Question

Water flows in a shallow semicircular channel with inner and outer radii of \(1 \mathrm{m}\) and \(2 \mathrm{m}\) (see figure). At a point \(P(r, \theta)\) in the channel, the flow is in the tangential direction (counterclockwise along circles), and it depends only on \(r\), the distance from the center of the semicircles. a. Express the region formed by the channel as a set in polar coordinates. b. Express the inflow and outflow regions of the channel as sets in polar coordinates. c. Suppose the tangential velocity of the water in \(\mathrm{m} / \mathrm{s}\) is given by \(v(r)=10 r,\) for \(1 \leq r \leq 2 .\) Is the velocity greater at \(\left(1.5, \frac{\pi}{4}\right)\) or \(\left(1.2, \frac{3 \pi}{4}\right) ?\) Explain. d. Suppose the tangential velocity of the water is given by \(v(r)=\frac{20}{r},\) for \(1 \leq r \leq 2 .\) Is the velocity greater at \(\left(1.8, \frac{\pi}{6}\right)\) or \(\left(1.3, \frac{2 \pi}{3}\right) ?\) Explain. e. The total amount of water that flows through the channel (across a cross section of the channel \(\theta=\theta_{0}\) ) is proportional to \(\int_{1}^{2} v(r) d r .\) Is the total flow through the channel greater for the flow in part (c) or (d)?

Short answer

Answer: For the velocity function \(v(r) = 10r\), the greater velocity is at point \((1.5, \frac{\pi}{4})\) with a value of \(15\,m/s\), and the total flow through the channel is \(15\). For the velocity function \(v(r) = \frac{20}{r}\), the greater velocity is at point \((1.3, \frac{2\pi}{3})\) with a value of approximately \(15.38\,m/s\), and the total flow through the channel is approximately \(13.86\).

Step-by-step solution

Writing the channel region in polar coordinates

To express the channel's region in polar coordinates, we need to consider the inner and outer radii, which are given as \(1\mathrm{m}\) and \(2\mathrm{m}\). A polar coordinate point \((r,\theta)\) lies in the channel if its radius \(r\) is between \(1\) and \(2\), and the angle \(\theta\) is between \(0\) and \(\pi\). Therefore, the channel is represented by \(S=\{(r,\theta) \mid 1 \leq r \leq 2, 0\leq \theta \leq \pi\}\).

Writing inflow and outflow regions in polar coordinates

The inflow region of the channel is where water enters. This region should be close to the starting point, where the radius is approaching \(1\). The outflow region is where water exits the channel. This region should be close to the ending point, where the radius is approaching \(2\). The inflow and outflow regions in polar coordinates can be represented as Inflow: \(A= \{(r,\theta) \mid r\approx 1, 0\leq \theta \leq \pi\}\) Outflow: \(B= \{(r,\theta) \mid r\approx 2, 0\leq \theta \leq \pi\}\).

Comparing velocities using \(v(r) = 10r\)

For velocity equation \(v(r) = 10r\), we need to find the velocities at the given points \((1.5, \frac{\pi}{4})\) and \((1.2, \frac{3\pi}{4})\). To do this, we only need to determine the \(r\)-values for both points and substitute them in the function \(v(r)\). \(v(1.5) = 10(1.5) = 15 \;\text{m/s}\) \(v(1.2) = 10(1.2) = 12 \;\text{m/s}\) For the velocity equation \(v(r) = 10r\), the velocity is greater at \((1.5, \frac{\pi}{4})\).

Comparing velocities using \(v(r) = \frac{20}{r}\)

For velocity equation \(v(r) = \frac{20}{r}\), we need to find the velocities at the given points \((1.8, \frac{\pi}{6})\) and \((1.3, \frac{2\pi}{3})\). To do this, we only need to determine the \(r\)-values for both points and substitute them in the function \(v(r)\). \(v(1.8) = \frac{20}{1.8} \approx 11.11 \;\text{m/s}\) \(v(1.3) = \frac{20}{1.3} \approx 15.38 \;\text{m/s}\) For the velocity equation \(v(r) = \frac{20}{r}\), the velocity is greater at \((1.3, \frac{2\pi}{3})\).

