Fundamental concept misunderstanding

[joeyjon]

Hi,

I hope you can help. I do well in calculus, but i have a fundamental misunderstanding of a concept, but I am able to get the right answers anyway.

So here is a problem - I can get the right answer, but my questions will be after the problem...

Anyway, the computer put the ds there, it should be dt.



let u = square root t
du= (1/2) t^(-1/2) dt
dt= 2t^(1/2)du


so...

(sin u times 2t^1/2 du )/ the square root of t cos square root t cubed =

(2 sin u du) / the square root of cos u cubed


Let v= cos u
dv= -sin u du
du= dv/-sin u

so...

2 sin u times (dv/-sin u) / v^3/2 =

-2 v^(-3/2) dv =

4 v^(-1/2) =

4 / square root of cos square root t (the answer)


Now my question deals with the derivatives thrown in there, the dt, du, and dv...

Why isn't the answer 4dv / square root of cos square root t??

How come you can just throw the dv away? How come we didn't just throw the du away earlier in the middle steps of the problem?

If, in the beginning of the problem, the problem didn't have dt in it, would it make the problem impossible, or do we just assume the dt and write it in there anyway?

Also, could the problem have been written with the dt in the denominator instead of the numerator, making a completely different problem?


I am just unsure about this concept and exactly what the dt, du, and dv mean...

 

[HallsofIvy]

Hi,

I hope you can help. I do well in calculus, but i have a fundamental misunderstanding of a concept, but I am able to get the right answers anyway.

So here is a problem - I can get the right answer, but my questions will be after the problem...

Anyway, the computer put the ds there, it should be dt.



let u = square root t
du= (1/2) t^(-1/2) dt
dt= 2t^(1/2)du


so...

(sin u times 2t^1/2 du )/ the square root of t cos square root t cubed =

(2 sin u du) / the square root of cos u cubed


Let v= cos u
dv= -sin u du
du= dv/-sin u

so...

2 sin u times (dv/-sin u) / v^3/2 =

-2 v^(-3/2) dv =

Every thing is good to here except you haven't written the integral sign!
You mean \(\displaystyle 2\int v^{-3/2}dv\)

4 v^(-1/2) =

And here, you have done the integral- technically, there should be a "+ C", the "constant of integration".

4 / square root of cos square root t (the answer)


Now my question deals with the derivatives thrown in there, the dt, du, and dv...

Why isn't the answer 4dv /

square root of cos square root t??

How come you can just throw the dv away? How come we didn't just throw the du away earlier in the middle steps of the problem?

If, in the beginning of the problem, the problem didn't have dt in it, would it make the problem impossible, or do we just assume the dt and write it in there anyway?

No, the "fundamental theorem of Calculus" says \(\displaystyle \int f'(x)dx= f(x)+ C\). The "\(\displaystyle \int\)" and "dx" go together. Once you have done the integral, they both are gone.

Also, could the problem have been written with the dt in the denominator instead of the numerator, making a completely different problem?

No, that would be meaningless. Given that the derivative of a function, F(x), is f(x), so that \(\displaystyle \frac{dF}{dx}= f(x)\) then in "differential form" \(\displaystyle dF= f(x)dx\). The "fundamental theorem of Calculus" then says that \(\displaystyle \int dF= F= \int f(x)dx\). The "dx" cannot be "in the denominator".

I am just unsure about this concept and exactly what the dt, du, and dv mean...

Then you should go back a review the Riemann sum definition of "integral". We divide an interval, say from a to b, into many segments, each of length \(\displaystyle \Delta x\), evaluate f at some point in each interval, constructing a rectangle of that height and so area \(\displaystyle f(x)\Delta x\), and sum, \(\displaystyle \sum f(x)\Delta x\) approximates the "area under the curve". The limit, as the number of segments goes to infinity, and \(\displaystyle \Delta x\) goes to 0, is the integral, \(\displaystyle \int f(x)dx\). The "dx" is effectively the limit of the \(\displaystyle \Delta x\).

 

[thepillow]

Like HallsofIvy said,

as the number of rectangles we use to approximate an area under a curve approaches infinity, the width of these rectangles will approach zero (because we have
a fixed area and we're fitting more and more rectangles into it, they must each get narrower and narrower).

This width (Δx) essentially gets infinitely small, and we represent this as a 'differential' (the dx), an unfathomably tiny change in x.

So once we've actually done our actual calculations we no longer have the dx lying around.​

What's amazing about integrals is that we are essentially adding up a series of infinitely small things and getting a meaningful result (kind of like how you and I are
composed of astonishingly small particles that together form a human being).

I should also say that this is just intended to give some intuition (some mathematicians might take issue with this sort of explanation, but I think it's helpful initially).






​