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Chapter 13: Problem 25

Find the four second partial derivatives of the following functions. $$h(x, y)=x^{3}+x y^{2}+1$$

Short Answer

Expert verified
Answer: The four second-order partial derivatives are: $$ \frac{\partial^2 h}{\partial x^2} = 6x, \frac{\partial^2h}{\partial x \partial y} = 2y, \frac{\partial^2h}{\partial y \partial x} = 2y, \text{ and } \frac{\partial^2h}{\partial y^2} = 0 $$

Step by step solution

01

Find the first-order partial derivatives of h(x, y)

To find the first-order partial derivatives, we'll differentiate the function h(x, y) with respect to x and y: $$ \frac{\partial h}{\partial x} = \frac{\partial (x^3 + xy^2 + 1)}{\partial x} $$ and $$ \frac{\partial h}{\partial y} = \frac{\partial (x^3 + xy^2 + 1)}{\partial y} $$
02

Calculate the first-order partial derivatives

Differentiate h(x, y) with respect to x and y: $$ \frac{\partial h}{\partial x} = 3x^2 + y^2 $$ and $$ \frac{\partial h}{\partial y} = 2xy $$
03

Find the second-order partial derivatives

Now, we'll differentiate the first-order partial derivatives obtained in step 2 with respect to x and y. This will give us the four second-order partial derivatives: $$ \frac{\partial^2 h}{\partial x^2} = \frac{\partial (\frac{\partial h}{\partial x})}{\partial x} $$ $$ \frac{\partial^2 h}{\partial x \partial y} = \frac{\partial (\frac{\partial h}{\partial x})}{\partial y} $$ $$ \frac{\partial^2 h}{\partial y \partial x} = \frac{\partial (\frac{\partial h}{\partial y})}{\partial x} $$ $$ \frac{\partial^2 h}{\partial y^2} = \frac{\partial (\frac{\partial h}{\partial y})}{\partial y} $$
04

Calculate the second-order partial derivatives

Differentiate the first-order partial derivatives with respect to x and y to obtain the second-order partial derivatives: $$ \frac{\partial^2 h}{\partial x^2} = 6x $$ $$ \frac{\partial^2h}{\partial x \partial y} = 2y $$ $$ \frac{\partial^2h}{\partial y \partial x} = 2y $$ $$ \frac{\partial^2h}{\partial y^2}=0 $$ The four second-order partial derivatives are: $$ \frac{\partial^2 h}{\partial x^2} = 6x, \frac{\partial^2h}{\partial x \partial y} = 2y, \frac{\partial^2h}{\partial y \partial x} = 2y, \text{ and } \frac{\partial^2h}{\partial y^2} = 0 $$

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Most popular questions from this chapter

In the advanced subject of complex variables, a function typically has the form \(f(x, y)=u(x, y)+i v(x, y),\) where \(u\) and \(v\) are real-valued functions and \(i=\sqrt{-1}\) is the imaginary unit. A function \(f=u+i v\) is said to be analytic (analogous to differentiable) if it satisfies the Cauchy-Riemann equations: \(u_{x}=v_{y}\) and \(u_{y}=-v_{x}\) a. Show that \(f(x, y)=\left(x^{2}-y^{2}\right)+i(2 x y)\) is analytic. b. Show that \(f(x, y)=x\left(x^{2}-3 y^{2}\right)+i y\left(3 x^{2}-y^{2}\right)\) is analytic. c. Show that if \(f=u+i v\) is analytic, then \(u_{x x}+u_{y y}=0\) and \(v_{x x}+v_{y y}=0\)

Identify and briefly describe the surfaces defined by the following equations. $$x^{2}+y^{2}+4 z^{2}+2 x=0$$

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1.\) \(u(x, t)=A e^{-a^{2} t} \cos a x,\) for any real numbers \(a\) and \(A\)

A function of one variable has the property that a local maximum (or minimum) occurring at the only critical point is also the absolute maximum (or minimum) (for example, \(f(x)=x^{2}\) ). Does the same result hold for a function of two variables? Show that the following functions have the property that they have a single local maximum (or minimum), occurring at the only critical point, but that the local maximum (or minimum) is not an absolute maximum (or minimum) on \(\mathbb{R}^{2}\). a. \(f(x, y)=3 x e^{y}-x^{3}-e^{3 y}\) b. \(f(x, y)=\left(2 y^{2}-y^{4}\right)\left(e^{x}+\frac{1}{1+x^{2}}\right)-\frac{1}{1+x^{2}}\) This property has the following interpretation. Suppose that a surface has a single local minimum that is not the absolute minimum. Then water can be poured into the basin around the local minimum and the surface never overflows, even though there are points on the surface below the local minimum.

Find an equation of the plane passing through (0,-2,4) that is orthogonal to the planes \(2 x+5 y-3 z=0\) and \(-x+5 y+2 z=8\)
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