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Let \(z\) be a function of \(x\) and \(y,\) while \(x\) and \(y\) are functions of \(t\) Explain how to find \(\frac{d z}{d t}\).

Short Answer

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Question: Given that \(z\) is a function of \(x\) and \(y\), and both \(x\) and \(y\) are functions of \(t\), find the derivative of \(z\) with respect to \(t\). Solution: To find \(\frac{dz}{dt}\), we apply the chain rule, which gives us the following general expression: $$\frac{dz}{dt} = \left(\left.\frac{\partial f}{\partial x}(x, y)\right|_{x=g(t),\ y=h(t)}\right) \cdot g'(t) + \left(\left.\frac{\partial f}{\partial y}(x, y)\right|_{x=g(t),\ y=h(t)}\right) \cdot h'(t)$$

Step by step solution

01

Identify the given functions

We are given that \(z\) is a function of \(x\) and \(y\), and both \(x\) and \(y\) are functions of \(t\), so we can write: $$z = f(x, y)$$ $$x = g(t)$$ $$y = h(t)$$
02

Apply the chain rule

To find \(\frac{dz}{dt}\), we apply the chain rule, which states that the derivative of a function with respect to an intermediate variable is the product of partial derivatives with respect to each of the intermediate variables. In this case, we have: $$\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot\frac{dy}{dt}$$
03

Find the partial derivatives of \(z\) with respect to \(x\) and \(y\)

To find \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\), we differentiate \(f(x, y)\) with respect to \(x\) and \(y\), holding the other variable constant: $$\frac{\partial z}{\partial x} = \frac{\partial f}{\partial x}(x, y)$$ $$\frac{\partial z}{\partial y} = \frac{\partial f}{\partial y}(x, y)$$
04

Substitute the functions of \(t\) and their derivatives

Now we substitute \(x = g(t)\) and \(y = h(t)\), as well as their derivatives \(\frac{dx}{dt} = g'(t)\) and \(\frac{dy}{dt} = h'(t)\), into the chain rule equation we derived in Step 2: $$\frac{dz}{dt} = \left.\frac{\partial f}{\partial x}(x, y)\right|_{x=g(t),\ y=h(t)} \cdot g'(t) + \left.\frac{\partial f}{\partial y}(x, y)\right|_{x=g(t),\ y=h(t)} \cdot h'(t)$$
05

General expression for \(\frac{dz}{dt}\)

The expression for \(\frac{dz}{dt}\) that we derived in Step 4 is our general solution to the problem: $$\frac{dz}{dt} = \left(\left.\frac{\partial f}{\partial x}(x, y)\right|_{x=g(t),\ y=h(t)}\right) \cdot g'(t) + \left(\left.\frac{\partial f}{\partial y}(x, y)\right|_{x=g(t),\ y=h(t)}\right) \cdot h'(t)$$

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Key Concepts

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

Partial Derivatives
When dealing with a function of more than one variable, like \(z = f(x, y)\), it's essential to understand the concept of partial derivatives. A partial derivative measures how a function changes as one variable changes while keeping other variables constant.
For example, if you want to understand how \(z\) changes with \(x\), while keeping \(y\) constant, you calculate the partial derivative \(\frac{\partial f}{\partial x}(x, y)\). Similarly, \(\frac{\partial f}{\partial y}(x, y)\) shows how \(z\) changes with \(y\), keeping \(x\) constant.

Partial derivatives are foundational in exploring areas like gradient, divergence, and curl, and they are crucial in methods such as optimization and approximation in multivariable calculus settings.
Multivariable Calculus
Multivariable calculus is an extension of calculus to functions of several variables, like \(f(x, y)\). This field focuses on calculating and understanding how changes in multiple variables affect the outcome of a function.
Unlike single-variable calculus, where you only differentiate or integrate with respect to one variable, multivariable calculus requires handling all variables simultaneously.
  • Chain rule: This rule helps differentiate composite functions and is vital in multivariable settings. In this exercise, the chain rule links the changes in \(z\) to changes in \(x\) and \(y\) as they depend on \(t\).
  • Double and triple integrals: These are used to integrate functions over a plane or a spatial region, respectively.
  • Vector fields: These functions assign vectors to points in space and are foundational for understanding flow and field lines.
Understanding these concepts enhances our ability to model real-world phenomena using mathematics.
Derivatives with Respect to a Parameter
Sometimes, you encounter functions where the variables themselves are functions of another variable, often called a parameter. In this problem, \(x\) and \(y\) are functions of \(t\), the parameter.
To find how \(z\) changes with \(t\), we don't just differentiate \(z\) with respect to \(x\) and \(y\). Instead, we use the chain rule to account for the dependency of \(x\) and \(y\) on \(t\).

