I did a mechanics sum in which I considered (dx)^2=0 exactly. My friend did it without calculus and got the same result. Does that mean (dx)^2 is exactly 0? How?The sum was to find the distance moved under closed path of x-co-ordinate as a function of time.X(t)=bt(1-at)?

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I did a mechanics sum in which I considered (dx)^2=0 exactly. My friend did it without calculus and got the same result. Does that mean (dx)^2 is exactly 0? How?The sum was to find the distance moved under closed path of x-co-ordinate as a function of time.X(t)=bt(1-at)?

I did a mechanics sum in which I considered (dx)^2=0 exactly. My friend did it without calculus and got the same result. Does that mean (dx)^2 is exactly 0? How?The sum was to find the distance moved under closed path of x-co-ordinate as a function of time.X(t)=bt(1-at)?

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I did a mechanics sum in which I considered (dx)^2=0 exactly. My friend did it without calculus and got the same result. Does that mean (dx)^2 is exactly 0? How?The sum was to find the distance moved under closed path of x-co-ordinate as a function of time.X(t)=bt(1-at)?

I did a mechanics sum in which I considered (dx)^2=0 exactly. My friend did it without calculus and got the same result. Does that mean (dx)^2 is exactly 0? How?The sum was to find the distance moved under closed path of x-co-ordinate as a function of time.X(t)=bt(1-at)?

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