Question

A trough in the shape of a half cylinder has length \(5 \mathrm{m}\) and radius \(1 \mathrm{m}\). The trough is full of water when a valve is opened, and water flows out of the bottom of the trough at a rate of \(1.5 \mathrm{m}^{3} / \mathrm{hr}\) (see figure). (Hint: The area of a sector of a circle of a radius \(r\) subtended by an angle \(\theta\) is \(r^{2} \theta / 2 .\) ) a. How fast is the water level changing when the water level is \(0.5 \mathrm{m}\) from the bottom of the trough? b. What is the rate of change of the surface area of the water when the water is \(0.5 \mathrm{m}\) deep?

Short answer

Answer: When the water level is 0.5 meters deep, the rate at which the water level is changing is -3/(5π) meters per hour, and the rate of change of the surface area of the water is 3/π square meters per hour.

Step-by-step solution

Determine Volume and Area Formulas

For half-cylinder trough, the volume of the water can be represented as: \(V = \dfrac{1}{2}\pi r^2 h L\) Where \(V\) is the volume, \(r\) is the radius, \(h\) is the height of the water, and \(L\) is the length of the trough. In this exercise, we are given that \(r=1\mathrm{m}\) and \(L=5\mathrm{m}\). The area of the sector of a circle can be represented as: \(A = \dfrac{r^2\theta}{2} \) We will use the area of the sector to find the surface area of the water. We have to find the relationship between \(\theta\) and \(h\) first.

Find the Relationship between Height and Angle

Use the right triangle formed by the height of the water, the radius, and the radius minus the height of the water. Let \(\theta\) be the angle formed by the radius and the horizontal side of the triangle (radius minus height). In this case, we can see that: \(\cos\theta = \dfrac{r-h}{r}\) Now, solve for \(\theta\): \(\theta = \arccos{\left(\dfrac{r-h}{r}\right)}\) Using this relationship, we can substitute \(\theta\) in the area formula.

Find the Surface Area Formula

Substitute the relationship between angle and height we found earlier into the area formula: \(A = \dfrac{r^2}{2}\cdot\arccos{\left(\dfrac{r-h}{r}\right)}\) As the surface area of the water is formed by a sector of a circle and the length of the trough, we can find the complete surface area: \(A = L \cdot \dfrac{r^2}{2}\cdot\arccos{\left(\dfrac{r-h}{r}\right)}\)

Finding the Rates

Now, we can use related rates to find the desired rates. The volume of water is decreasing at a rate of \(1.5 \mathrm{m^3 / hr}\), i.e., \(\dfrac{dV}{dt}=-1.5\). Using the chain rule, we can find \(\dfrac{dh}{dt}\) when \(h = 0.5\mathrm{m}\): \(\dfrac{dV}{dt} = \dfrac{1}{2}\pi r^2 L \dfrac{dh}{dt}\) Substitute the given values and solve for \(\dfrac{dh}{dt}\): \(-1.5 = \dfrac{1}{2}\pi(1)^2(5) \dfrac{dh}{dt}\) \(\dfrac{dh}{dt} = -\dfrac{3}{5\pi} \mathrm{m / hr}\) Now we will find the rate of change of the surface area when the water is \(0.5 \mathrm{m}\) deep. We'll differentiate the surface area formula with respect to time and solve for \(\dfrac{dA}{dt}\): \(\dfrac{dA}{dt} = L\cdot\dfrac{r^2}{2}\cdot\dfrac{-1}{\sqrt{1 - \left(\dfrac{r-h}{r}\right)^2}}\cdot \dfrac{dh}{dt}\) Substitute the given values and \(\dfrac{dh}{dt}\) we found earlier: \(\dfrac{dA}{dt} = 5\cdot\dfrac{1^2}{2}\cdot \dfrac{-1}{\sqrt{1-\left(\dfrac{1-0.5}{1}\right)^2}}\cdot \left(-\dfrac{3}{5\pi}\right)\) \(\dfrac{dA}{dt} = \dfrac{3}{\pi} \mathrm{m^2 / hr}\) The final answers are: a. The rate at which the water level is changing when the water level is \(0.5\mathrm{m}\) from the bottom of the trough is \(-\dfrac{3}{5\pi} \mathrm{m / hr}\). b. The rate of change of the surface area of the water when the water is \(0.5 \mathrm{m}\) deep is \(\dfrac{3}{\pi} \mathrm{m^2 / hr}\).

Key concepts

Volume of a Cylinder

Understanding the volume of a cylinder is crucial for solving problems involving solids, especially in related rates. The volume formula for a full cylinder is given by \( V = \pi r^2 h \). Here, \( V \) is the volume, \( r \) is the radius of the cylinder, and \( h \) is the height. In the context of a half-cylinder, like our trough, the volume is calculated as half of a full cylinder:

• Volume of half-cylinder: \( V = \frac{1}{2} \pi r^2 h L \)
• Where \( L \) is the length of the trough.

In this exercise, with a radius \( r = 1 \) meter and a length \( L = 5 \) meters, the volume formula simplifies to: \( V = \frac{1}{2} \pi (1)^2 h (5) \) or \( V = \frac{5}{2} \pi h \). Understanding this form is essential because as water flows out, changes in volume affect the water height, which we need to calculate how fast or slow the height is changing.

Surface Area Calculation

Calculating the surface area of different shapes is often required to understand how certain characteristics, like the water surface area in this problem, change. A half-cylinder presents a unique challenge as it combines linear and curved surfaces. The surface area involved with the water in the trough is the area of the circular segment formed by the water's height.

• The formula for the area of a circle segment or sector is \( A = \frac{r^2 \theta}{2} \), where \( \theta \) is the central angle in radians.
• To relate \( \theta \) to the water height \( h \), a trigonometric approach involving \( \cos \theta = \frac{r-h}{r} \) is used.
• This forms \( \theta = \arccos\left(\frac{r-h}{r}\right) \).
• Substituting \( \theta \) back into the surface area formula gives us a dynamic, height-based area calculation.

By understanding this, you can calculate the change in surface area as water flows out by differentiating with respect to \( h \), giving insights into the increasing or decreasing trend of the exposed surface as the fluid level changes.

Trigonometric Relationships

Trigonometry frequently ties into real-world applications such as this exercise. Here, we have a half-cylinder trough, and the relationship established between the water height and the angle \( \theta \) formed in the circle segment is crucial.

• Using basic trigonometry, \( \cos \theta = \frac{r-h}{r} \) determines how the water's height changes relate to the angle.
• This formula demonstrates that when water height, \( h \), varies, the angle \( \theta \) adjusts correspondingly.
• Solving for \( \theta \) with \( \theta = \arccos\left(\frac{r-h}{r}\right) \) allows us to link physical measurements with mathematical angles and calculate aspects like surface area changes.

Understanding this trigonometric relationship helps provide a fuller picture of changes in volume and surface area over time. It's essential for interpreting related rates, aiding in discovering how one changing element in a geometric related rate problem leads to changes in another aspect of the system.