Math Problem Statement

Answer

Show that for a function z of two variables x and y, with u and v defined as u = ax + by and v = cx + dy, the second-order partial derivatives of z satisfy the given equation.

Solution

The problem shown in the image appears to be about proving a formula involving partial derivatives. The goal is to show that for a function zz which is a function of two variables xx and yy, where uu and vv are new variables expressed as:

u=ax+byandv=cx+dy,u = ax + by \quad \text{and} \quad v = cx + dy,

the relationship between the second partial derivatives of zz with respect to xx and yy can be written in a specific form. The objective is to prove:

2zx2+2zy2=(ux)22zu2+2uxvx2zuv+(vx)22zv2\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = \left( \frac{\partial u}{\partial x} \right)^2 \frac{\partial^2 z}{\partial u^2} + 2 \frac{\partial u}{\partial x} \frac{\partial v}{\partial x} \frac{\partial^2 z}{\partial u \partial v} + \left( \frac{\partial v}{\partial x} \right)^2 \frac{\partial^2 z}{\partial v^2}

To proceed with the proof, I would apply the chain rule to the transformation from variables xx and yy to uu and vv, and compute the partial derivatives accordingly. Let's begin by first calculating the first-order derivatives and then proceed to the second-order derivatives.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Chain Rule
Multivariable Calculus

Formulas

u = ax + by
v = cx + dy
∂²z/∂x² + ∂²z/∂y² = (∂u/∂x)² ∂²z/∂u² + 2(∂u/∂x)(∂v/∂x) ∂²z/∂u∂v + (∂v/∂x)² ∂²z/∂v²

Theorems

Chain Rule for Partial Derivatives

Suitable Grade Level

Undergraduate (Calculus)