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Magazine Chapter 12: Problem 9
Find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\). $$ w=e^{x^{2}+y^{2}} ; x=s \sin t, y=t \sin s $$
Short Answer
Expert verified
\(\frac{\partial w}{\partial t} = e^{(s \sin t)^2 + (t \sin s)^2} \left[ 2s^2 \sin t \cos t + 2t \sin s (\sin s + t \cos s) \right]\)
Step by step solution
01
Identify Variables and Given Equations
First, recognize that the function given is \( w = e^{x^2 + y^2} \). The variables \( x \) and \( y \) are expressed in terms of \( s \) and \( t \): \( x = s \sin t \) and \( y = t \sin s \). Our task is to find \( \frac{\partial w}{\partial t} \).
02
Differentiate w with respect to x and y
Apply the chain rule to find the partial derivatives of \( w \) with respect to \( x \) and \( y \):\[ \frac{\partial w}{\partial x} = \frac{d}{dx}(e^{x^2 + y^2}) = 2xe^{x^2 + y^2} \] \[ \frac{\partial w}{\partial y} = \frac{d}{dy}(e^{x^2 + y^2}) = 2ye^{x^2 + y^2} \]
03
Differentiate x and y with respect to t
Differentiate the expressions for \( x \) and \( y \) with respect to \( t \): \[ \frac{\partial x}{\partial t} = s \cos t \] \[ \frac{\partial y}{\partial t} = \sin s + t \cos s \]
04
Apply the Chain Rule to Find \(\frac{\partial w}{\partial t}\)
Using the chain rule, express \( \frac{\partial w}{\partial t} \) as:\[ \frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t} \] Substitute the previously found derivatives:\[ = (2x e^{x^2 + y^2}) (s \cos t) + (2y e^{x^2 + y^2}) (\sin s + t \cos s) \]
05
Substitute x and y
Substitute the expressions for \( x \) and \( y \) back into the equation:\[ = 2(s \sin t) e^{(s \sin t)^2 + (t \sin s)^2} \cdot (s \cos t) + 2(t \sin s) e^{(s \sin t)^2 + (t \sin s)^2} \cdot (\sin s + t \cos s) \] This expression gives \( \frac{\partial w}{\partial t} \) purely in terms of \( s \) and \( t \).
06
Simplify the Expression
Combine and simplify the expression as much as possible:\[ \frac{\partial w}{\partial t} = e^{(s \sin t)^2 + (t \sin s)^2} \left[ 2s^2 \sin t \cos t + 2t \sin s (\sin s + t \cos s) \right] \] Now the partial derivative is fully expressed in terms of \( s \) and \( t \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Derivative
In calculus, a partial derivative represents the rate of change of a function with respect to one variable while keeping other variables constant. This concept is crucial when dealing with functions of more than one variable, like in multivariable calculus. For a function such as \( w = f(x, y) \), the partial derivative with respect to \( t \), denoted as \( \frac{\partial w}{\partial t} \), is found by using the chain rule. The chain rule helps in computations involving composite functions by breaking them down into simpler parts. To apply this rule, we consider how \( x \) and \( y \) change with \( t \). Key considerations:
- Identify intermediary functions and how they relate to each other, e.g., \( x(t) \) and \( y(t) \).
- Compute the derivative of these variables with respect to \( t \).
- Apply the chain rule to connect these derivatives back to the original function \( w \).
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions of several variables. It is essential when exploring phenomena in higher dimensions or multiple parameters. Here, we are working with a function \( w = e^{x^2 + y^2} \), which depends on two variables \( x \) and \( y \). These variables are functions themselves, \( x(s, t) = s \sin t \) and \( y(s, t) = t \sin s \). This introduces the aspect of multiple dependencies where each input can affect the output via different pathways.Steps in handling such problems:
- Identify the primary function and its dependencies.
- Determine the derivatives with respect to the independent variables using the rules of calculus.
- Use the chain rule to evaluate how changes in input variables affect the output function \( w \).
Exponential Function
An exponential function, notably expressed as \( f(x) = e^{x} \), where \( e \) is a mathematical constant approximately equal to 2.71828, is significant in calculus due to its unique properties. Its derivative is proportional to the function itself, making it a key component when differentiating functions that involve compounded exponents.In the given problem, the function \( w = e^{x^2 + y^2} \) demonstrates how an exponential function can encapsulate the sum of variables. The role of exponential functions here underscores their utility in transforming variable-dependent expressions for differentiation purposes.Consider the following:
- The exponential term \( e^{x^2 + y^2} \) can be seen as a composition of functions, requiring careful application of the chain rule.
- The repeated appearance of the same exponential base in derivatives suggests a pattern that can simplify calculations.
- Interpreting these results through the lens of exponential growth or decay helps in understanding real-world applications, such as population growth or radioactive decay.
