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Chapter 10: Problem 5

If $$z=\sin \left(x^{2}+y^{2}\right), \quad x=v \cos (u), \quad y=v \sin (u)$$ find \(\partial z / \partial u\) and \(\partial z / \partial v\). The variables are restricted to domains on which the functions are defined. \(\partial z / \partial u=\) _________. \(\partial z / \partial v=\) _________.

Short Answer

Expert verified
\(\frac{\partial z}{\partial u} = -2v^2\cos(u)\sin(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin(u)\cos(u)\cos(v^2)\) \(\frac{\partial z}{\partial v} = 2v\cos^2(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin^2(u)\cos(v^2)\)

Step by step solution

01

Calculate preliminary partial derivatives

First, calculate \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\): \[\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} (\sin({x^2 + y^2})) = 2x \cos(x^2 + y^2)\] \[\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} (\sin({x^2 + y^2})) = 2y \cos(x^2 + y^2)\]
02

Replace x and y with given relations in terms of u and v

Now we know the partial derivatives of z with respect to x and y, let's replace x and y with the given relationships between x, y, u, and v: \[x = v\cos(u)\] \[y = v\sin(u)\] Substitute x and y into \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\): \[\frac{\partial z}{\partial x} = 2(v\cos(u)) \cos((v \cos(u))^2 + (v \sin(u))^2)\] \[\frac{\partial z}{\partial y} = 2(v\sin(u)) \cos((v \cos(u))^2 + (v \sin(u))^2)\]
03

Find partial derivatives of x and y with respect to u and v

Next, we need to find the following partial derivatives: \[\frac{\partial x}{\partial u}\] \[\frac{\partial x}{\partial v}\] \[\frac{\partial y}{\partial u}\] \[\frac{\partial y}{\partial v}\] Using the given relations between x, y, u, and v, we find: \[\frac{\partial x}{\partial u} = -v \sin(u)\] \[\frac{\partial x}{\partial v} = \cos(u)\] \[\frac{\partial y}{\partial u} = v \cos(u)\] \[\frac{\partial y}{\partial v} = \sin(u)\]
04

Calculate the required partial derivatives using the chain rule

Now we have all the required partial derivatives, let's find the partial derivatives of z with respect to u and v using the chain rule: \(\frac{\partial z}{\partial u}\) = \(\frac{\partial z}{\partial x}\) \(\frac{\partial x}{\partial u}\) + \(\frac{\partial z}{\partial y}\) \(\frac{\partial y}{\partial u}\) \(\frac{\partial z}{\partial v}\) = \(\frac{\partial z}{\partial x}\) \(\frac{\partial x}{\partial v}\) + \(\frac{\partial z}{\partial y}\) \(\frac{\partial y}{\partial v}\) Plug in all the derivatives we calculated in Steps 2 and 3, and we get: \[\frac{\partial z}{\partial u} = 2(v\cos(u)) \cos((v \cos(u))^2 + (v \sin(u))^2) ( -v \sin(u)) + 2(v\sin(u)) \cos((v \cos(u))^2 + (v \sin(u))^2) (v \cos(u))\] \[\frac{\partial z}{\partial v} = 2(v\cos(u)) \cos((v \cos(u))^2 + (v \sin(u))^2) (\cos(u)) + 2(v\sin(u)) \cos((v \cos(u))^2 + (v \sin(u))^2) (\sin(u))\] So, we have: \(\partial z / \partial u\)= \(-2v^2\cos(u)\sin(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin(u)\cos(u)\cos(v^2)\) \(\partial z / \partial v\)= \(2v\cos^2(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin^2(u)\cos(v^2)\) Now we have the requested derivatives in terms of u and v: \(\partial z / \partial u\) = \(-2v^2\cos(u)\sin(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin(u)\cos(u)\cos(v^2)\) \(\partial z / \partial v\) = \(2v\cos^2(u) \cos((v \cos(u))^2 + (v \sin(u))^2) + 2v\sin^2(u)\cos(v^2)\)

