Integral of a vector with respect to another vector.

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The discussion centers on evaluating the integral of work done by a varying force, defined as the dot product of the force vector and the differential displacement vector. The integral is expressed as the sum of forces integrated along a curve parameterized by time, where the differential displacement is represented as d\vec{r} = (d\vec{r}/dt) dt. The correct approach involves calculating the dot product of the force vector and the derivative of the position vector, followed by integrating this product with respect to time. This method allows for the evaluation of work done along a specified path. Understanding this process is essential for applying the concept of work in physics.
Ludwig
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My physics text gives the following as a general definition of work done by a varying force on a system:
## \sum W = \int (\sum \vec{F}) \cdot d \vec{r} ##
Unfortunately, I haven't the faintest idea how to evaluate this. I know how to evaluate an integral with respect to some parameter, but not with respect to another vector. Help?

(Note: I would be particularly interested to know how to evaluate this if given ##\vec{F}(t)## and ##\vec{r}(t)## ).
 
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The integral is evaluated along a curve ##\vec r(t)## with some curve parameter ##t##. It holds that ##d\vec r = (d\vec r/dt) dt## and you can integrate between whatever parameter values you are interested in.
 
Orodruin said:
The integral is evaluated along a curve ##\vec r(t)## with some curve parameter ##t##. It holds that ##d\vec r = (d\vec r/dt) dt## and you can integrate between whatever parameter values you are interested in.

So, does that mean I would evaluate this like so? ## \int (\sum \vec{F} \cdot \frac{d\vec{r}}{dt})dt ##
I.e., evaluate the dot product of the the force and derivative of curve vectors, then integrate with respect to t.
 
Yes
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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