By substituting \(x = g(t)\) and \(y = h(t)\), and their respective derivatives \(g'(t)\) and \(h'(t)\), into the chain rule formula:
  • \(\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}\)
we find the total derivative of \(z\) with respect to the parameter \(t\).
This method is compelling because it allows us to see the direct effect \(t\) has on \(z\), through its influence on \(x\) and \(y\). Such calculations are beneficial in dynamic systems, where variables change over time or other parameters.

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Most popular questions from this chapter

A classical equation of mathematics is Laplace's equation, which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials, and the steadystate distribution of heat in a conducting medium. In two dimensions, Laplace's equation is $$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0.$$ Show that the following functions are harmonic; that is, they satisfy Laplace's equation. $$u(x, y)=e^{-x} \sin y$$

Extending Exercise \(62,\) when three electrical resistors with resistance \(R_{1}>0, R_{2}>0,\) and \(R_{3}>0\) are wired in parallel in a circuit (see figure), the combined resistance \(R,\) measured in ohms \((\Omega),\) is given by \(\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}\) Estimate the change in \(R\) if \(R_{1}\) increases from \(2 \Omega\) to \(2.05 \Omega, R_{2}\) decreases from \(3 \Omega\) to \(2.95 \Omega,\) and \(R_{3}\) increases from \(1.5 \Omega\) to \(1.55 \Omega.\)

In its many guises, the least squares approximation arises in numerous areas of mathematics and statistics. Suppose you collect data for two variables (for example, height and shoe size) in the form of pairs \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) The data may be plotted as a scatterplot in the \(x y\) -plane, as shown in the figure. The technique known as linear regression asks the question: What is the equation of the line that "best fits" the data? The least squares criterion for best fit requires that the sum of the squares of the vertical distances between the line and the data points is a minimum. Let the equation of the best-fit line be \(y=m x+b,\) where the slope \(m\) and the \(y\) -intercept \(b\) must be determined using the least squares condition. First assume that there are three data points \((1,2),(3,5),\) and \((4,6) .\) Show that the function of \(m\) and \(b\) that gives the sum of the squares of the vertical distances between the line and the three data points is $$ \begin{aligned} E(m, b)=&((m+b)-2)^{2}+((3 m+b)-5)^{2} \\ &+((4 m+b)-6)^{2} \end{aligned}. $$ Find the critical points of \(E\) and find the values of \(m\) and \(b\) that minimize \(E\). Graph the three data points and the best-fit line.

Consider the following functions \(f\). a. Is \(f\) continuous at (0,0)\(?\) b. Is \(f\) differentiable at (0,0)\(?\) c. If possible, evaluate \(f_{x}(0,0)\) and \(f_{y}(0,0)\). d. Determine whether \(f_{x}\) and \(f_{y}\) are continuous at (0,0). e. Explain why Theorems 12.5 and 12.6 are consistent with the results in parts \((a)-(d)\). $$f(x, y)=\sqrt{|x y|}$$

Let \(x, y,\) and \(z\) be non-negative numbers with \(x+y+z=200\) a. Find the values of \(x, y,\) and \(z\) that minimize \(x^{2}+y^{2}+z^{2}\) b. Find the values of \(x, y,\) and \(z\) that minimize \(\sqrt{x^{2}+y^{2}+z^{2}}\). c. Find the values of \(x, y,\) and \(z\) that maximize \(x y z\) d. Find the values of \(x, y,\) and \(z\) that maximize \(x^{2} y^{2} z^{2}\).

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