Problem 45 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ R=r s^{2} t^{4} ; r=u e^{v^{2}}, s=v e^{-u^{2}}, t=e^{u^{2} v^{2}} ; \frac{\partial R}{\partial u}, \frac{\partial R}{\partial v} $$

Step-by-Step Solution

Answer

The chain rule is applied to find \( \frac{\partial R}{\partial u} \) and \( \frac{\partial R}{\partial v} \); both involve combining partial derivatives of \( r, s, t \). This results in expressions involving \( s^2 t^4 e^{v^2}, \) etc.

Step 1: Understand the Chain Rule for Partial Derivatives

The chain rule for partial derivatives is used to differentiate a function of several variables that are themselves functions of other variables. It combines the effect of the inner and outer derivatives. For example, to find \( \frac{\partial R}{\partial u} \), we need to consider how each component \( r, s, \) and \( t \) changes with respect to \( u \) and how they affect \( R \).

Step 2: Compute Partial Derivatives of r, s, t with respect to u

To find \( \frac{\partial r}{\partial u} \): \( r = u e^{v^2} \), so differentiate with respect to \( u \) to get \( \frac{\partial r}{\partial u} = e^{v^2} \). For \( s = v e^{-u^2} \), \( \frac{\partial s}{\partial u} = -2uv e^{-u^2} \). For \( t = e^{u^2 v^2} \), \( \frac{\partial t}{\partial u} = 2uv^2 e^{u^2 v^2} \).

Step 3: Compute Partial Derivatives of r, s, t with respect to v

For \( r = u e^{v^2} \), \( \frac{\partial r}{\partial v} = 2v u e^{v^2} \). For \( s = v e^{-u^2} \), \( \frac{\partial s}{\partial v} = e^{-u^2} \). For \( t = e^{u^2 v^2} \), \( \frac{\partial t}{\partial v} = 2u^2 v e^{u^2 v^2} \).

Step 4: Apply Chain Rule to Find \( \frac{\partial R}{\partial u} \)

Using the chain rule: \[ \frac{\partial R}{\partial u} = \frac{\partial R}{\partial r} \cdot \frac{\partial r}{\partial u} + \frac{\partial R}{\partial s} \cdot \frac{\partial s}{\partial u} + \frac{\partial R}{\partial t} \cdot \frac{\partial t}{\partial u} \] Where \( \frac{\partial R}{\partial r} = s^2 t^4 \), \( \frac{\partial R}{\partial s} = 2rs t^4 \), \( \frac{\partial R}{\partial t} = 4r s^2 t^3 \). Combine all partial derivatives: \[ s^2 t^4 e^{v^2} - 2uv r t^4 s e^{-u^2} + 4v^2 u r s^2 t^3 e^{u^2 v^2} \].

Step 5: Apply Chain Rule to Find \( \frac{\partial R}{\partial v} \)

Again, using the chain rule: \[ \frac{\partial R}{\partial v} = \frac{\partial R}{\partial r} \cdot \frac{\partial r}{\partial v} + \frac{\partial R}{\partial s} \cdot \frac{\partial s}{\partial v} + \frac{\partial R}{\partial t} \cdot \frac{\partial t}{\partial v} \] Substituting the previously calculated partial derivatives: \[ s^2 t^4 (2vu e^{v^2}) + 2rs t^4 e^{-u^2} + 4r s^2 t^3 u^2 v e^{u^2 v^2} \].

Step 6: Final Simplification and Combination

Simplify the expressions from steps 4 and 5. This may involve factoring out common terms and simplifying exponents. Combine like terms to get the simplest expression form for \( \frac{\partial R}{\partial u} \) and \( \frac{\partial R}{\partial v} \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with partial derivatives. It helps us differentiate functions that are composed of other functions—a perfect fit for multivariable scenarios. In essence, the chain rule allows us to understand how a function changes as its input variables change, by considering how each intermediate function contributes to the overall rate of change.

When using the chain rule for partial derivatives, we do not differentiate in a straightforward manner. Instead, we calculate the partial derivative of the outer function with respect to an inner variable, multiplied by the derivative of the inner variable itself with respect to the primary variable.

• Consider a function, such as several layers of functions: think of this process like peeling an onion layer by layer.
• The process involves forming and summing products of derivatives for each layer.

This method is indispensable in multivariable scenarios where changes in one variable affect many others through complex relationships.

Multivariable Calculus

Multivariable calculus extends the basic principles of differentiation and integration from single-variable calculus to functions of several variables. This branch of mathematics is all about working with functions that have more than one input, requiring us to consider how changes in each input affect the function's output.

Working in a multivariable context, we encounter concepts like partial derivatives, gradients, and Jacobians, which give insights into the behavior of these functions. Importantly, in multivariable calculus:

• Functions can depend on two or more variables.
• We compute derivatives with respect to one variable while considering the others as constants.
• Derivatives can be taken for any combination of variables.

This makes multivariable calculus vital for analyzing scenarios in physics, engineering, and economics, where relationships between multiple factors need to be understood and optimized.

Differentiation

Differentiation is a key technique in calculus that lets us determine how a function changes as its input variables change. In the context of partial derivatives, differentiation involves computing the derivative of a multivariable function with respect to one variable, keeping other variables fixed. This process shows how that specific variable contributes to the function's rate of change.

In our exercise, the differentiation process involves finding partial derivatives such as \( \frac{\partial R}{\partial u} \) and \( \frac{\partial R}{\partial v} \). A few steps in this process include:

• Identifying each component function in the expression.
• Calculating derivatives of these components with respect to each variable of interest.
• Using these derivatives to form the final derivatives of the overall function regarding each variable involved.

Differentiation, therefore, provides a granular insight into the behavior of functions, essential for predictions and optimizations in many fields.