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Chapter 14: Problem 45

Explain what is wrong with the statement. $$\begin{aligned} &\text { If } z=f(g(t), h(t)), \text { then } d z / d t=f\left(g^{\prime}(t), h(t)\right)+\\\ &f\left(g(t), h^{\prime}(t)\right) \end{aligned}$$

Short Answer

Expert verified
The problem misuses the chain rule; it should involve partial derivatives, not \( f \) evaluated at \( g' \) and \( h' \).

Step by step solution

01

Understand the Function Composition

We have a function \( z \) that is expressed as a composition of functions \( f \), \( g \), and \( h \). Specifically, \( z = f(g(t), h(t)) \). It means that \( z \) depends on \( t \) indirectly through \( g(t) \) and \( h(t) \).
02

Recall the Chain Rule

The chain rule of calculus is used to differentiate composite functions. For a function \( z = f(u, v) \), where \( u = g(t) \) and \( v = h(t) \), the derivative \( \frac{dz}{dt} \) is given by:\[\frac{dz}{dt} = \frac{\partial f}{\partial u} \cdot \frac{du}{dt} + \frac{\partial f}{\partial v} \cdot \frac{dv}{dt}\]This means you need the partial derivatives of \( f \) with respect to \( u \) and \( v \), and then multiply by the derivatives of \( u \) and \( v \) with respect to \( t \).
03

Identify the Mistake in the Statement

The given statement misinterprets the application of the chain rule. Instead of using partial derivatives of \( f \) (\( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \)), it incorrectly replaces them with \( f \) evaluated at the derivative of \( g \) and \( h \). The correct application should involve partial derivatives, not ordinary differentiation of the function at different points.
04

Correct the Statement

The correct formula should be:\[\frac{dz}{dt} = \frac{\partial f}{\partial x}(g(t), h(t)) \cdot \frac{dg}{dt} + \frac{\partial f}{\partial y}(g(t), h(t)) \cdot \frac{dh}{dt}\]This ensures that the derivatives account for the changes in \( g(t) \) and \( h(t) \), while respecting the multivariable nature of \( f \).

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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 functions of multiple variables, partial derivatives become a powerful tool. They help you understand how a function changes as one of the variables changes, while keeping the others constant. For example, if you have a function \( f(x, y) \), then:
  • \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) changes, while \( y \) is kept constant.
  • \( \frac{\partial f}{\partial y} \) measures how \( f \) changes as \( y \) changes, while \( x \) is constant.
This concept is crucial in multivariable calculus because it allows for a detailed exploration of how different variables influence the function's output. Understanding partial derivatives is essential when applying the chain rule to composite functions.
Composite Functions
A composite function is essentially a function made up of other functions. When you have a function like \( z = f(g(t), h(t)) \), \( z \) is dependent on \( t \) through two intermediary functions \( g(t) \) and \( h(t) \). Each of these functions affects \( z \) in different ways.
  • \( g(t) \) and \( h(t) \) are the inner functions that provide inputs to the outer function \( f \).
  • Changes in \( g(t) \) or \( h(t) \) directly affect \( z \), hence the term 'composite function'.
Understanding composite functions is vital because many real-world problems involve multiple dependent variables. Knowing how these are structured through composition allows us to use differentiation techniques to solve complex problems effectively.
Differentiation Techniques
Differentiation techniques, especially when dealing with multivariable functions, include using the chain rule. The chain rule is a powerful tool to differentiate composite functions. It states that the derivative of \( z = f(u, v) \) where \( u = g(t) \) and \( v = h(t) \) is:\[\frac{dz}{dt} = \frac{\partial f}{\partial u} \cdot \frac{du}{dt} + \frac{\partial f}{\partial v} \cdot \frac{dv}{dt}\]Breaking it down:
  • The partial derivatives \( \frac{\partial f}{\partial u} \) and \( \frac{\partial f}{\partial v} \) capture how \( f \) changes with respect to \( u \) and \( v \).
  • The derivatives \( \frac{du}{dt} \) and \( \frac{dv}{dt} \) measure the rate of change for \( u \) and \( v \) with respect to \( t \).
Correct application of these principles is crucial, particularly when correcting misstatements like the one in the exercise, ensuring the proper derivatives are used in conjunction with the chain rule.

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Most popular questions from this chapter

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

In Problems \(58-64,\) find the quantity. Assume that \(g\) is a smooth function and that $$ \nabla g(2,3)=-2 \vec{i}+\vec{j} \quad \text { and } \quad \nabla g(2.4,3)=4 \vec{i} $$ $$g_{y}(2.4,3)$$

The period, \(T\), of oscillation in seconds of a pendulum clock is given by \(T=2 \pi \sqrt{l / g},\) where \(g\) is the acceleration due to gravity. The length of the pendulum, \(l\) depends on the temperature, \(t\), according to the formula \(I=l_{0}\left(1+\alpha\left(t-t_{0}\right)\right)\) where \(l_{0}\) is the length of the pendulum at temperature \(t_{0}\) and \(a\) is a constant which characterizes the clock. The clock is set to the correct period at the temperature \(t_{0} .\) How many seconds a day does the clock gain or lose when the temperature is \(t_{0}+\Delta t ?\) Show that this gain or loss is independent of \(l_{0}\)

Find the gradient of the function. Assume the variables are restricted to a domain on which the function s defined. $$f(m, n)=m^{2}+n^{2}$$

Let \(f(100,100)=500\) and \(\operatorname{grad} f(100,100)=2 \vec{t}+3 \vec{j}\) (a) Find the directional derivative of \(f\) at the point (100,100) in the direction \(\vec{i}+\vec{j}\) (b) Use the directional derivative to approximate \(f(102,102)\)
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