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For Activities 9 through \(16,\) write formulas for the indicated partial derivatives for each of the multivariable functions. \(b(s, t, r)=\frac{s}{t}+\frac{t}{r}-(s t-t r)^{2}\) a. \(\frac{\partial b}{\partial s}\) b. \(\frac{\partial b}{\partial t}\) c. \(\frac{\partial h}{\partial r}\) d. \(\left.\frac{\partial h}{\partial r}\right|_{(s, t, r)=(1,2,-1)}\)

Short Answer

Expert verified
The partial derivatives are: a. \( \frac{1}{t} - 2t(st - tr) \) b. \( -\frac{s}{t^2} + \frac{1}{r} - 2(st-tr)(s-r) \) c. \( -\frac{t}{r^2} + 2t(st-tr) \) d. \( 30 \).

Step by step solution

01

Understand the Function

The given function is a multivariable function of three variables: \( b(s, t, r) = \frac{s}{t} + \frac{t}{r} - (st - tr)^2 \). We will find the partial derivatives of this function with respect to each variable as requested.
02

Calculate \( \frac{\partial b}{\partial s} \)

To find the partial derivative of \( b \) with respect to \( s \), treat \( t \) and \( r \) as constants. The derivative of \( \frac{s}{t} \) with respect to \( s \) is \( \frac{1}{t} \), and the derivative of \( -(st - tr)^2 \) with respect to \( s \) is \( -2(st-tr)t \). The complete partial derivative is: \[ \frac{\partial b}{\partial s} = \frac{1}{t} - 2t(st - tr). \]
03

Calculate \( \frac{\partial b}{\partial t} \)

To find the partial derivative of \( b \) with respect to \( t \), treat \( s \) and \( r \) as constants. Use the quotient rule for \( \frac{s}{t} \) and \( \frac{t}{r} \), and apply the chain rule to \( -(st - tr)^2 \). The derivatives are:\[ \frac{d}{dt}\left(\frac{s}{t}\right) = -\frac{s}{t^2}, \quad \frac{d}{dt}\left(\frac{t}{r}\right) = \frac{1}{r}, \quad \text{and} \quad \frac{d}{dt}(-(st-tr)^2) = -2(st-tr)(s-r). \]Thus, combine these to get:\[ \frac{\partial b}{\partial t} = -\frac{s}{t^2} + \frac{1}{r} - 2(st-tr)(s-r). \]
04

Calculate \( \frac{\partial b}{\partial r} \)

For the partial derivative of \( b \) with respect to \( r \), treat \( s \) and \( t \) as constants. The derivative of \( \frac{t}{r} \) with respect to \( r \) is \( -\frac{t}{r^2} \), and the derivative of \(-(st-tr)^2\) with respect to \(r\) is \(2t(st-tr)\). Therefore:\[ \frac{\partial b}{\partial r} = -\frac{t}{r^2} + 2t(st-tr). \]
05

Evaluate \( \left.\frac{\partial b}{\partial r}\right|_{(s, t, r)=(1,2,-1)} \)

Substitute \( s = 1, t = 2, r = -1 \) into the expression for \( \frac{\partial b}{\partial r} \):\[ \frac{\partial b}{\partial r} = -\frac{2}{(-1)^2} + 2 \cdot 2 \cdot (1 \cdot 2 - 2 \cdot (-1)) = -2 + 8 \cdot 4 = 30. \]Thus, \[ \left.\frac{\partial b}{\partial r}\right|_{(s, t, r)=(1,2,-1)} = 30. \]

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Key Concepts

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

Multivariable Functions
A multivariable function is an important concept in calculus and involves functions with more than one input variable. In this context, our function is \( b(s, t, r) = \frac{s}{t} + \frac{t}{r} - (st - tr)^2 \), which is dependent on three variables: \( s \), \( t \), and \( r \). This means that the value of the function changes based on the values of these three variables.

Multivariable functions are useful in a variety of fields because they allow us to model and analyze situations where several factors are influencing an outcome. By understanding how the output of the function is affected by changes in each variable, we can make informed decisions or predictions.

When dealing with these types of functions, it is crucial to understand the role and interaction of each variable within the equation. Each variable can impact the function in distinct ways, and being able to isolate these effects through concepts like partial derivatives (as we do here) provides valuable insights.
Mathematical Notation
Mathematical notation is a standardized set of symbols used to express mathematical ideas succinctly. In this exercise, we primarily used symbols like \( \partial \) to denote partial derivatives. This is a special derivative symbol because, unlike regular derivatives that handle single-variable functions, partial derivatives work with multivariable functions.

For example, the notation \( \frac{\partial b}{\partial s} \) signifies the partial derivative of the function \( b \) with respect to variable \( s \), treating all other variables as constants. Similarly, expressions like \( \frac{s}{t} \) and \( (st - tr)^2 \) show the operations performed within the function. Mathematical notation serves as a universal language, allowing mathematicians and scientists to communicate complex ideas clearly and efficiently.

Understanding this notation is essential for solving calculus problems, as it tells you exactly what operation to perform. It's important to get familiar with it, as it becomes the foundation for more advanced study in mathematics and related fields.
Chain Rule
The chain rule is a powerful tool in calculus that allows us to differentiate composite functions. It is essential when functions are nested within one another, which happens often in multivariable contexts. For instance, in the exercise, we encounter composite expressions like \( (st - tr)^2 \).

When using the chain rule, we differentiate the outer function while multiplying it by the derivative of the inner function. In simplified terms, this all boils down to a single formula: if you have a function \( g(f(x)) \), then the derivative is \( g'(f(x)) \cdot f'(x) \). In our example, the outer function is "squared," and the inner function is \( st - tr \), so we differentiate accordingly.

Using the chain rule ensures that we accurately compute derivatives of more complex expressions, reflecting how changes in one part of the function affect the larger whole. It's a cornerstone concept for differentiation techniques in calculus and crucial for finding accurate solutions to problems involving nested functions.
Quotient Rule
The quotient rule comes into play when differentiating functions that are ratios of other functions. Given two functions \( u(x) \) and \( v(x) \), the quotient rule helps find the derivative of their ratio. It is formulated as:
  • \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \).
In this exercise, when finding the derivative of terms like \( \frac{s}{t} \) or \( \frac{t}{r} \), the quotient rule is essential.

To apply it, you determine the derivative of the numerator and the derivative of the denominator, then follow the quotient rule formula. For instance, with \( \frac{s}{t} \), we treat \( s \) as \( u \) and \( t \) as \( v \), leading to the partial derivative expression \( -\frac{s}{t^2} \) when differentiated with respect to \( t \).

The quotient rule is vital for handling division in differentiation, ensuring accurate results when tackling calculus problems that involve ratio terms. Whether you're dealing with single-variable or multivariable functions, this rule provides a systematic way to understand how ratio changes impact overall function behavior.

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Most popular questions from this chapter

Sketch the contour curve indicated in the activity. Sketch the tangent line indicated and calculate its slope. $$ \begin{aligned} &f(s, t)=s \ln (2 t)+e^{-1.34 t}\\\ &f(s, t)=2 \text { for } 1 \leq t \leq 20\\\ &\text { tangent line at } t=10 \end{aligned} $$

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