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Chapter 10: Problem 11

Find the indicated derivative. In each case, state the version of the Chain Rule that you are using. a. \(\frac{d f}{d t},\) if \(f(x, y)=2 x^{2} y, x=\cos (t),\) and \(y=\ln (t) .\) b. \(\frac{\partial f}{\partial w},\) if \(f(x, y)=2 x^{2} y, x=w+z^{2},\) and \(y=\frac{2 z+1}{w}\) c. \(\frac{\partial f}{\partial v},\) if \(f(x, y, z)=2 x^{2} y+z^{3}, x=u-v+2 w, y=w 2^{v}-u^{3},\) and \(z=u^{2}-v\)

Short Answer

Expert verified
\(\frac{\partial f}{\partial w} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial w} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial w}\) \(\frac{\partial f}{\partial w}=(4xy)(1)+(2x^{2})(-\frac{2z+1}{w^2})\) Substitute \(x = w + z^2\) and \(y = \frac{2z + 1}{w}\): \(\frac{\partial f}{\partial w}=(4(w+z^2)(\frac{2z + 1}{w}))(1)+(2(w+z^2)^{2})(-\frac{2z+1}{w^2})\) #Step 5: Simplify the expression \(\frac{\partial f}{\partial w}=4(w+z^2)(2z+1)-2(w+z^2)^{2}\frac{2z+1}{w^2}\) c. Find \(\frac{\partial f}{\partial v}\) with \(f(x, y, z) = 2x^2y + z^3\), \(x = u -v + 2w\), \(y = 2^vw - u^3\), and \(z = u^2 - v\). #Step 1: Identify the variables and functions In this section, we want to find \(\frac{\partial f}{\partial v}\). The variables are \(x\), \(y\), \(z\), \(u\), \(v\), and \(w\), and \(f(x, y, z) = 2x^2y + z^3\). We have functions of \(u\), \(v\), and \(w\): \(x(u, v, w) = u -v + 2w\), \(y(u, v, w) = 2^vw - u^3\), and \(z(u, v, w) = u^2 - v\). #Step 2: Apply the Chain Rule We need to use the appropriate version of the Chain Rule for this case. In this case, since \(f\) is a function of three variables, \(x\), \(y\), and \(z\), we will apply the Chain Rule for multivariable functions, which is: \(\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial v}\) #Step 3: Compute the partial derivatives First, let's compute the necessary partial derivatives: 1. \(\frac{\partial f}{\partial x} = 4xy\) 2. \(\frac{\partial f}{\partial y} = 2x^2\) 3. \(\frac{\partial f}{\partial z} = 3z^2\) 4. \(\frac{\partial x}{\partial v} = -1\) 5. \(\frac{\partial y}{\partial v} = 2^vw\ln(2)\) 6. \(\frac{\partial z}{\partial v} = -1\) #Step 4: Plug in the values Now, we will substitute the values of \(x(u, v, w)\), \(y(u, v, w)\), and \(z(u, v, w)\) along with their respective derivatives into the Chain Rule formula: \(\frac{\partial f}{\partial v} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial f}{\partial z}\frac{\partial z}{\partial v}\) \(\frac{\partial f}{\partial v}=(4xy)(-1)+(2x^2)(2^vw\ln(2))+(3z^2)(-1)\) Substitute \(x = u - v + 2w\), \(y = 2^vw - u^3\), and \(z = u^2 - v\): \(\frac{\partial f}{\partial v}=4(u - v + 2w)(2^vw - u^3)(-1)+2(u - v + 2w)^2(2^vw\ln(2))+3(u^2 - v)^2(-1)\) #Step 5: Simplify the Expression \(\frac{\partial f}{\partial v}=-4(u - v + 2w)(2^vw - u^3) + 2(u - v + 2w)^2(2^vw\ln(2)) - 3(u^2 - v)^2\) In summary, the short answers for each part are: a. \(\frac{df}{dt}=-4\cos(t)\ln(t)\sin(t)+\frac{2\cos^2(t)}{t}\) b. \(\frac{\partial f}{\partial w}=4(w+z^2)(2z+1)-2(w+z^2)^{2}\frac{2z+1}{w^2}\) c. \(\frac{\partial f}{\partial v}=-4(u - v + 2w)(2^vw - u^3) + 2(u - v + 2w)^2(2^vw\ln(2)) - 3(u^2 - v)^2\)

