Problem 43 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ w=\left(u^{2}+v^{2}\right)^{3 / 2} ; u=e^{-t} \sin \theta, v=e^{-t} \cos \theta ; \frac{\partial w}{\partial t}, \frac{\partial w}{\partial \theta} $$

Step-by-Step Solution

Answer

\(\frac{\partial w}{\partial t} = -3e^{-t}(u \sin \theta + v \cos \theta)(u^2 + v^2)^{1/2}\) \(\frac{\partial w}{\partial \theta} = 3e^{-t}(u \cos \theta - v \sin \theta)(u^2 + v^2)^{1/2}\)

Step 1: Identify the Composite Function

We need to find the partial derivatives with respect to both \(t\) and \(\theta\). First, recognize that \(w\) is a function of \(u\) and \(v\), which are in turn functions of \(t\) and \(\theta\). This makes \(w\) a composite function, which requires the use of the chain rule for partial differentiation.

Step 2: Find the Partial Derivative \(\frac{\partial w}{\partial u}\) and \(\frac{\partial w}{\partial v}\)

Apply the power rule and chain rule to differentiate \(w\) with respect to \(u\) and \(v\):\[\frac{\partial w}{\partial u} = \frac{3}{2} (u^2 + v^2)^{\frac{1}{2}} \cdot 2u = 3u(u^2 + v^2)^{\frac{1}{2}}\]\[\frac{\partial w}{\partial v} = \frac{3}{2} (u^2 + v^2)^{\frac{1}{2}} \cdot 2v = 3v(u^2 + v^2)^{\frac{1}{2}}\]

Step 3: Differentiate \(u\) and \(v\) with Respect to \(t\) and \(\theta\)

Calculate the partial derivatives needed to apply the chain rule:\[\frac{\partial u}{\partial t} = -e^{-t} \sin \theta, \quad \frac{\partial u}{\partial \theta} = e^{-t} \cos \theta\]\[\frac{\partial v}{\partial t} = -e^{-t} \cos \theta, \quad \frac{\partial v}{\partial \theta} = -e^{-t} \sin \theta\]

Step 4: Apply the Chain Rule for \(\frac{\partial w}{\partial t}\)

Combine the partial derivatives using the chain rule:\[\frac{\partial w}{\partial t} = \frac{\partial w}{\partial u}\frac{\partial u}{\partial t} + \frac{\partial w}{\partial v}\frac{\partial v}{\partial t}\]Substitute the expressions derived in the previous steps:\[\frac{\partial w}{\partial t} = 3u(u^2 + v^2)^{\frac{1}{2}}(-e^{-t} \sin \theta) + 3v(u^2 + v^2)^{\frac{1}{2}}(-e^{-t} \cos \theta)\]Simplify to obtain:\[\frac{\partial w}{\partial t} = -3e^{-t}(u \sin \theta + v \cos \theta)(u^2 + v^2)^{\frac{1}{2}}\]

Step 5: Find \(\frac{\partial w}{\partial \theta}\) Using the Chain Rule

Apply the chain rule to find the partial derivative with respect to \(\theta\):\[\frac{\partial w}{\partial \theta} = \frac{\partial w}{\partial u}\frac{\partial u}{\partial \theta} + \frac{\partial w}{\partial v}\frac{\partial v}{\partial \theta}\]Substitute the previous results:\[\frac{\partial w}{\partial \theta} = 3u(u^2 + v^2)^{\frac{1}{2}}(e^{-t} \cos \theta) + 3v(u^2 + v^2)^{\frac{1}{2}}(-e^{-t} \sin \theta)\]Simplify the expression:\[\frac{\partial w}{\partial \theta} = 3e^{-t}(u \cos \theta - v \sin \theta)(u^2 + v^2)^{\frac{1}{2}}\]

Key Concepts

Chain Rule

In multivariable calculus, the chain rule is a fundamental tool for finding the derivative of a composite function. It helps us differentiate a function depending on multiple inputs, each being a function of other variables.
Think of it as following a trail through a dependency chain to find out how a change in one variable affects another.
When working with composite functions, the chain rule allows us to break down complex expressions. It enables differentiation with respect to different variables by following successive dependencies.
For instance, if you have a function like \(w\), which depends on \(u\) and \(v\), and these depend on \(t\) and \(\theta\), the chain rule links these derivatives together smoothly.
Apply the chain rule in a step-by-step manner:

• Differentially break down your function into smaller, manageable parts.
• Find the partial derivatives of each part.
• Use the chain rule by substituting these partials in the required expressions.

Partial Derivatives

Partial derivatives delve into understanding how multivariable functions change one variable at a time. It isolates how changing one input of a function affects the output while keeping other inputs constant. This concept is essential in optimization and sensitivity analysis.
For example, to find \( \frac{\partial w}{\partial t} \), where \( w = (u^2 + v^2)^{\frac{3}{2}} \), we first determine how \( u \) and \( v \) change with \( t \) and how \( w \) relies on \( u \) and \( v \).
The notation \( \frac{\partial w}{\partial u} \) indicates deriving \( w \) with respect to \( u \) while considering \( v \) as a constant. It's like asking, "What happens to \( w \) when only \( u \) increases?"
Partial derivatives use the Leibniz notation, indicating the variable held constant to avoid confusion. Key steps are:

• Differentiating functions while assuming that all but one variable remain unchanged.
• Understanding each variable's direct influence on the whole function.

Composite Function

Composite functions are combinations of two or more functions, where the output of one function becomes the input of another. This structure allows for constructing complex expressions from simpler parts.
Consider our function \( w(u, v) = (u^2 + v^2)^{\frac{3}{2}} \). Here, \( u \) and \( v \) are themselves dependent on \( t \) and \( \theta \). This indicates that \( w \) isn't just a function of \( u \) and \( v \), but indirectly of \( t \) and \( \theta \) as well.
Composite functions follow these patterns:

• Always think of the input-output relationship. Identify what goes into the function and what comes out.
• Analyze each layer of function dependency.
• Use functions within functions explicitly, with clear mapping from inputs to outputs.

Understanding composite functions helps in untangling complex relationships in multivariable calculus scenarios.