Question

[Calculus] Converting from multiplication into integrals

Answer 1

Yeah so, if you have a function
\[P = I * V\] in a constant case (P, I, V are all constants) and you want to convert it into a function of time for example...
\[P(t) = V(t) * I(t)\]
really nothing changes, P is still I * V at every instant.
BUT in cases like
\[displacement = velocity * time\]
in order to measure this with d(t) and v(t) you need to integrate
\[d(t) = \int\limits v(t) dt\]
I know why it's not simply.... d(t) = v(t) * t because this doesn't sum all the values from v(0) up to v(t) as it goes along........

I'm trying to check if this pattern is correct, or can be proven:
\[A * B \rightarrow A(x) * B(x)\]
\[A*B \rightarrow \int\limits A(B) dB\]

Sorry this was long, Thank you!!

Answer 2

Trying to catch on to what you're saying. Sounds like you're trying to relate everything together through parameterization.

Answer 3

Hmm. Can you elaborate on how this relates to parametric equations? That sounds interesting.

I'm just trying to see when a constant-case multiplication turns into an integral or not when the variables become dependent

\[d = v * t\]
\[d(t) = \int\limits v(t) dt\]
So the pattern is:
\[x = y * z \rightarrow x(z) = \int\limits y(z)dz\]

I have no idea how to go about proving this, I still didn't find a counter example either.
I hope this helps. What do you need clarifying on?

Answer 4

@legomyego180 oh oops I dunno why I said parametric equations there. But yeah you're correct. I'm trying to see how parameterization affects multiplication generally i guess

Answer 5

d = vt only works when velocity is constant. Integral is used when velocity is not.

Linear equations of the form d = vt can be solved without integrals. Non linear equations of the same form cannot.

Answer 6

And also with P = IV, both current and voltage in a circuit are dependent on one another - this is not at all like d = vt; time in no way depends on velocity.

Answer 7

You can't use d(t) = v(t) * t to find displacement, with a velocity that is changing over time.

Answer 8

Since velocity is changing with respect to time, over a very small interval of time it is true that \( v(t) \approx \large \frac{\Delta s}{\Delta t} \) as \( \Delta t \to 0 \), provided \( v(t) \) is a differentiable function.

From this we can get a small increment of the displacement by rearranging that approximation.
$$ \Delta s \approx v(t) \cdot \Delta t $$ And that is basically where the integral comes in , it sums up many small increments of displacement

Answer 9

If you scroll down on this page, there is an example of a definite integral with displacement.
http://v-fedun.staff.shef.ac.uk/Integration%20and%20Differential%20Equations/ACS123_lecture_4.html
They use the notation \( \Delta x \) instead of \( \Delta s \), but it means the same thing.

Answer 10

Sums need to have a very specific form before you can cast them into a continuous integral. One such sum that can be turned into an integral is called a Riemann sum.

Answer 11

Yes I am aware! Thank you guys. I'm aware of what integrals are, and also in relation to the small increments building up.

So I guess based on what you guys are saying, I am correct?

P = I * V works as well as P(t) = I(t) * V(t) because they're independent of each others?

while x = y * z would turn into an integral if both x and y were dependent on z?
x(y) = y(z) * z << WRONG x(y) = int(y(z) dz)

Also, @perl Thank you! Yeah I'm aware of the conditions that have to be met before I can actually turn an equation into an integral of some sort!

Answer 12

I'm having trouble trying to understand what you're trying to do.

\( P = I(t) \cdot V(t) \) is not analogous to \( d(t) = v(t) \cdot t
\) since the \(V(t)\) in power is not the same as time.

A better example would be work in physics. For a variable force
\( dW = F(x) dx\)

Answer 13

@perl Oh umm, yeah I might have worded that poorly.
Yeah P(t) = I(t) * V(t) is not analogous to the time displacement equation. I wasn't trying to imply that. I guess this may be more theoretical than procedural but I just wanted to show that for some multiplicative equations where the variables are constant, what we really are trying to show is the accumulation effect of speed over a time period to get displacement for example. But in other cases, it is just a stretch, like in V = I * R for example

That was my line of thought, so I thought that any equation in the form
A = B * C would be written as \[A(C) = \int\limits B(C) dC\] IF A was meant to be the accumulation of B over C.

So it's just more of "What is meant by this in a constant quantities-case and how do I extend it to be dependent on a variable while preserving its meaning"

That probably is more application-based, but I think the Idea of multiplying dependent vs independent quantities helps me not attempt to write \[P(t)=\int\limits V(t)dI(t)\]
or something like that. THANK YOU VERY MUCH!!!