Step by step solution

01

Identify the variables and functions

In this case, we want to find \(\frac{df}{dt}\). The variables are \(x\), \(y\), and \(t\), and \(f(x, y) = 2x^2y\). We have two functions of \(t\), \(x(t) = \cos(t)\) and \(y(t) = \ln(t)\).
02

Apply the Chain Rule

We need to use the appropriate version of the Chain Rule. In this case, since \(f\) is a function of two variables \(x\) and \(y\), we will apply the Chain Rule for implicit functions which is given by: \(\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\)
03

Compute the Partial Derivatives

First, let's compute the necessary partial derivatives: 1. \(\frac{\partial f}{\partial x} = 4xy\) 2. \(\frac{\partial f}{\partial y} = 2x^{2}\) 3. \(\frac{dx}{dt} = -\sin(t)\) 4. \(\frac{dy}{dt} = \frac{1}{t}\)
04

Plug in the values

Now, we will substitute the values of \(x(t)\) and \(y(t)\) along with their respective derivatives into the Chain Rule formula: \(\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}\) \(\frac{df}{dt}=(4xy)(-\sin(t))+(2x^{2})(\frac{1}{t})\) Substitute \(x = \cos(t)\) and \(y = \ln(t)\): \(\frac{df}{dt}=(4(\cos(t))(\ln(t)))(-\sin(t))+(2(\cos(t))^2)(\frac{1}{t})\)
05

Simplify the expression

Finally, simplify the expression: \(\frac{df}{dt}=-4\cos(t)\ln(t)\sin(t)+\frac{2\cos^2(t)}{t}\) b. Find \(\frac{\partial f}{\partial w}\) with \(f(x, y) = 2x^2y\), \(x = w + z^2\), and \(y = \frac{2z + 1}{w}\).
06

Identify the variables and functions

In this part, we want to find \(\frac{\partial f}{\partial w}\). The variables are \(x\), \(y\), \(w\), and \(z\), and \(f(x, y) = 2x^2y\). We have two functions of \(w\) and \(z\), \(x(w, z) = w + z^2\) and \(y(w, z) = \frac{2z + 1}{w}\).
07

Apply the Chain Rule

We need to use the appropriate version of the Chain Rule for this case. In this part, since \(f\) is a function of two variables, \(x\) and \(y\), we will apply the Chain Rule for implicit functions which is: \(\frac{\partial f}{\partial w} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial w} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial w}\)
08

Compute the Partial Derivatives

First, let's compute the necessary partial derivatives: 1. \(\frac{\partial f}{\partial x} = 4xy\) 2. \(\frac{\partial f}{\partial y} = 2x^{2}\) 3. \(\frac{\partial x}{\partial w} = 1\) 4. \(\frac{\partial y}{\partial w} = -\frac{2z+1}{w^2}\)
09

Plug in the values

Now, we will substitute the values of \(x(w, z)\) and \(y(w, z)\) along with their respective derivatives into the Chain Rule formula: \(...

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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 fundamental to tackling problems in multivariable calculus, especially when it comes to functions of multiple variables like in our textbook exercise. When you have a function f(x, y), a partial derivative represents how f changes as one variable changes, holding the other variable constant. For example, \(\frac{\partial f}{\partial x}\) measures the rate at which f changes with respect to x, while treating y as a constant.

When calculating partial derivatives, we apply the usual differentiation rules as if the other variables were simply constants. In our exercise, for the function f(x, y) = 2x^2y, we calculate the partial derivative with respect to x by treating y as a constant, which yields \(\frac{\partial f}{\partial x} = 4xy\). Similarly, when differentiating with respect to y, x is taken as a constant, resulting in \(\frac{\partial f}{\partial y} = 2x^2\). These partial derivatives are then used in the Chain Rule to find derivatives with respect to another variable.
Implicit Differentiation
Implicit differentiation is a technique used to find derivatives when the function is not explicitly solved for one variable in terms of others, hence the term 'implicit'. It comes into play particularly with equations that define one variable in terms of others in a complicated way that isn't easy to separate. Instead of solving for y to get y=f(x), we differentiate both sides of the equation with respect to x, and solve for \(\frac{dy}{dx}\).

