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Chapter 12: Problem 38

Find the four second partial derivatives of the following functions. $$H(x, y)=\sqrt{4+x^{2}+y^{2}}$$

Short Answer

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Question: Find the four second partial derivatives of the function $$H(x, y) = \sqrt{4 + x^{2} + y^{2}}$$. Answer: The four second partial derivatives of the function $$H(x, y)$$ are: 1. $$\frac{\partial^2 H}{\partial x^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ 2. $$\frac{\partial^2 H}{\partial y^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ 3. $$\frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$ 4. $$\frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$

Step by step solution

01

Find the first partial derivatives

First, let's find the first partial derivatives of the function $$H(x, y)$$: $$ \frac{\partial H}{\partial x} = \frac{d (\sqrt{4 + x^{2} + y^{2}})}{dx} $$ $$ \frac{\partial H}{\partial y} = \frac{d (\sqrt{4 + x^{2} + y^{2}})}{dy} $$ By applying the chain rule, we get: $$ \frac{\partial H}{\partial x} = \frac{x}{\sqrt{4 + x^{2} + y^{2}}} $$ and $$ \frac{\partial H}{\partial y} = \frac{y}{\sqrt{4 + x^{2} + y^{2}}} $$
02

Find the second partial derivatives

Now that we have the first partial derivatives, we can find the second partial derivatives by taking their derivatives with respect to $$x$$ and $$y$$ again. 1. $$\frac{\partial^2 H}{\partial x^2} = \frac{d (\frac{x}{\sqrt{4 + x^{2} + y^{2}}})}{dx}$$ 2. $$\frac{\partial^2 H}{\partial y^2} = \frac{d (\frac{y}{\sqrt{4 + x^{2} + y^{2}}})}{dy}$$ 3. $$\frac{\partial^2 H}{\partial x \partial y} = \frac{\partial (\frac{y}{\sqrt{4 + x^{2} + y^{2}}})}{\partial x}$$ 4. $$\frac{\partial^2 H}{\partial y \partial x} = \frac{\partial (\frac{x}{\sqrt{4 + x^{2} + y^{2}}})}{\partial y}$$ After calculating the derivatives, we get: 1. $$\frac{\partial^2 H}{\partial x^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ 2. $$\frac{\partial^2 H}{\partial y^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ 3. $$\frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$ 4. $$\frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$ So, the four second partial derivatives of the function $$H(x, y)$$ are: $$\frac{\partial^2 H}{\partial x^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ $$\frac{\partial^2 H}{\partial y^2} = \frac{4}{(4 + x^{2} + y^{2})^{3/2}}$$ $$\frac{\partial^2 H}{\partial x \partial y} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$ $$\frac{\partial^2 H}{\partial y \partial x} = \frac{-xy}{(4 + x^{2} + y^{2})^{3/2}}$$

