MHB Area under the curve of a projectile

[Vedant97]

I am trying to calculate the area under the curve of a projectile for a school project.
A simple way to do this is to integrate the following equation of the trajectory:

View attachment 3263

However I've tried to use another method. Since we have the two equations for the horizontal and vertical displacements (respectively):

View attachment 3264

View attachment 3265

As it can be seen from the above graph, we can integrate Sy with respect to Sx to get the area under the curve. However since Sy and Sx themselves are in terms of 't' (other variables are considered constant), we multiply them by (dt/dt) which will give us:

View attachment 3266

(where Tt is the time when the projectile hits the ground)

Both methods ultimately simplify to give the equation:
View attachment 3267However, my teacher says that my method of multiplying the (dt/dt) is 'mathematically incorrect' even though both give the same results. But I believe that what I have done is correct. Can anyone help me come up with a justification as to why this is mathematically correct?

 

[Euge]

I am trying to calculate the area under the curve of a projectile for a school project.
A simple way to do this is to integrate the following equation of the trajectory:

https://www.physicsforums.com/attachments/3263

However I've tried to use another method. Since we have the two equations for the horizontal and vertical displacements (respectively):

View attachment 3264

View attachment 3265

As it can be seen from the above graph, we can integrate Sy with respect to Sx to get the area under the curve. However since Sy and Sx themselves are in terms of 't' (other variables are considered constant), we multiply them by (dt/dt) which will give us:

View attachment 3266

(where Tt is the time when the projectile hits the ground)

Both methods ultimately simplify to give the equation:
View attachment 3267However, my teacher says that my method of multiplying the (dt/dt) is 'mathematically incorrect' even though both give the same results. But I believe that what I have done is correct. Can anyone help me come up with a justification as to why this is mathematically correct?

Actually, $$\frac{dt}{dt}$$, the $t$-derivative of $t$, is equal to one. So it makes sense mathematically. Based on your argument, I suspect that your teacher thought you were treating $$\frac{dt}{dt}$$ as a fraction, not a derivative. In that case, your method is mathematically incorrect. To avoid confusion, remove the expression

$$\int_0^{t_T} S_y \cdot dS_x \cdot \frac{dt}{dt}\, dt$$

and argue carefully by substitution that the area of under your curve is

$$\int_0^{t_T} S_y \cdot \frac{dS_x}{dt}\, dt$$.

Then you can proceed with the calculation.

Here is a link that covers a method for computing areas of parametric curves.

Pauls Online Notes : Calculus II - Area with Parametric Equations

 

[magneto1]

That notation of $\frac{dy}{dx}$ is very misleading in that it is not really a fraction, but it is very convenient notation for things like differential equation that you can separate out the variables.

However, it is not something you need in your work. If $S_x = ut \cos a$ where $u$ and $a$ are constants, then since you are integrating against $t$,
$dS_x = d(ut\cos a) = u\cos a \cdot dt$. You get a $dt$ just out of that expression without introducing another $\frac{dt}{dt}$ into the work.

 

[Vedant97]

That notation of $\frac{dy}{dx}$ is very misleading in that it is not really a fraction, but it is very convenient notation for things like differential equation that you can separate out the variables.

However, it is not something you need in your work. If $S_x = ut \cos a$ where $u$ and $a$ are constants, then since you are integrating against $t$,
$dS_x = d(ut\cos a) = u\cos a \cdot dt$. You get a $dt$ just out of that expression without introducing another $\frac{dt}{dt}$ into the work.

Thanks for the reply.
Okay I could show that. But the main problem is that she says this cannot be done since Sy is not in terms of or related to Sx.
How can I justify my method?

 

[magneto1]

What did your teach say cannot be done? the multiplication of $\frac{dt}{dt}$? Or that you are not allowed to use the parametric equation formula?

If you see the link provided by Euge, you see that given the parametric equations $x = f(t)$ and $y= g(t)$, the area under the curve is given by:
\[
A = \int_a^b g(t) f'(t) dt,
\]
where $a,b$ are appropriate bound. The formula is sound even when $f$ and $g$ are not related other than the fact they both depends on $t$.