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Key Concepts

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

Chain Rule
In calculus, the chain rule is a fundamental technique for finding the derivative of a function. It simplifies the process of differentiating compositions of functions. When you have a function of several variables, like in this problem, the chain rule becomes even more crucial.
To use the chain rule here, imagine that you want to find how a change in one variable (say, \(u\)) affects another (\(z\)), given that \(z\) indirectly depends on \(u\) through other intermediate variables (\(x\) and \(y\)).
  • We first find the partial derivatives of \(z\) with respect to \(x\) and \(y\).
  • Next, we find the partial derivatives of \(x\) and \(y\) with respect to \(u\) and \(v\).
  • Finally, we plug these derivatives into the chain rule formula, which combines these partial derivatives to give us \(\frac{\partial z}{\partial u}\) and \(\frac{\partial z}{\partial v}\).
This method ensures that we consider all the paths through which \(u\) and \(v\) can influence \(z\). This interconnectedness is at the heart of multivariable calculus.
Trigonometric Substitution
Trigonometric substitution is a calculation tool typically used to simplify expressions involving trigonometric functions by introducing new variables.
In this problem, trigonometric substitution is cleverly used to define a systematic way of expressing \(x\) and \(y\) in terms of polar coordinates \(u\) and \(v\).
Note the following:
  • The variable \(x\) is given by \(x = v \cos(u)\), which means it relates the vector \(v\) to the angle \(u\) through the cosine function.
  • Similarly, \(y = v \sin(u)\) connects \(v\) and \(u\) using the sine function.
Using these relations simplifies our differentiation work, especially when we handle functions of two variables. This substitution allows us to leverage the chain rule more effectively, transforming our calculations into more manageable trigonometric identities.
Multivariable Calculus
Multivariable calculus extends the concepts and techniques of calculus to functions of more than one variable. In our problem, we're dealing with a function of three variables: \(x\), \(y\), and \(z\).
This branch of calculus studies how these variables interact and vary together. It's frequently used in fields like physics, engineering, and economics.
Here, key concepts include:
  • Partial Derivatives: These focus on differentiating a multivariable function with respect to one variable, treating others as constants.
  • Chain Rule for Multivariable Functions: Integrates partial derivatives to connect changes in dependent and independent variables.
  • Trigonometric Substitution: Helps to simplify variable transitions between different coordinate systems, like Cartesian and polar.
All of these concepts play critical roles in the problem, allowing us to analyze how changing \(u\) and \(v\) affects \(z\). Whether it's optimizing processes or simulating systems, multivariable calculus provides the tools necessary for understanding and solving complex real-world problems.
Jacobian Determinants
Although our problem here doesn't explicitly involve Jacobian determinants, these determinants are essential in the broader scope of multivariable calculus. They help in understanding transformations involving multiple variables.
Essentially, a Jacobian determinant is a matrix derivative that represents how much a function involving several variables stretches/shrinks space locally.
  • Informally, it's a measure of how volume elements change under the transformation governed by a set of functions.
  • Each element of the Jacobian matrix is a partial derivative of one variable with respect to another.
Being comfortable with Jacobian determinants helps solve complex transformations and integrals, especially when moving between coordinate systems. For those diving deeper into topics like optimization and economics, understanding Jacobians adds a powerful tool to their problem-solving repertoire. They can tell us crucial information about the behavior of dynamic systems and the interdependence between variables.

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Most popular questions from this chapter

Find the maximum and minimum values of \(f(x, y)=x y\) on the ellipse \(5 x^{2}+y^{2}=3\) maximum value \(=\) __________. minimum value \(=\) __________.

The Cobb-Douglas production function is used in economics to model production levels based on labor and equipment. Suppose we have a specific Cobb-Douglas function of the form $$f(x, y)=50 x^{0.4} y^{0.6}$$ where \(x\) is the dollar amount spent on labor and \(y\) the dollar amount spent on equipment. Use the method of Lagrange multipliers to determine how much should be spent on labor and how much on equipment to maximize productivity if we have a total of 1.5 million dollars to invest in labor and equipment.

Find the partial derivatives of the function $$f(x, y)=\int_{y}^{x} \cos \left(7 t^{2}+8 t-2\right) d t$$ \(f_{x}(x, y)=\) ___________. \(f_{y}(x, y)=\) ___________.

(a) Check the local linearity of \(f(x, y)=e^{-x} \cos (y)\) near \(x=1, y=1.5\) by filling in the following table of values of \(f\) for \(x=0.9,1,1.1\) and \(y=1.4,1.5,1.6 .\) Express values of \(f\) with 4 digits after the decimal point. (b) Next, fill in the table for the values \(x=0.99,1,1.01\) and \(y=\) 1.49,1.5,1.51 , again showing 4 digits after the decimal point. Notice if the two tables look nearly linear, and whether the second looks more linear than the first (in particular, think about how you would decide if they were linear, or if the one were more closely linear than the other). (c) Give the local linearization of \(f(x, y)=e^{-x} \cos (y)\) at (1,1.5) : Using the second of your tables: \(f(x, y) \approx\) ____________. Using the fact that \(f_{x}(x, y)=-e^{-x} \cos (y)\) and \(f_{y}(x, y)=-e^{-x} \sin (y):\) \(f(x, y) \approx\) ___________.

Consider the three points \((5,1),(7,3),\) and (9,4) . (a) Supposed that at (5,1) , we know that \(f_{x}=f_{y}=0\) and \(f_{x x}=0\), \(f_{y y}=0\), and \(f_{x y}>0\). What can we conclude about the behavior of this function near the point (5,1)\(? \quad(\square(5,1)\) is a local maximum \(\square(5,1)\) is a local minimum \(\square(5,1)\) is a saddle point \(\square(5,1)\) is a none of these) (b) Supposed that at \((7,3),\) we know that \(f_{x}=f_{y}=0\) and \(f_{x x}>0\), \(f_{y y}<0,\) and \(f_{x y}=0 .\) What can we conclude about the behavior of this function near the point (7,3)\(? \quad(\square(7,3)\) is a local maximum \(\square(7,3)\) is a local minimum \(\square(7,3)\) is a saddle point \(\square(7,3)\) is a none of these) (c) Supposed that at (9,4) , we know that \(f_{x}=f_{y}=0\) and \(f_{x x}<0\), \(f_{y y}=0,\) and \(f_{x y}>0 .\) What can we conclude about the behavior of this function near the point (9,4)\(? \quad(\square(9,4)\) is a local maximum \(\square(9,4)\) is a local minimum \(\square(9,4)\) is a saddle point \(\square(9,4)\) is a none of these) Using this information, on a separate sheet of paper sketch a possible contour diagram for \(f\).
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