I want to ask a question about solving a first order differential.
I received the following question in a lecture regarding thermodynamics, and this was one of the parts of a question:
Consider a tank of cylindrical shape. The radius of the circular cross-section of the tank is 40 cm. At time $t$ minutes, the depth of liquid in the tank is $h$ centimetres. The liquid leaks from a hole P at the bottom of the tank.
The liquid leaks from the tank at a rate of $32π\sqrt{h}$ cm$^3$ min$^{–1}$.
At time $t$ minutes, the height $h$ cm of liquid in the tank satisfies the differential equation: $\frac{dh}{dt}= – 0.02\sqrt{h}$Find the time taken, to the nearest minute, for the depth of liquid in the tank to decrease from 100 cm to 50 cm.
Now, I started by solving the differential equation above:
$$\int h^{-\frac{1}{2}}dh = \frac{1}{50} \int dt$$ $$2\sqrt{h}^\frac{1}{2} = \frac{1}{50}t + C$$ $$100h^\frac{1}{2} = t + C$$
Now, I thought I should substitute $t = 0$ and $h = 100$ to solve for C.
Then, I should substitute $h = 50$ and then find $t$ for this value.
However, when listening to the professor in the lecture, he stated there would be no need for this and instead evaluate the integral between 100 and 50 as such:
$\int_{50}^{100} 100\sqrt{h} = t$
as the value of the these two limits give the time.
I'm confused as to how this is the case.
Can someone explain why the integral works to finding the solution ($=1000 - 100\sqrt{50}$)
Let $\varphi \colon I \to J$, where $I, J \subset \mathbb{R}$ are open intervals, be a solution of $x' = f(x)$, where $f \colon J \to \mathbb{R}$ is continuous, with $f(x) \ne 0$ for all $x \in J$. Take $t_1, t_2 \in I$, $t_1 < t_2$, and put $x_1 := \varphi(t_1)$, $x_2 := \varphi(t_2)$. In the definite integral $$ \int\limits_{x_1}^{x_2} \frac{dx}{f(x)} $$ substitute $x = \varphi(t)$: $$ \int\limits_{x_1}^{x_2} \frac{dx}{f(x)} = \int\limits_{t_1}^{t_2} \frac{\varphi'(t)}{f(\varphi(t))}. $$ But $\varphi'(t) = f(\varphi(t))$ for all $t \in [t_1, t_2]$, so $$ \int\limits_{t_1}^{t_2} \frac{\varphi'(t)}{f(\varphi(t))} = \int\limits_{t_1}^{t_2} dt, $$ which is equal to $t_2 - t_1$.