Comparing the total flow for equation \(v(r) = 10r\) and \(v(r) = \frac{20}{r}\)

The total flow is proportional to \(\int_{1}^{2} v(r) dr\). We need to evaluate the integrals for both \(v(r) = 10r\) and \(v(r) = \frac{20}{r}\). For \(v(r) = 10r\): \(\int_{1}^{2} 10r \,dr = 10\int_{1}^{2} r \,dr = 10\left[\frac{r^2}{2}\right]_{1}^{2} = 5(4-1) = 15\) For \(v(r) = \frac{20}{r}\): \(\int_{1}^{2} \frac{20}{r} \,dr = 20\int_{1}^{2} \frac{1}{r} \,dr = 20\left[\ln |r|\right]_{1}^{2} = 20\ln \frac{2}{1} = 20\ln 2 \approx 13.86\) The total flow through the channel is greater for the flow in part (c) with the velocity equation \(v(r) = 10r\).

Key concepts

Semicircular Channel

A semicircular channel is a type of water channel shaped like a half-circle, or semicircle. This shape is practical for efficiently directing the flow of liquids, such as water. In our exercise, the channel is defined in polar coordinates, which use the radius (\( r \)) and angle (\( \theta \)) to determine the location of any point.

The semicircular channel we are examining has an inner radius of 1 meter and an outer radius of 2 meters. This means any point within the channel must have a radius between these two values. As for the angle, points are located anywhere from the start of the semicircle (\( \theta = 0 \)) to the end (\( \theta = \pi \)), essentially covering a 180-degree arc. This can be written as a set of points:

• The radius \( r \) is between 1 and 2 meters: \( 1 \leq r \leq 2 \).
• The angle \( \theta \) is between 0 and \( \pi \): \( 0 \leq \theta \leq \pi \).

Tangential Velocity

Tangential velocity refers to the velocity of an object moving along a circular path. In this context, the water flow in the semicircular channel moves in the tangential (or circular) direction around a point. The speed of this tangential flow changes depending on how far out the water is from the center of the semicircle.

The formula given for tangential velocity in the exercise illustrates two cases:

• First case: \( v(r) = 10r \), where velocity increases linearly as the radius increases.
• Second case: \( v(r) = \frac{20}{r} \), which means velocity decreases as the radius increases.

Tangential velocity is distinct from the linear speed we typically think about because it incorporates the direction in which the object is moving along a curve or circular path.

Integrals in Calculus

In calculus, integrals are used to find quantities like area under a curve, total accumulated value, or total flow in our case. In the exercise, we see how the integral helps calculate the total flow through the channel.

When we integrate a function over a certain interval, we effectively add up all tiny pieces (infinitesimal quantities) over that interval to find the whole. Here, the functions \( v(r) = 10r \) and \( v(r) = \frac{20}{r} \) are integrated from the inner to the outer radius of the channel: \( 1 \) to \( 2 \).

• For \( v(r) = 10r \), the integral calculates to 15, representing the total flow for that scenario.
• For \( v(r) = \frac{20}{r} \), it results in approximately 13.86, slightly less total flow than the first scenario.

This integration shows how calculus can be applied to real-world problems to derive meaningful values.

Velocity Comparison

Velocity comparison involves checking which among two or more speeds is higher at certain points. This can help determine where fluid dynamics, like water flow, varies at different locations within the channel.

In the exercise, you compare the tangential velocities at specific points using the velocity functions.

• For \( v(r) = 10r \), the comparisons are made at \( (r = 1.5, \theta = \frac{\pi}{4}) \) and \( (r = 1.2, \theta = \frac{3\pi}{4}) \); the velocity is greater at the first point because \( 15 \text{ m/s} > 12 \text{ m/s} \).
• For \( v(r) = \frac{20}{r} \), the comparisons are at \( (r = 1.8, \theta = \frac{\pi}{6}) \) and \( (r = 1.3, \theta = \frac{2\pi}{3}) \), where the second point is faster because \( 15.38 \text{ m/s} > 11.11 \text{ m/s} \).

This comparison is crucial in understanding how velocity distribution can affect fluid systems.