Find the curvature ? of the plane curve y = 2x^2 + 3x - 1 at x = 2. ? =
Added by James C.
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- The first derivative \( y' \) (or \( \frac{dy}{dx} \)) is calculated as follows: \[ y' = \frac{d}{dx}(2x^2 + 3x - 1) = 4x + 3 \] - The second derivative \( y'' \) (or \( \frac{d^2y}{dx^2} \)) is calculated as follows: \[ y'' = \frac{d}{dx}(4x + 3) = Show more…
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