Chapter 12: Problem 7
In Problems \(7-12\), find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\) $$ w=x^{2} y ; x=s t, y=s-t $$
Short Answer
Expert verified
\(\frac{\partial w}{\partial t} = 2s^3 t - 3s^2 t^2\)
Step by step solution
01
Understanding the Problem
We need to find the partial derivative of \(w\) with respect to \(t\), that is \(\frac{\partial w}{\partial t}\), using the chain rule. The function \(w\) is given as \(w = x^2 y\), where \(x = st\) and \(y = s-t\). We need to express \(\frac{\partial w}{\partial t}\) in terms of \(s\) and \(t\).
02
Express w in Terms of s and t
Substitute \(x = st\) and \(y = s-t\) into \(w = x^2 y\):\[ w = (st)^2 (s-t) \]Simplify \(w\):\[ w = s^2 t^2 (s - t) = s^3 t^2 - s^2 t^3 \].
03
Differentiate w with Respect to t
To find \(\frac{\partial w}{\partial t}\), differentiate \(w = s^3 t^2 - s^2 t^3\) with respect to \(t\):\[ \frac{\partial w}{\partial t} = \frac{\partial}{\partial t}(s^3 t^2) - \frac{\partial}{\partial t}(s^2 t^3) \].
04
Apply the Power Rule
Differentiate each term separately using the power rule:- For \(s^3 t^2\), the derivative is:\[ \frac{\partial}{\partial t}(s^3 t^2) = 2s^3 t \]- For \(-s^2 t^3\), the derivative is:\[ \frac{\partial}{\partial t}(-s^2 t^3) = -3s^2 t^2 \].
05
Combine Results
Combine the results from Step 4:\[ \frac{\partial w}{\partial t} = 2s^3 t - 3s^2 t^2 \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Chain Rule
The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. When dealing with nested functions, the chain rule helps us find how a change in one variable affects another variable indirectly. This rule is particularly useful when we have functions of multiple variables, as it allows us to break down complex relationships into simpler parts.
For example, when we're given that - \(w = x^2 y\) - \(x = st\) - \(y = s-t\) we use the chain rule to express how changes in \(t\) affect \(w\) without needing to work directly with \(x\) and \(y\). By expressing \(w\) entirely in terms of \(s\) and \(t\), it is easier to calculate partial derivatives. This approach makes it manageable to analyze the dependency of the function \(w\) on its variables, turning a potentially intricate differentiation task into straightforward calculations.
For example, when we're given that - \(w = x^2 y\) - \(x = st\) - \(y = s-t\) we use the chain rule to express how changes in \(t\) affect \(w\) without needing to work directly with \(x\) and \(y\). By expressing \(w\) entirely in terms of \(s\) and \(t\), it is easier to calculate partial derivatives. This approach makes it manageable to analyze the dependency of the function \(w\) on its variables, turning a potentially intricate differentiation task into straightforward calculations.
Functions of Multiple Variables
A function of multiple variables is an equation that involves two or more independent variables. These functions are common in mathematical modeling and real-world applications where multiple factors interact to influence outcomes. In our context, the function \(w = x^2 y\) depends not only on \(x\) but also on \(y\). Furthermore, \(x\) and \(y\) themselves depend on \(s\) and \(t\).
Understanding how to work with multiple variables is key to solving complex problems in calculus and other fields. Through substitution, like expressing \(x\) and \(y\) in terms of \(s\) and \(t\), we simplify the function into a form \(w = s^3 t^2 - s^2 t^3\). This makes it straightforward to apply calculus techniques, allowing us to focus on varying one variable at a time while considering the others as constants.
Understanding how to work with multiple variables is key to solving complex problems in calculus and other fields. Through substitution, like expressing \(x\) and \(y\) in terms of \(s\) and \(t\), we simplify the function into a form \(w = s^3 t^2 - s^2 t^3\). This makes it straightforward to apply calculus techniques, allowing us to focus on varying one variable at a time while considering the others as constants.
Partial Differentiation
Partial differentiation extends the concept of taking a derivative from single-variable functions to functions of multiple variables. By taking the partial derivative, we compute the rate of change of a function with respect to one variable, while keeping other variables constant. This is an essential technique in multivariable calculus.
In our exercise, after substituting the expressions for \(x\) and \(y\) into \(w\), we find ourselves with \(w = s^3 t^2 - s^2 t^3\). To discover how \(w\) changes with respect to \(t\), we perform partial differentiation. This involves applying the power rule to differentiate each individual term regarding \(t\): - \(\frac{\partial}{\partial t}(s^3 t^2) = 2s^3 t\) - \(\frac{\partial}{\partial t}(-s^2 t^3) = -3s^2 t^2\). This process gives us the partial derivative \(\frac{\partial w}{\partial t} = 2s^3 t - 3s^2 t^2\). This expression illustrates the sensitivity of the function \(w\) in relation to changes in \(t\).
In our exercise, after substituting the expressions for \(x\) and \(y\) into \(w\), we find ourselves with \(w = s^3 t^2 - s^2 t^3\). To discover how \(w\) changes with respect to \(t\), we perform partial differentiation. This involves applying the power rule to differentiate each individual term regarding \(t\): - \(\frac{\partial}{\partial t}(s^3 t^2) = 2s^3 t\) - \(\frac{\partial}{\partial t}(-s^2 t^3) = -3s^2 t^2\). This process gives us the partial derivative \(\frac{\partial w}{\partial t} = 2s^3 t - 3s^2 t^2\). This expression illustrates the sensitivity of the function \(w\) in relation to changes in \(t\).
Mathematical Problem-Solving
Problem-solving in mathematics is not just about finding the answer but understanding the process that leads there. This involves a series of logical steps that include identifying the correct method, such as the chain rule, applying applicable mathematical rules, and simplifying where necessary.
By breaking the problem into smaller tasks, such as substituting known relationships like \(x = st\) and \(y = s-t\) into \(w = x^2 y\), a more approachable function is provided. Solving involves: - Simplification to get \(w = s^3 t^2 - s^2 t^3\) - Differentiating each term separately - Combining these to find \(\frac{\partial w}{\partial t}\)
Each of these steps has its rationale, and by going through them systematically, you reinforce not only your understanding but your ability to tackle new problems. Mastering these techniques enhances critical thinking and your capacity to apply mathematical principles in diverse contexts.
By breaking the problem into smaller tasks, such as substituting known relationships like \(x = st\) and \(y = s-t\) into \(w = x^2 y\), a more approachable function is provided. Solving involves: - Simplification to get \(w = s^3 t^2 - s^2 t^3\) - Differentiating each term separately - Combining these to find \(\frac{\partial w}{\partial t}\)
Each of these steps has its rationale, and by going through them systematically, you reinforce not only your understanding but your ability to tackle new problems. Mastering these techniques enhances critical thinking and your capacity to apply mathematical principles in diverse contexts.
