I am currently learning about the finding the area under the curve via integration using parametric equations. I was looking at this website http://tutorial.math.lamar.edu/Classes/CalcII/ParaArea.aspx and understood the material up till this line:
"So, if this is going to be a substitution we’ll need, dx = f'(t) dt".
Separately, I was given an easy way to remember this which is that "(dx/dt) dt = dx" is the way it is since "the two dt cancel each other out". Granted it's easy to remember but it does nothing in telling me why it should be the case, let alone the fact that it's mathematically improper.
So…can someone prove to me why dx = (dx/dt) dt?
Best Answer
If you want to avoid Leibniz notation altogether (as I tend to prefer doing), you can derive the area for a parametric curve using simple Riemann approximations.
Recall the definition of the definite integral of $f(x)$ on some continuous interval $[a,b]$: $$\int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*) \Delta x$$ where $x_i^*$ lies on the subinterval $[x_{i-1}, x_i]$ after subdividing $[a,b]$ into $n$ subregions of equal length $\Delta x$, such that $\Delta x = \frac{b-a}{n}$.
Now, consider a parametric curve $(f(t), g(t))$ for $t$ on some interval $[\alpha, \beta]$. We will show that the area under the parametric curve can be approximated by adding up rectangles, where rectangle $i$ will have a width of $\Delta x_i = f(t_i) - f(t_{i-1})$ and a height of $y_i = g(t_i)$. For instance, the $1^{st}$ subinterval will have a width of $\Delta x_1 = f(t_1) - f(t_0)$, where $t_0 = \alpha$. (Each rectangle's area is approximated using right endpoints, although we could just as easily have used left endpoints or midpoints.)
First, we subdivide the interval $[\alpha, \beta]$ into $n$ subregions, each of length $\Delta t = \frac{\beta - \alpha}{n}$.
Next, consider the $i^{th}$ subinterval $[t_i, t_{i-1}]$. The Mean Value Theorem guarantees that there is some value $t_i^*$ in this interval whose slope is the secant line of the endpoints. In symbols, we have $$\frac{f(t_i) - f(t_{i-1})}{t_i - t_{i-1}} = f'(t_i^*)$$ We can rewrite this equation in the form $$f(t_i) - f(t_{i-1}) = f'(t_i^*)(t_i - t_{i-1})$$ or, equivalently, $$\Delta x_i = f'(t_i^*) \Delta t$$ (Notice how closely this equation resembles the form $dx = f'(t)dt$. I will say more about this after the proof.)
Now, we have a formula for the width of an approximating rectangle. The height can then be taken to be $g(t_i^*)$, and we have the area is approximately equal to: $$\sum_{i=1}^n g(t_i^*)f'(t_i^*) \Delta t$$ Take the limit and we have the definite integral $$\lim_{n \to \infty}\sum_{i=1}^n g(t_i^*)f'(t_i^*) \Delta t = \int_{\alpha}^{\beta}g(t)f'(t) \Delta t$$ Of course, $f'(t)$ must be differentiable and $g(t)$ must be continuous on the interval $[\alpha, \beta]$ for this equation to make sense. So, if you're considering the actual curve drawn out by the parameter, it only makes sense to find the area under portions of the graph that are actual functions (you can't calculate area over a region that involves the graph curving around over itself, since $x'(t)$ would be undefined at the point of curvature). Also, if the graph retraces itself for some t on the interval $[\alpha, \beta]$, your area calculations will be repeated, and you will end up with more (or less) area than you bargained for.
To address your original question about why dx = (dx/dt)dt: the short answer is that this is the definition of a differential. To give you some insight into why this is the case, consider a normal slope defined by two points, $\frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x}$. In particular, we have that $\frac{\Delta y}{\Delta x} \approx f'(x)$. Notice here that $\frac{\Delta y}{\Delta x}$ refers to a normal, everyday fraction, so if we wanted to approximate $\Delta y$, we could rearrange the equation to $\Delta y \approx f'(x) \Delta x$ without fuss.
This can be useful in cases where we know some easily-calculable value of $f$ (say $f(a)$), and want to calculate values of $f$ for $x$ very close to $a$. For instance, if $f(x)=\sqrt x$, it is easy to calculate $f(4)$, but not so much $f(4.0034)$. Thus, we can use equation $\Delta y \approx f'(x) \Delta x$ to approximate: $$f(a + \Delta x) = f(a) + \Delta y$$ $$f(a + \Delta x) \approx f(a) + f'(a) \Delta x$$ $$f(4 + 0.0034) \approx 2 + (0.25)(0.0034) = 2.00085$$ which is a good approximation for the actual value, $\approx 2.0008498$.
The differential, defined $dy = y'(x) dx$, is an expression of the fact that, for EXTREMELY small values (infinitely small) of $\Delta x$, the change in $y$ is exactly equal to the linear approximation. In other words, $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \frac{dy}{dx} = f'(x)$$ which is the definition of the derivative, is essentially the same thing as saying that the linear approximation for $f(x + \Delta x)$ is exactly accurate when $\Delta x$ goes to 0, which is what $dy = f'(x)dx$ is saying.
The reason why you are allowed to multiply by differentials when using substitution: It is true that $\frac{dy}{dx}$ is not a fraction; it is the derivative. A proof of why it is okay to treat $\frac{dy}{dx}$ like a fraction lies in the proof of the Substitution Rule for integration.
If $F$ is a differentiable function of $g(x)$, and $F' = f$, then $$[F(g(x))]' = F'(g(x))g'(x)$$ by the chain rule. By the Fundamental Theorem of Calculus, we can integrate both sides to obtain $$\int F'(g(x))g'(x) dx = \int f(g(x))g'(x) dx = F(g(x)) + C$$ If we let $u = g(x)$, and we assume it's okay to operate with differentials on their own, then $$\frac{du}{dx} = g'(x) \Rightarrow du = g'(x)dx$$ Plugging this in, $$\int f(g(x))g'(x) dx = \int f(u) du = F(u) + C = F(g(x)) + C$$ which we know is the correct result. Thus, if we assume it's okay to operate with differentials in the context of substitution, we arrive at the same result, meaning that it's "basically" okay to do so.
This answer was longer than I planned, but hopefully it has helped in some way.