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So on Khan Academy where this derivation is performed you are given $\dfrac{dT}{dt} = -k(T-T_a)$, where $T$ is the final temperature and $T_a$ is the, constant, environmental temperature.

From here you would rearrange this equation into $\dfrac{dT}{-k(T-T_a)} = -k\,dt$. However, here is where I am confused. From here the next step seems to be to integrate both sides, but I don't understand why this is possible. On the left side you are integrating with respect to temperature whereas on the right side you are integrating with respect to time.

I always assumed that the reason why we could perform any function like this was because one would always do the exact same thing to both sides of the equation, and I believe that reasoning still holds true. What am I missing here?

Semiclassical
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Copywright
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3 Answers3

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The reason this works is that the integral is the anti-differential. Just like you can differentiate both sides, you can do the reverse process and integrate both sides.

Let's look at a simple example:

$x^2 = \sin(y)$.

We can differentiate both sides and get:

$$2x\,dx = \cos(y)\,dy$$

A lot of people want to differentiate "with respect to a variable" but that's actually unnecessary. The "with respect to a variable" part just means dividing by that differential after-the-fact. So, if we differentiated "with respect to q", then we would just divide both sides by $dq$:

$$2x\,\frac{dx}{dq} = \cos(y)\,\frac{dy}{dq}$$

But, there is nothing special about "with respect to" a variable. It just means that after differentiating, we will divide by the differential of that variable.

So, if the differential of our equation is just $2x\,dx = \cos(y)\,dy$, then, we can go the other way by anti-differentiating (also known as integrating):

$$ \int 2x\,dx = \int \cos(y)\,dy \\ x^2 + C_1 = \sin(y) + C_2 $$

It's just the reverse process. And, like differentiating, it doesn't require that you integrate "with respect to" any variable. It just requires that the forms match.

Here's another example that's a little more trippy:

$$ \int (x\,dz + z\,dx) = \int dy $$

Here, we can see the left-hand side is the result of the product rule. So, this becomes:

$$ xz + C_1 = y + C_2 $$.

This all works because we are performing the exact same operation to both sides - anti-differentiation. Since neither differentiation nor integrations needs to be "with respect to a variable", it is essentially "symmetric" to both sides, even when they are using different variables.

johnnyb
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This is a technique called "separation of variables". I'll start with the simple case that you are used to from the fundamental theorem of calculus.

$$\int f'(x)dx=f(x)+c$$ Now I will substitute in Liebniz notation: $$\int \frac{dy}{dx}dx=f(x)+c$$

Notice we are using the convention $y=f(x)$, so subbing that in $$\int \frac{dy}{dx}dx=y+c$$.

But just like $$\int dx=x+c$$ we have $$\int dy=y+c$$

so replacing the RHS we have

$$\int \frac{dy}{dx}dx=\int dy$$

note we are integrating a function of $x\cdot dx$ on the left and a function of $y$ (in this case 1) times $dy$ on the right.

Now, if we have a function of more than one variably involved in a differential equation, it can get hard to figure out what to do: $$\frac {dy} {dx}=e^{x-2y}+3x-2y$$

However, there's something nice that happens if we can get it to the case that we can "Separate" the variables involved in the function in the RHS to single variable functions multiplied by each other, in other words, if we can write it in the form

$$\frac {dy} {dx}=f(x)g(y)$$

Then we can do the same change of notation we had above by moving the $dx$ to the other side via "multiplication" (not really, but the notation is suggestive and it works) and dividing by $g(y)$ to et $$\frac 1 {g(y)} dy=f(x)dx$$

Now we are just like we were in before, only instead of integrating just $dx$ or $dy$ on one side, we have a function next to it. The idea Liebniz had was that we would be doing little rectangles of "infinitesimal" widths $dy$, $dx$ times the function heights, and integrating would be adding them all up. It turns out that if we live in the real numbers, or most other "nice" (i.e. Archimedean) fields, infinitesimals can't exist, so all of calculus got rebuilt around limits. However, the notation is so suggestive and it turns out to work that we kind of pretend to go along with the old way of thinking.

Alan
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Your equation in differential form tells you how corresponding changes in $t$ and $T$ are related. To give you an example, suppose you have two discrete quantities $x_i$ and $y_i$ whose increments are related like this $$ y_i-y_{i-1}=2(x_i-x_{i-1}) $$ for $i=1,2,...$, and $x_0=0$, $y_0=1$. Then you can add the left hand sides and the right hand sides to conclude (via cancellation) that $$ y_i=1+2x_i $$ for every $i$. So we performed "integration" w.r.t. $x$ on one side and with respect to $y$ on the other.

GReyes
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