Problem 104 - Advanced Calculus

Question

If \(\mathrm{G}(\mathrm{s}, \mathrm{t})=\mathrm{F}\left(\mathrm{e}^{\mathrm{s}} \cos \mathrm{t}, \mathrm{e}^{\mathrm{s}} \sin \mathrm{t}\right)\), show that \(\mathrm{G}_{11}+\mathrm{G}_{22}=\mathrm{e}^{2 \mathrm{~s}}\left(\mathrm{~F}_{11}+\mathrm{F}_{22}\right)\), where \(\mathrm{G}_{11}\) and \(\mathrm{G}_{22}\) are evaluated at \((\mathrm{s}, \mathrm{t})\) and \(\mathrm{F}_{11}\) and \(\mathrm{F}_{22}\) are evaluated at \(\left(\mathrm{e}^{3} \cos \mathrm{t}, \mathrm{e}^{\mathrm{s}} \sin \mathrm{t}\right)\). Note: \(\mathrm{G}_{11}, \mathrm{G}_{22}\) are abbreviations for \(\left(\partial^{2} \mathrm{G} / \partial \mathrm{s}^{2}\right),\left(\partial^{2} \mathrm{G} / \partial \mathrm{t}^{2}\right)\) respectively and \(F_{11}, F_{22}\) are abbreviations for \(\left(\partial^{2} F / \partial s^{2}\right)\) and \(\left(\partial^{2} \mathrm{~F} / \partial \mathrm{t}^{2}\right)\), respectively.

Step-by-Step Solution

Answer

The sum of the second order partial derivatives of G with respect to s and t (G11 and G22) is given by \(G11 + G22 = e^{2s} \left(F_{11} + F_{22}\right)\), where \(F_{11}\) and \(F_{22}\) are the second order partial derivatives of F with respect to x and y, evaluated at \(\left(e^{3} \cos t, e^s \sin t\right)\).

Step 1: Calculate the first partial derivatives of G

Let's start by calculating the first-order partial derivatives of G with respect to s and t: \(\frac{\partial G}{\partial s} = \frac{\partial F}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial s}\) where \(x = e^s \cos t\) and \(y = e^s \sin t\). We have: \(\frac{\partial x}{\partial s} = e^s \cos t\) \(\frac{\partial y}{\partial s} = e^s \sin t\) So, \(\frac{\partial G}{\partial s} = \frac{\partial F}{\partial x} e^s \cos t + \frac{\partial F}{\partial y} e^s \sin t\) Now let's find the partial derivative of G with respect to t: \(\frac{\partial G}{\partial t} = \frac{\partial F}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial t}\) We have: \(\frac{\partial x}{\partial t} = -e^s \sin t\) \(\frac{\partial y}{\partial t} = e^s \cos t\) So, \(\frac{\partial G}{\partial t} = -\frac{\partial F}{\partial x} e^s \sin t + \frac{\partial F}{\partial y} e^s \cos t\)

Step 2: Calculate the second partial derivatives of G (G11 and G22)

Next, we need to calculate the second-order partial derivatives of G with respect to s and t. G11 = \(\frac{\partial^2 G}{\partial s^2}\) G22 = \(\frac{\partial^2 G}{\partial t^2}\) Using the previously found first-order partial derivatives, we can find G11 and G22: G11 = \(\frac{\partial}{\partial s} \left(\frac{\partial F}{\partial x} e^s \cos t + \frac{\partial F}{\partial y} e^s \sin t \right)\) G22 = \(\frac{\partial}{\partial t} \left(-\frac{\partial F}{\partial x} e^s \sin t + \frac{\partial F}{\partial y} e^s \cos t \right)\)

Step 3: Simplify G11 and G22 using product rule

To simplify G11 and G22, we can use the product rule for differentiation. The product rule states that: \(\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}\) Apply the product rule for G11 derivation: G11 = \(\frac{\partial F}{\partial x} \frac{\partial}{\partial s}(e^s \cos t) + \frac{\partial F}{\partial y} \frac{\partial}{\partial s}(e^s \sin t) + e^s \cos t \frac{\partial^2 F}{\partial x^2} + e^s \sin t \frac{\partial^2 F}{\partial y^2}\) Similarly, apply the product rule for G22 derivation: G22 = \(-\frac{\partial F}{\partial x} \frac{\partial}{\partial t}(e^s \sin t) + \frac{\partial F}{\partial y} \frac{\partial}{\partial t}(e^s \cos t) - e^s \sin t \frac{\partial^2 F}{\partial x^2} + e^s \cos t \frac{\partial^2 F}{\partial y^2}\)

Step 4: Compute the sum of G11 and G22

Now, we can calculate the sum G11 + G22: \(G11 + G22 = e^s \cos t \left(\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2}\right) + e^s \sin t \left(\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2}\right)\) Factoring out the common term \(e^s\) gives us: \((G11 + G22) = e^{2s} \left(\frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2}\right)\) This matches the given expression, so we have shown that: \(G11 + G22 = e^{2s} \left(F_{11} + F_{22}\right)\)

Key Concepts

Calculus

Calculus is a mathematical discipline that deals with rates of change and the accumulation of quantities. It's traditionally divided into two branches: differential calculus and integral calculus. Differential calculus heavily involves understanding derivatives, which measure how a function changes as its input changes. This is essential when we delve into more complex areas like multivariable calculus, where we consider functions with more than one variable.

Multivariable Calculus

Moving beyond the scope of single-variable calculus, multivariable calculus takes us into the world of functions with several inputs. In this realm, partial derivatives come into play, allowing us to probe the change of a function along one variable while keeping others fixed. This gives us the tools needed to map out the landscape of functions in higher dimensions and to identify critical points, optimize functions, and much more. Calculating the first and second-order partial derivatives is pivotal in many disciplines including physics, engineering, and economics, where they are applied to model real-world phenomena.

Chain Rule

The chain rule is a formula used to compute the derivative of a composition of two or more functions. In single-variable calculus, the chain rule is straightforward, but it becomes a bit more intricate in multivariable calculus. Notably, when we want to differentiate a function of a function, like in our problem where G(s, t) is a composition involving exponentials and trigonometric functions of s and t, the chain rule allows us to untangle this relationship into partial derivatives that are easier to manage. It's a powerful tool that enables us to break down complex derivatives into simpler components.

Second-Order Derivatives

Second-order derivatives are essentially 'derivatives of derivatives,' giving us information about the curvature or concavity of a function's graph. They play a crucial role in optimization problems, helping us distinguish between local minima, maxima, and points of inflection. In multivariable calculus, second-order partial derivatives can be particularly interesting. They can be mixed (involving different variables) or pure (involving the same variable). Depending on their signs and relationships, they contribute to our understanding of a function's behavior in space—much like studying the curvature of the landscape.