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In Exercises \(13-24\) , draw a branch diagram and write a Chain Rule formula for each derivative. \(\frac{d z}{d t} \text { for } z=f(x, y), \quad x=g(t), \quad y=h(t)\)

Short Answer

Expert verified
Use the chain rule: \( \frac{dz}{dt} = \frac{\partial f}{\partial x} g'(t) + \frac{\partial f}{\partial y} h'(t) \).

Step by step solution

01

Understand the Problem

We need to find the derivative \( \frac{dz}{dt} \) for a function \( z = f(x, y) \) where \( x = g(t) \) and \( y = h(t) \). This requires using the Chain Rule for multivariable functions.
02

Visualize with a Branch Diagram

Draw a branch diagram to represent how \( z \) depends on \( t \). Start with \( t \) at the base branching to \( x \) and \( y \), since \( x = g(t) \) and \( y = h(t) \). Then branch \( x \) and \( y \) to \( z \) as \( z = f(x, y) \).
03

Apply the Chain Rule

According to the Chain Rule for functions of several variables, the derivative \( \frac{dz}{dt} \) is given by:\[\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}\]This formula combines the partial derivatives of \( f(x, y) \) with respect to \( x \) and \( y \), and the derivatives of \( x \) and \( y \) with respect to \( t \).
04

Calculate Partial Derivatives

Find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), which are the rates at which \( z \) changes with a small change in \( x \) and \( y \), keeping the other variable constant.
05

Determine Derivatives of x and y with respect to t

Calculate \( \frac{dx}{dt} = g'(t) \) and \( \frac{dy}{dt} = h'(t) \). These express how \( x \) and \( y \) change with respect to \( t \).
06

Substitute and Simplify

Substitute \( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{dx}{dt}, \) and \( \frac{dy}{dt} \) into the Chain Rule formula to find \( \frac{dz}{dt} \). The formula becomes:\[\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot g'(t) + \frac{\partial f}{\partial y} \cdot h'(t)\]

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Key Concepts

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

Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of more than one variable. It is essential in understanding how changes in several variables affect a particular outcome or response, often represented by a function. In the context of the given problem, the function \( z = f(x, y) \) illustrates a case where the output \( z \) depends on two independent input variables, \( x \) and \( y \).

These input variables, \( x \) and \( y \), are themselves dependent on another variable \( t \), represented by \( x = g(t) \) and \( y = h(t) \). This nested dependency is typical in multivariable calculus, where one must analyze and compute how variations or derivatives in one level translate through the entire system.

Comprehending this flow of dependencies allows us to use tools like the Chain Rule to discover how an overall change in the initial independent variables (like \( t \)) manifests in the dependent variables (like \( z \)). This understanding forms the backbone for modeling and solving complex real-world problems such as in physics and engineering that require tracking multiple changes in interrelated variables.
Partial Derivatives
Partial derivatives are central to multivariable calculus, serving as a way to examine the rate of change of a function with regard to one of its variables while holding the other variables constant. For the function \( z = f(x, y) \), we use partial derivatives to determine how \( z \) changes with respect to \( x \) and \( y \).

When we write \( \frac{\partial f}{\partial x} \), we are assessing the change in \( z \) when only \( x \) varies and \( y \) stays fixed. Similarly, \( \frac{\partial f}{\partial y} \) indicates the change in \( z \) due to a variation in \( y \) while keeping \( x \) constant. These derivatives are critical in applying the Chain Rule, helping to determine the overall rate of change of \( z \) as a function of another variable \( t \).

Calculating partial derivatives allows us to understand sensitivity within systems: which variable contributes more significantly to the change of \( z \) and how small perturbations in input values might impact the output. Therefore, mastering partial derivatives is fundamental to predicting outcomes in functions involving multiple variables.
Functional Dependencies
Functional dependencies describe the relationships between variables within a mathematical model. In this exercise, understanding how each variable depends on others helps formulate a correct application of the Chain Rule.

The dependency structure here begins with \( z = f(x, y) \), indicating that \( z \) directly depends on \( x \) and \( y \). Further, \( x \) and \( y \) are influenced by \( t \) through \( x = g(t) \) and \( y = h(t) \). This forms a chain of dependencies where a change in \( t \) affects both \( x \) and \( y \), and consequently, \( z \).

Drawing a branch diagram can significantly aid in visualizing these dependencies, as it maps out each pathway from \( t \) to \( z \). By tracing these pathways, it becomes clearer how a shift in one variable propagates through to affect the others.

Understanding these functional dependencies is also vital in fields such as economics and biology, where changes in one factor might ripple through a network of interactions, influencing the entire system's behavior. Recognizing these linkages helps predict system responses and strategize accordingly.

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Most popular questions from this chapter

Suppose that \begin{equation}w=x^{2}-y^{2}+4 z+t \quad \text { and } \quad x+2 z+t=25.\end{equation} Show that the equations \begin{equation}\frac{\partial w}{\partial x}=2 x-1 \quad \text { and } \quad \frac{\partial w}{\partial x}=2 x-2\end{equation} each give \(\partial w / \partial x,\) depending on which variables are chosen to be dependent and which variables are chosen to be independent. Identify the independent variables in each case.

The condition \(\nabla f=\lambda \nabla g\) is not sufficient Although \(\nabla f=\lambda \nabla g\) is a necessary condition for the occurrence of an extreme value of \(f(x, y)\) subject to the conditions \(g(x, y)=0\) and \(\nabla g \neq 0,\) it does not in itself guarantee that one exists. As a case in point, try using the method of Lagrange multipliers to find a maximum value of \(f(x, y)=x+y\) subject to the constraint that \(x y=16 .\) The method will identify the two points \((4,4)\) and \((-4,-4)\) as candidates for the location of extreme values. Yet the sum \((x+y)\) has no maximum value on the hyperbola \(x y=16\) . The farther you go from the origin on this hyperbola in the first quadrant, the larger the sum \(f(x, y)=x+y\) becomes.

Find the linearization \(L(x, y, z)\) of the function \(f(x, y, z)\) at \(P_{0} .\) Then find an upper bound for the magnitude of the error \(E\) in the approximation \(f(x, y, z) \approx L(x, y, z)\) over the region \(R\) $$ \begin{array}{l}{f(x, y, z)=\sqrt{2} \cos x \sin (y+z) \text { at } P_{0}(0,0, \pi / 4)} \\ {R :|x| \leq 0.01, \quad|y| \leq 0.01, \quad|z-\pi / 4| \leq 0.01}\end{array} $$

Maximizing a utility function: an example from economics In economics, the usefulness or utility of amounts \(x\) and \(y\) of two capital goods \(G_{1}\) and \(G_{2}\) is sometimes measured by a function \(U(x, y) .\) For example, \(G_{1}\) and \(G_{2}\) might be two chemicals a pharmaceutical company needs to have on hand and \(U(x, y)\) the gain from manufacturing a product whose synthesis requires different amounts of the chemicals depending on the process used. If \(G_{1}\) costs \(a\) dollars per kilogram, \(G_{2}\) costs \(b\) dollars per kilogram, and the total amount allocated for the purchase of \(G_{1}\) and \(G_{2}\) together is \(c\) dollars, then the company's managers want to maximize \(U(x, y)\) given that \(a x+b y=c .\) Thus, they need to solve a typical Lagrange multiplier problem. Suppose that $$U(x, y)=x y+2 x$$ and that the equation \(a x+b y=c\) simplifies to $$2 x+y=30.$$ Find the maximum value of \(U\) and the corresponding values of \(x\) and \(y\) subject to this latter constraint.

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=x^{2}+y^{2} $$

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