Find the Jacobian given x = ucos(theta) - vsin(theta), y = usin(theta) + vcos(theta) | Summary and Q&A

TL;DR

Learn how to find the Jacobian with ease using a determinant formula, resulting in a clean and straightforward answer.

Key Insights

• 💄 The Jacobian can be found using a determinant formula, making it easier to calculate partial derivatives systematically.
• 🚚 From the given functions, del x del u simplifies to cosine theta, del x del v simplifies to negative sine theta, del y del u simplifies to sine theta, and del y del v simplifies to cosine theta.
• ✖️ The resulting Jacobian of x and y with respect to u and v is equal to one, confirming a commonly used trigonometric identity.

Transcript

hi in this problem we're going to find the jacobian so we have to find the jacobian of x and y with respect to u and v and we're given x equals u cosine theta minus v sine theta and y equals u sine theta plus v cosine theta so there is a very nice formula that we can use to memorize this it's basically a determinant and in the first row we simply h... Read More

Questions & Answers

Q: What is the formula for finding the Jacobian of x and y with respect to u and v?

The formula is a determinant with del x del u, del x del v, del y del u, and del y del v as its elements.

Q: How does the partial derivative del x del u simplify to cosine theta?

Since the derivative of v is zero, del x del u only involves the derivative of u, which is one, resulting in cosine theta.

Q: What is the value of del y del v in the Jacobian?

Del y del v simplifies to cosine theta because the derivative of u is zero, and the derivative of v is one.

Q: How does the formula for the Jacobian simplify to cosine squared theta plus sine squared theta?

By evaluating each partial derivative and applying the formula, the Jacobian simplifies to the trigonometric identity cosine squared theta plus sine squared theta, which equals one.

Summary & Key Takeaways

• To find the Jacobian of x and y with respect to u and v, use a determinant formula: del x del u, del x del v, del y del u, and del y del v.

• By applying the formula, the partial derivatives can be calculated systematically, resulting in cosine squared theta plus sine squared theta.

• Contrary to initial expectations, finding the Jacobian is cleaner and simpler than anticipated.