In the context of our exercise, even though our function f(x, y) is given explicitly, we are looking for the derivative with respect to another variable like t, which x and y themselves depend on. This is where we use the Chain Rule in conjunction with implicit differentiation. By applying the Chain Rule, as done in the solution, we effectively perform implicit differentiation with respect to t. The calculated derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) are the results of implicit differentiation of x(t) and y(t).
Multivariable Calculus
The realm of multivariable calculus extends concepts of single-variable calculus to functions of several variables. It includes the study of partial derivatives, double and triple integrals, gradient vectors, and more. When dealing with functions like f(x, y, z), we are not just looking at how the function changes along one path, but in an entire space, each direction represented by a different variable.

In our initial problem, we approach a function f(x, y) that depends on two different functions of t. This interdependence between variables requires the use of multivariable calculus concepts such as the Chain Rule for multiple variables. The Chain Rule, in this case, allows us to combine the effects of changing x and y with respect to t to find out how f ultimately changes as t varies. This exercise perfectly exemplifies the application of multivariable calculus principles to solve practical problems involving related rates of change.

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Most popular questions from this chapter

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)

Let \(E(x, y)=\frac{100}{1+(x-5)^{2}+4(y-2.5)^{2}}\) represent the elevation on a land mass at location \((x, y)\). Suppose that \(E, x,\) and \(y\) are all measured in meters. a. Find \(E_{x}(x, y)\) and \(E_{y}(x, y)\). b. Let \(\mathbf{u}\) be a unit vector in the direction of \(\langle-4,3\rangle .\) Determine \(D_{\mathbf{u}} E(3,4)\). What is the practical meaning of \(D_{\mathbf{u}} E(3,4)\) and what are its units? c. Find the direction of greatest increase in \(E\) at the point (3,4) . d. Find the instantaneous rate of change of \(E\) in the direction of greatest decrease at the point \((3,4) .\) Include units on your answer. e. At the point \((3,4),\) find a direction \(\mathbf{w}\) in which the instantaneous rate of change of \(E\) is \(0 .\)

In the following questions, we investigate two different applied settings using the differential. a. Let \(f\) represent the vertical displacement in centimeters from the rest position of a string (like a guitar string) as a function of the distance \(x\) in centimeters from the fixed left end of the string and \(y\) the time in seconds after the string has been plucked. A simple model for \(f\) could be $$f(x, y)=\cos (x) \sin (2 y)$$ Use the differential to approximate how much more this vibrating string is vertically displaced from its position at \((a, b)=\left(\frac{\pi}{4}, \frac{\pi}{3}\right)\) if we decrease \(a\) by \(0.01 \mathrm{~cm}\) and increase the time by 0.1 seconds. Compare to the value of \(f\) at the point \(\left(\frac{\pi}{4}-0.01, \frac{\pi}{3}+0.1\right)\). b. Resistors used in electrical circuits have colored bands painted on them to indicate the amount of resistance and the possible error in the resistance. When three resistors, whose resistances are \(R_{1}, R_{2},\) and \(R_{3},\) are connected in parallel, the total resistance \(R\) is given by $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$$ Suppose that the resistances are \(R_{1}=25 \Omega, R_{2}=40 \Omega,\) and \(R_{3}=\) \(50 \Omega\). Find the total resistance \(R\). If you know each of \(R_{1}, R_{2}\), and \(R_{3}\) with a possible error of \(0.5 \%\), estimate the maximum error in your calculation of \(R\).

The dimensions of a closed rectangular box are measured as 60 centimeters, 60 centimeters, and 80 centimeters, respectively, with the error in each measurement at most .2 centimeters. Use differentials to estimate the maximum error in calculating the surface area of the box. __________ square centimeters.

Let \(z=g(u, v)\) and \(u(r, s), v(r, s)\). How many terms are there in the expression for \(\partial z / \partial r ?\) __________ terms
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