Answer 14

@LannyXX you've touched on some pretty important and often misunderstood concepts in integral calculus.

The case P(t) = V(t) I(t) gives you power at a point in time but doesn't give you the dissipated power overall unless power is related uniformly over time. (more specifically Power = energy/time = charge x voltage / time = qV/t = IV, so only current by technical definition is influenced by time.

The integrand variable does matter when you integrate.
Here's an easy example:

let v(t) = f(t) = dx/dt where x is the displacement.
Then we say
\[x = \int\limits v(t) ~dt = \int\limits f(t) dt \]

But now suppose v(x) = f(x) i.e. velocity was written in terms of displacment, x, rather than time. Then we couldn't use v(x) = f(x) = dx/dt as the variables for the integrand wouldn't match the differential.
\[x = \int\limits f(x) dt ~ NOT ~possible\]
However if you sufficiently adjust the equation you might be able to solve it but will come up with t as the subject, rather than displacement.
We may write instead dt/dx = 1/v(x) = 1/f(x)
\[\frac{ dt }{ dx } = \frac{ 1 }{ f(x) } \rightarrow t = \int\limits f(x)~dx \]

Acceleration the rate of change of velocity over time may be expressed in a number of ways.
\[a = \frac{ dv }{ dt } = \frac{ dx }{ dt } \times \frac{ dv }{ dx } = v(t) \frac{ dv }{ dx } = \frac{ d }{ dv }(\frac{ 1 }{ 2 }v^2) \times \frac{ dv }{ dx } = \frac{ d }{ dx }(\frac{ 1 }{ 2 }v^2)\]

Thus we can write acceleration in terms of t, v or x and this will affect the corresponding integrals - notice the variables involved in the integrand compared to the differential:

\[v = \int\limits a(t) ~dt \]
\[a(v) = v \frac{ dv }{ dx } \rightarrow \frac{ dx }{ dv }=\frac{ v }{ a(v) } \rightarrow x = \int\limits \frac{ v ~dv }{ a(v) }\]
\[a(x) = \frac{ d }{ dx } (\frac{ 1 }{ 2 } v^2) \rightarrow v^2 = 2\int\limits a(x) dx\]

This is studied in kinematics. Key point is that you may only integrate with respect to the variable to which the integrand function is expressed in. However there are double integrals etc. to deal with 3 dimensional/multivariables beyond the scope of your general calculus.

Answer 15

@mww Whoa, that's interesting. Yeah, I can manipulates differentials to form different sort of integrals based on different variables that still MEAN the same thing, sure.

So two questions:
1) "The case P(t) = V(t) I(t) gives you power at a point in time but doesn't give you the dissipated power overall unless power is related uniformly over time. (more specifically Power = energy/time = charge x voltage / time = qV/t = IV, so only current by technical definition is influenced by time. "
Are you saying that P(t) calculated through V(t) I(t) is actually different from P(t) = dE/dt ?
What do you mean by P being related "uniformly" over time? how do you check for that?

2) \[\frac{ dt }{ dx } = \frac{ 1 }{ f(x) } \rightarrow t = \int\limits f(x) dx\]
Shouldn't this be \[dt = \frac{ dx }{ f(x) }\] \[t = \int\limits \frac{ 1 }{ f(x) } dt\]

Thank you very much!

Answer 16

@LannyXX "Are you saying that P(t) calculated through V(t) I(t) is actually different from P(t) = dE/dt ?"
You end up with the same result based on the final set of units. P(t) = dE/dt is the rawest form. Voltage is fundamentally defined as the work per unit charge and current as charge per unit time. So 1V is more accurately defined as 1J per coulomb charge.

It would be more helpful to explain integration thinking about areas and starting off a rate.
as you know velocity is the change in displacement over time
it is possible to find the the displacement travelled over a certain period by integrating over time, and this is represented by the area below the velocity-time graph.



\[v = v_o = \frac{ dx }{ dt } \rightarrow x = \int\limits v_0 dt = v_0t + C\]

\[v = kt =\frac{ dx }{ dt } \rightarrow x = \int\limits v dt = \int\limits kt dt = \frac{ 1 }{ 2 } kt^2 + C\] (actually area of the triangle)

Here is becomes clearer why we have a different form if the equation is not uniform constant for our rate of change.

And yes my second equation should be reciprocal but with a dx on the end.

Answer 17

Yes that makes sense!! P(t) = I(t) * V(t) is the same as P(t) = dE/dt in the same sense that acceleration due to gravity can be written as different equations it's all algebraic manipulations with the differentials right? it's so cool

Answer 18

@mww