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Kepler's second law can also be derived from vector calculus. This derivation follows that given at [MIT OpenCourseWare](https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-c-parametric-equations-for-curves/session-21-keplers-second-law/MIT18_02SC_MNotes_k.pdf) and [OpenCourseWare](https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/index.htm).
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Kepler's second law can also be derived from vector calculus. This derivation follows that given at MIT OpenCourseWare (link defunct); a succinct approach is given by [Strang](https://ocw.mit.edu/courses/res-18-001-calculus-fall-2023/mitres_18_001_f17_ch12.pdf).
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The second law states that the area being swept out during a time duration only depends on the duration of time, not the time. Let $\Delta t$ be this duration. Then if $\vec{x}(t)$ is the position vector, as above, we have the area swept out between $t$ and $t + \Delta t$ is visualized along the lines of:
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