Chapter 13: Q. 60 (page 1080)

Let $Ω$ and $Ω'$ be subsets of $R^{2}$ . Use the results of Exercise 59 to prove that if a transformation $T : Ω \to Ω'$ is invertible, and if both T and $T^{- 1}$ are differentiable, then role="math" $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = 1$ .

Short Answer

Expert verified

It is proved that $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = 1$ .

Step by step solution

Given information

Consider a chain of functions is defined as,

$x = x (u, v); y = y (u, v) u = u (s, t); v = v (s, t)$

The chain rule of Jacobian is defined as,

role="math" $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (s, t)} = \frac{\partial (x, y)}{\partial (s, t)}$

Proof

Consider a transformation $T : (x, y) \to (u, v) a n d T^{- 1} : (u, v) \to (x, y)$ .

Use the above-given definition of the chain rule of Jacobians to derive the relation.

$\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = \frac{\partial (x, y)}{\partial (x, y)} = 1$

Hence, proved.

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Let $Ω$ and $Ω'$ be subsets of $R^{2}$ . Use the results of Exercise 59 to prove that if a transformation $T : Ω \to Ω'$ is invertible, and if both T and $T^{- 1}$ are differentiable, then role="math" $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = 1$ .

Short Answer

Step by step solution

Given information

Proof

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Let Ωand Ω'be subsets of R2. Use the results of Exercise 59 to prove that if a transformation T:Ω→Ω'is invertible, and if both T and T-1are differentiable, then role="math" ∂(x,y)∂(u,v)∂(u,v)∂(x,y)=1.

Short Answer

Step by step solution

Given information

Proof

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

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Applied Mathematics

Study anywhere. Anytime. Across all devices.

Let $Ω$ and $Ω'$ be subsets of $R^{2}$ . Use the results of Exercise 59 to prove that if a transformation $T : Ω \to Ω'$ is invertible, and if both T and $T^{- 1}$ are differentiable, then role="math" $\frac{\partial (x, y)}{\partial (u, v)} \frac{\partial (u, v)}{\partial (x, y)} = 1$ .