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			<h3 id="back-link"><a href="../../#1.2">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty.
			</p>

			<p>
			b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero.

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			<figcaption>
			For problem $1.2.11$
			</figcaption>
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			<h3>Solution</h3>
			<p>

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			<img src="drawing1.png"
			loading="lazy" width="200" />
			<figcaption>
			Cone vessel
			</figcaption>
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			</center>

			$a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to:
			$$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$
			Where $dx$ is the change in water level:

		@@ -89,13 +89,13 @@
			$$\frac{dV}{dt} = \pi r(t)^2 v_0$$

			From Geometry,
		−	$$r(t) = v_0 t \cdot \tan\alpha $$
		+	$$r(t) = v_0 t \cdot \tan\alpha$$

			By definition, the velocity of incoming water is equal to
			$$v = \frac{dV}{sdt}$$

			Substituting the previous expressions:
		−	$$v = {\pi v_0^3 t^2 \cdot \tan^2(\alpha)}/s$$
		+	$$v = {\pi v_0^3 t^2 \cdot \tan^2(\alpha )}/s$$

			$b)$ For a small time interval $dt$, the volume changes by
			$$dV = q dt$$

			Also the volume increment can be written as
			$$dV = 2\pi r dr \cdot h $$

			Thus:

			$$ q dt = 2\pi r dr \cdot h$$

			Considering $v = \frac{dr}{dt}$,
			$$ \fbox{$v = \frac{q}{2\pi r h}$}$$
			</p>

			<h4>Answer</h4>
			<p>
			$$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$
			$$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$
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	<h3 id="back-link"><a href="../../#1.2">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.2">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty.		$1.2.11.$ a. In a conical vessel, the water level rises at a constant rate $v_0$. How does the rate of water entering a vessel through a tube of section $s$ depend on time? At time zero, the vessel is empty.
	</p>		</p>

	<p>		<p>
	b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero.		b. A jet of oil hitting the surface of the water spreads over it in a round spot of thickness $h$. How does the speed of movement of the spot boundary depend on time, if the volume of oil $q$ enters per unit of time? At the initial time, the spot radius is zero.

	</p>		</p>
	<center>		<center>
	<figure>		<figure>
	<img src="statement.png"		<img src="statement.png"
	loading="lazy" width="160" />		loading="lazy" width="160" />
	<figcaption>		<figcaption>
	For problem $1.2.11$		For problem $1.2.11$
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>
	<p>		<p>
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>

	<center>		<center>
	<figure>		<figure>
	<img src="drawing1.png"		<img src="drawing1.png"
	loading="lazy" width="200" />		loading="lazy" width="200" />
	<figcaption>		<figcaption>
	Cone vessel		Cone vessel
	</figcaption>		</figcaption>
	</figure>		</figure>
	</center>		</center>

	$a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to:		$a)$ By time $t$, the water level will be $v_0t$. And the rate of change of volume will be equal to:
	$$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$		$$\frac{dV}{dt} = \frac{\pi r(t)^2 dx}{dt}$$
	Where $dx$ is the change in water level:		Where $dx$ is the change in water level:

	@@ -89,13 +89,13 @@
	$$\frac{dV}{dt} = \pi r(t)^2 v_0$$		$$\frac{dV}{dt} = \pi r(t)^2 v_0$$

	From Geometry,		From Geometry,
	$$r(t) = v_0 t \cdot \tan\alpha $$		$$r(t) = v_0 t \cdot \tan\alpha$$

	By definition, the velocity of incoming water is equal to		By definition, the velocity of incoming water is equal to
	$$v = \frac{dV}{sdt}$$		$$v = \frac{dV}{sdt}$$

	Substituting the previous expressions:		Substituting the previous expressions:
	$$v = {\pi v_0^3 t^2 \cdot \tan^2(\alpha)}/s$$		$$v = {\pi v_0^3 t^2 \cdot \tan^2(\alpha )}/s$$

	$b)$ For a small time interval $dt$, the volume changes by		$b)$ For a small time interval $dt$, the volume changes by
	$$dV = q dt$$		$$dV = q dt$$

	Also the volume increment can be written as		Also the volume increment can be written as
	$$dV = 2\pi r dr \cdot h $$		$$dV = 2\pi r dr \cdot h $$

	Thus:		Thus:

	$$ q dt = 2\pi r dr \cdot h$$		$$ q dt = 2\pi r dr \cdot h$$

	Considering $v = \frac{dr}{dt}$,		Considering $v = \frac{dr}{dt}$,
	$$ \fbox{$v = \frac{q}{2\pi r h}$}$$		$$ \fbox{$v = \frac{q}{2\pi r h}$}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$		$$v=\frac{\pi v_0^3t^2\operatorname{tg}^2\alpha}{s}$$
	$$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$		$$v=\frac{1}{2}\sqrt{\frac{q}{\pi ht}}$$
	</p>		</p>


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