, economics, and any field where change occurs over more than one dimension. Understanding this helps model and predict complex systems by exploring interactions between variables.
Differentiation with Respect to a Variable
Differentiation with respect to a variable involves taking the derivative of a function concerning one variable, while treating others as constants. For example, in a function with three variables, like \(h(x, y, z) = ae^{xy/z^2}\), partial differentiation allows you to see how changing one variable will affect the function.
Here鈥檚 how it works:
  • Choose the variable you want to differentiate with respect to.
  • Treat the other variables as constants.
  • Apply standard differentiation rules, like the power rule, product rule, or chain rule, as if the function only depended on this one variable.
This method simplifies the calculation and reveals the rate of change with respect to one variable without considering others. It's a powerful technique used for functions where multiple factors are at play, offering a deeper insight into how variables interact.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Verify that the gravitational force \(-G M m \hat{\mathbf{r}} / r^{2}\) on a point mass \(m\) at \(\mathbf{r},\) due to a fixed point mass \(M\) at the origin, is conservative and calculate the corresponding potential energy.

Which of the following forces is conservative? (a) \(\mathbf{F}=k(x, 2 y, 3 z)\) where \(k\) is a constant. (b) \(\mathbf{F}=k(y, x, 0) .\) (c) \(\mathbf{F}=k(-y, x, 0)\). For those which are conservative, find the corresponding potential energy \(U,\) and verify by direct differentiation that \(\mathbf{F}=-\nabla U\).

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

Consider the Atwood machine of Figure \(4.15,\) but suppose that the pulley has radius \(R\) and moment of inertia \(I\). (a) Write down the total energy of the two masses and the pulley in terms of the coordinate \(x\) and \(\dot{x}\). (Remember that the kinetic energy of a spinning wheel is \(\frac{1}{2} I \omega^{2}\).) (b) Show (what is true for any conservative one-dimensional system) that you can obtain the equation of motion for the coordinate \(x\) by differentiating the equation \(E=\) const. Check that the equation of motion is the same as you would obtain by applying Newton's second law separately to the two masses and the pulley, and then eliminating the two unknown tensions from the three resulting equations.
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Geometrical and Physical Optics

Read Explanation

Translational Dynamics

Read Explanation

Nuclear Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.