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I'm following Professor R. Shankar's Fundamentals of Physics course on YouTube.

There I saw him doing manipulations of Calculus I never saw before.

Here it goes, $$\newcommand\deriv[2]{\frac{\mathrm d #1}{\mathrm d #2}} \deriv{v_t}{t}=a$$ $$\implies v_t \newcommand\deriv[2]{\frac{\mathrm d #1}{\mathrm d #2}} \deriv{v_t}{t}=a v_t$$ $$\implies \newcommand\deriv[2]{\frac{\mathrm d #1}{\mathrm d #2}} \deriv{\dfrac{{v_t}^2}{2}}{t}=a \newcommand\deriv[2]{\frac{\mathrm d #1}{\mathrm d #2}} \deriv{x_t}{t}$$

Then he cancels out $\mathrm dt$ from both sides and obtains, $\mathrm d(\dfrac{{v_t}^2}{2})=a\cdot \mathrm d(x_t)$

And proceeds by, $\dfrac{1}{2} \mathrm d({v_t}^2)=a\cdot \mathrm d(x_t)$

Thus deriving, $\dfrac{1}{2} ({v_t}^2-{v_0}^2)=a (x_t-x_0)$ $\implies {v_t}^2-{v_0}^2=2a(x_t-x_0)$

So, ${v_t}^2={v_0}^2+2a(x_t-x_0)$

How do you justify his calculations using Mathematical rigour?

Is it legitimate to cancel out $\mathrm dt$ from both sides? Aren't we considering $\newcommand\deriv[2]{\frac{\mathrm d #1}{\mathrm d #2}} \deriv{v_t}{t}$ as a fraction then? And Isn't there a better way of deriving this using Integral calculus?

Qmechanic
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ARahman
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1 Answers1

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I had the same doubt(can we cancel dt or any differential).I asked in a math forum.They told me that for basic calculations these can be be cancelled.But another interpretation is, if the rate of change two things with respect to time are propotional or equal then the change in their value for a given period of time will also be equal or propotional.

Mohan
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