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Key Concepts

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

Second Partial Derivatives
Second partial derivatives give us valuable information about the curvature of a function along different directions. When dealing with the function \( H(x, y) = \sqrt{4 + x^2 + y^2} \), we are interested in how it changes as we move slightly from a point in both the \( x \) and \( y \) directions. In two-variable calculus, finding second partial derivatives means calculating the rate of change of the first partial derivatives themselves.
To find these, we look at both the second derivatives with respect to the same variable, like \( \frac{\partial^2 H}{\partial x^2} \), and the mixed partial derivatives, such as \( \frac{\partial^2 H}{\partial x \partial y} \).
  • The terms \( \frac{\partial^2 H}{\partial x^2} \) and \( \frac{\partial^2 H}{\partial y^2} \) show how the function's slope changes purely along each coordinate axis.
  • Mixed partials like \( \frac{\partial^2 H}{\partial x \partial y} \) reveal how the function's slope changes in a diagonal direction moving across both \( x \) and \( y \).
It is also important to check if the mixed partials are equal, which is a result of Clairaut's theorem, confirming the function's smoothness.
Chain Rule
The chain rule in calculus helps in differentiating composite functions. When a function is expressed as an outer function working on an inner function, like \( H(x, y) = \sqrt{4 + x^2 + y^2} \), the chain rule becomes essential for finding derivatives.
In partial derivatives, we often need the chain rule because changes in one variable impact a nested function. For instance, in the first partial derivative \( \frac{\partial H}{\partial x} \), the dependency is on the expression under the square root. Here's how it works step-by-step:
  • First, find the derivative of the outer function \( \sqrt{u} \) with respect to \( u \): \( \frac{1}{2\sqrt{u}} \).
  • Next, compute the derivative of the inner function \( 4 + x^2 + y^2 \) relative to \( x \), yielding \( 2x \).
  • Combine these by multiplying: \( \frac{x}{\sqrt{4 + x^2 + y^2}} \).
This rule allows us to break down complex derivatives, applying separate steps for each function component, making the process manageable.
Partial Derivatives of Functions
Partial derivatives capture how a function changes with respect to one variable while keeping other variables fixed. Let's look at the function \( H(x, y) = \sqrt{4 + x^2 + y^2} \). Here, we have two variables, \( x \) and \( y \), and need to explore how \( H \) changes as each variable varies independently.
Initially, we find the first partial derivatives:
  • \( \frac{\partial H}{\partial x} \): This measures how \( H \) changes as \( x \) changes, holding \( y \) constant. It simplifies understanding when moving in the \( x \) direction only.
  • \( \frac{\partial H}{\partial y} \): Similarly, it tells us the rate of change of \( H \) along the \( y \) axis, with \( x \) held fixed.
These derivatives create a mathematical foundation to evaluate how local changes around a point affect the function's value, guiding us to compute second partial derivatives later. Both concepts are pivotal in many mathematical and engineering fields, enabling detailed modeling of response to varying conditions.

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Most popular questions from this chapter

Let \(w=f(x, y, z)=2 x+3 y+4 z\) which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\) \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\)

Potential functions arise frequently in physics and engineering. A potential function has the property that \(a\) field of interest (for example, an electric field, a gravitational field, or a velocity field is the gradient of the potential (or sometimes the negative of the gradient of the potential). (Potential functions are considered in depth in Chapter 14 .) The electric field due to a point charge of strength \(Q\) at the origin has a potential function \(\varphi=k Q / r,\) where \(r^{2}=x^{2}+y^{2}+z^{2}\) is the square of the distance between a variable point \(P(x, y, z)\) and the charge, and \(k>0\) is a physical constant. The electric field is given by \(\mathbf{E}=-\nabla \varphi,\) where \(\nabla \varphi\) is the gradient in three dimensions. a. Show that the three-dimensional electric field due to a point charge is given by $$ \mathbf{E}(x, y, z)=k Q\left\langle\frac{x}{r^{3}}, \frac{y}{r^{3}}, \frac{z}{r^{3}}\right\rangle $$ b. Show that the electric field at a point has a magnitude \(|\mathbf{E}|=k Q / r^{2} .\) Explain why this relationship is called an inverse square law.

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=\cos (2(x+c t))$$

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=5 \cos (2(x+c t))+3 \sin (x-c t)$$

Let \(f\) be a differentiable function of one or more variables that is positive on its domain. a. Show that \(d(\ln f)=\frac{d f}{f}.\) b. Use part (a) to explain the statement that the absolute change in \(\ln f\) is approximately equal to the relative change in \(f.\) c. Let \(f(x, y)=x y,\) note that \(\ln f=\ln x+\ln y,\) and show that relative changes add; that is, \(d f / f=d x / x+d y / y.\) d. Let \(f(x, y)=x / y,\) note that \(\ln f=\ln x-\ln y,\) and show that relative changes subtract; that is \(d f / f=d x / x-d y / y.\) e. Show that in a product of \(n\) numbers, \(f=x_{1} x_{2} \cdots x_{n},\) the relative change in \(f\) is approximately equal to the sum of the relative changes in the variables.
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