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Let \(z=f(x, y), x=g(s, t),\) and \(y=h(s, t) .\) Explain how to find \(\partial z / \partial t\)

Short Answer

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Question: Given a function z = f(x, y), where x = g(s, t) and y = h(s, t), find the partial derivative of z with respect to t, ∂z/∂t, using the chain rule.

Step by step solution

01

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

Using the given function f(x, y), find the partial derivative of z with respect to x (∂z/∂x) and the partial derivative of z with respect to y (∂z/∂y).
02

Find the partial derivatives of x and y with respect to t

Using the given function g(s, t) and h(s, t), find the partial derivative of x with respect to t (∂x/∂t) and the partial derivative of y with respect to t (∂y/∂t).
03

Apply the chain rule

To find the partial derivative of z with respect to t, we must apply the chain rule. This states: $$ \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t} $$ Here, we substitute our expressions from Steps 1 and 2 into the chain rule formula. This will give us the partial derivative of z with respect to t, ∂z/∂t.

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Key Concepts

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

Chain Rule in Multivariable Calculus
The chain rule is a powerful tool in multivariable calculus that helps us find how changing one variable impacts another in a system of interdependent variables. Imagine you have a function, say \( z = f(x, y) \), where \( x \) and \( y \) themselves are functions of other variables, like \( s \) and \( t \). Here, the chain rule lets us determine how a small change in \( t \) affects \( z \). We express this mathematically as:- \( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \)Each component of this formula represents a part of the chain reaction:- \( \frac{\partial z}{\partial x} \): how \( z \) changes with \( x \).- \( \frac{\partial x}{\partial t} \): how \( x \) changes with \( t \).- \( \frac{\partial z}{\partial y} \): how \( z \) changes with \( y \).- \( \frac{\partial y}{\partial t} \): how \( y \) changes with \( t \).Together, they allow us to weave through the variables and stitch them into the final derivative \( \frac{\partial z}{\partial t} \).
Understanding Partial Derivatives
Partial derivatives are essential in understanding functions of several variables. They're like regular derivatives but for functions that depend on more than one variable. When we take a partial derivative, we focus on how the function changes in relation to one variable while keeping other variables constant. For example, suppose we have a function \( z = f(x, y) \). The partial derivative \( \frac{\partial z}{\partial x} \) tells us how \( z \) changes as \( x \) changes with \( y \) held constant.In the step-by-step solution, we
  • First calculated \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) to see how \( z \) changes with \( x \) and \( y \)
  • Then found \( \frac{\partial x}{\partial t} \) and \( \frac{\partial y}{\partial t} \), which show how \( x \) and \( y \) change with respect to \( t \)
This information is then combined using the chain rule to find \(\frac{\partial z}{\partial t} \). Partial derivatives thus form the backbone of understanding how different variables influence each other in multivariable functions.
Functions of Several Variables
Functions of several variables, like \( f(x, y) \), \( g(s, t) \), and \( h(s, t) \), are more complex than single-variable functions because they depend on multiple inputs. These functions can represent various phenomena, such as temperature distribution across a surface or the behavior of a physical system influenced by multiple factors.In these functions, the value of the dependent variable (like \( z \) in \( z = f(x, y) \)) depends on several independent variables. Each independent variable can change independently, allowing the function to capture the interconnectedness of different influences.When dealing with these functions, we often make use of partial derivatives and the chain rule to explore how changes in one or more independent variables affect the dependent variable. This approach helps us
  • Understand complex systems
  • Create models that simulate real-world scenarios
  • Predict the outcome of changes within those systems
By mastering functions of several variables, we gain the analytical tools necessary to tackle challenges across various fields, from physics and engineering to economics and beyond.

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Most popular questions from this chapter

Traveling waves (for example, water waves or electromagnetic waves) exhibit periodic motion in both time and position. In one dimension, some types of wave motion are governed by the one-dimensional wave equation $$\frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u(x, t)\) is the height or displacement of the wave surface at position \(x\) and time \(t,\) and \(c\) is the constant speed of the wave. Show that the following functions are solutions of the wave equation. $$u(x, t)=\cos (2(x+c t))$$

Identify and briefly describe the surfaces defined by the following equations. $$-y^{2}-9 z^{2}+x^{2} / 4=1$$

Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}-4 y^{2}+x y ; R=\left\\{(x, y): 4 x^{2}+9 y^{2} \leq 36\right\\}$$

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 \(1.55 \Omega.\)

Among all triangles with a perimeter of 9 units, find the dimensions of the triangle with the maximum area. It may be easiest to use Heron's formula, which states that the area of a triangle with side length \(a, b,\) and \(c\) is \(A=\sqrt{s(s-a)(s-b)(s-c)},\) where \(2 s\) is the perimeter of the triangle.

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