How to Find the Curvature using the Cross Product Formula for r(t) = ti + t^2j + (t^2/2)k | Summary and Q&A

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May 24, 2020
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The Math Sorcerer
How to Find the Curvature using the Cross Product Formula for r(t) = ti + t^2j + (t^2/2)k

TL;DR

Find the curvature of a vector valued function using a formula that works only for space curves, involving cross product and derivatives.

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Q: What is the formula for finding the curvature of a vector valued function in three dimensions?

The formula is the magnitude of R prime of T cross R double prime of T divided by the magnitude of R prime cubed.

Q: What does the vector valued function need to have in order to use the curvature formula?

The vector valued function needs to have the components T, T^2, and T^2/2.

Q: How do you find the derivative and second derivative of the vector valued function?

The derivative is found by taking the derivative of each component, resulting in 1, 2T, and T. The second derivative is found similarly, resulting in 0, 2, and 1.

Q: How do you calculate the cross product of the derivatives?

The cross product is calculated using determinants, resulting in -J hat - K hat.

Q: What is the magnitude of the cross product?

The magnitude of the cross product is the square root of 5.

Q: What is the magnitude of the derivative cubed?

The magnitude of the derivative cubed is (1 + 5t^2)^(3/2).

Q: What is the final formula for curvature?

The curvature is given as the square root of 5 divided by (1 + 5t^2)^(3/2).

Summary & Key Takeaways

• The curvature of a vector valued function in three dimensions can be found using the formula: magnitude of R prime of T cross R double prime of T divided by magnitude of R prime cubed.

• To use the formula, the vector valued function must have the components T, T^2, and T^2/2.

• The derivative of the vector valued function can be found to be 1, 2T, and T, and the second derivative is 0, 2, and 1.

• The cross product of the derivatives is calculated using determinants, resulting in -J hat - K hat.

• The magnitude of the cross product is found to be the square root of 5, and the magnitude of the derivative cubed is (1 + 5t^2)^(3/2).

• The curvature is then given as the square root of 5 divided by (1 + 5t^2)^(3/2) or (1 + 5t^2)^(3/2).