Treating the derivative notation as a fraction?

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The discussion centers on the treatment of derivative notation, specifically dy/dx, as a fraction in solving physics problems involving integrals. While dy/dx is not technically a quotient, it behaves like one in many contexts, particularly when applying separation of variables in differential equations. Historical perspectives show that early mathematicians viewed differentials as real numbers, which contributes to this fractional treatment. The chain rule is also referenced, illustrating how derivatives can be manipulated to yield valid mathematical results. Ultimately, while the notation is not a fraction, it can be used as such in practical applications.
A.MHF
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Homework Statement


I was working on a physics problem that involves integrals and I stumbled upon this:

m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

Notice that we multiplied by dx on both sides and dx has been canceled from the right one.
I have learned before and it has been emphasised on me that the derivative notation isn't a fraction and has nothing to do with it, yet sometimes we treat it like one to solve a problem.
How do I make sense of this? Does it make a mathematical sense?

Homework Equations


m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

The Attempt at a Solution


-[/B]
 
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Historically speaking the quantities like dx were treated by Leibniz, Euler, and others as real numbers; very small, non zero numbers. These pioneers did not formulate calculus on the most rigorous of footings, but this was eventually done by other giants like Riemann, Cauchy, and Weierstrass. Yes, you are correct that dy/dx is not technically a quotient. However, the when solving differential equations with separation of variables one does treat dy/dx as a quotient. It is easy to show that the result is the same regardless of whether this technique is used or not (see http://www.math-cs.gordon.edu/courses/ma225/handouts/sepvar.pdf).

So... No, dy/dx is not a quotient, but it behaves like one in many different circumstances.
 
A.MHF said:

Homework Statement


I was working on a physics problem that involves integrals and I stumbled upon this:

m*g*sin θ − k*x = m*dv/dt =m*v*dv/dx
→ (m*g*sin θ − k*x)/m =v*dv/dx
→∫[(m*g*sin θ − k*x)/m]dx =∫v*dv

Notice that we multiplied by dx on both sides and dx has been canceled from the right one.
I have learned before and it has been emphasised on me that the derivative notation isn't a fraction and has nothing to do with it, yet sometimes we treat it like one to solve a problem.
How do I make sense of this? Does it make a mathematical sense?

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Think of this as using the chain rule.
dv/dt = (dv/dx)⋅(dx/dt) = v⋅(dv/dx), because dx/dt = v

If v is a differentiable function of x, then the differential dx is given by dx = (dv/dx)⋅dx .

Then ##\displaystyle\ \int \left(v\frac{dv}{dx}\right)dx=\int v\,dv\ ##
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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