Coordinated Calculus

1. Find a linear function \(B(h)\) that satisfies \(B(0) = 200 \cdot 0.3\) and \(B(1) = 199 \cdot 0.3\)

2. The weight of the bucket is given by a linear function.

3. Recall that work is given by a certain definite integral.

1. \(W = \int_0^{200} 0.3(200-h) \, dh = 6000 \text{ foot-pounds}\text{.}\)

2. \(W = \int_0^{100} (40-0.1h) \, dh = 3500 \text{ foot-pounds}\text{.}\)

3. \(B_{\text{AVG} [0,100]} \approx 25.9798 \text{ pounds}\text{.}\)

1. When the full rope is extended over the cliff, the weight of the hanging rope is 200 pounds. Every foot of rope weighs 0.3 pounds; in addition, the quantity \((200-h)\) measures the number of feet of rope hanging over the cliff when \(h\) feet of rope have been pulled in. This means that the function \(F(h)\) whose value is the weight of the rope hanging over the cliff after \(h\) feet have been pulled in is given by \(F(h) = 0.3(200-h)\text{.}\) When a small amount of rope, \(\Delta h\text{,}\) is pulled in, the work to move that slice of rope is given by

   \begin{equation*} W_{\text{slice}} = F(h) \Delta h = 0.3(200-h) \Delta h\text{.} \end{equation*}

   Thus, the total work to pull in the 200 feet of rope is given by

   \begin{equation*} W = \int_0^{200} 0.3(200-h) \, dh = 6000 \text{ foot-pounds}\text{.} \end{equation*}

2. Since \(B(h)\) changes at a constant rate, \(B\) is a linear function. We know that \(B(0) = 40\) and that \(B\) loses weight at a rate of 0.1 pounds per foot of rope hauled in. Thus, \(B(h) = 40 - 0.1h\text{.}\) It follows that the total work to life the leaky bucket 100 feet is

   \begin{equation*} W = int_0^{100} (40-0.1h) \, dh = 3500 \text{ foot-pounds}\text{.} \end{equation*}

3. Given that \(B(h) = 25 + 15e^{-0.05h}\) is the weight of the bucket when \(h\) feet of rope have been pulled in, it follows that the total work to move the bucket 100 feet is

   \begin{equation*} W = \int_0^{100} (25 + 5e^{-0.05h})\, dh \approx 2597.98 \text{ foot-pounds}\text{.} \end{equation*}

   In addition, the average force that was exerted on the bucket is precisely the average weight of the bucket, which is

   \begin{equation*} B_{\text{AVG} [0,100]} = \frac{1}{100-0} \int_0^{100} (25 + 5e^{-0.05h})\, dh \approx 25.9798 \text{ pounds}\text{.} \end{equation*}

Follow the rule for Hooke's Law and then use \(W = \int_a^b F(x) \, dx\text{.}\)

For the given spring,

1. \(k = 15\text{.}\)

2. \(W = \int_0^1 15x \, dx = \frac{15}{2} \text{ foot-pounds}\text{.}\)

3. \(W = \int_1^{1.5} 15x \, dx = 9.375 \text{ foot-pounds}\text{.}\)

Now considering the given spring,

1. Since \(F(1/3) = 5\) and \(F(x) = kx\) by Hoooke's Law, it follows that \(5 = k \cdot \frac{1}{3}\text{,}\) and thus \(k = 15\text{.}\)

2. Because the work to stretch the spring 1 foot is given by \(W = \int_0^1 F(x) \, dx\text{,}\) it follows \(W = \int_0^1 15x \, dx = \frac{15}{2} \text{ foot-pounds}\text{.}\)

3. Here, \(W = \int_1^{1.5} F(x) \, dx = \int_1^{1.5} 15x \, dx = 9.375 \text{ foot-pounds}\text{.}\)

It is helpful to think of the depth below the surface of the crock as being the independent variable, so we let the positive \(x\)-axis point down, and the positive \(y\)-axis to the right, as pictured in the figure. Because the pump sits on the surface of the water, it makes sense to think about the pump moving the water one slice at a time, where it takes a thin slice from the surface, pumps it out of the tank, and then proceeds to pump the next slice below.

Each slice of water is cylindrical in shape. We see that the radius of each slice varies according to the linear function \(y = f(x)\) that passes through the points \((0,1.5)\) and \((4,0.75)\text{,}\) where \(x\) is the depth of the particular slice in the tank; it is a straightforward exercise to find that \(f(x) = 1.5 - 0.1875x\text{.}\) Now we think about the problem in several steps:

1. determining the volume of a typical slice;
2. finding the weight⁴We assume that the weight density of water is 62.4 pounds per cubic foot. of a typical slice (and thus the force that must be exerted on it);
3. deciding the distance that a typical slice moves;
4. and computing the work to move a representative slice.

Once we know the work it takes to move one slice, we use a definite integral over an appropriate interval to find the total work.

Consider a representative cylindrical slice at a depth of \(x\) feet below the top of the crock. The approximate volume of that slice is given by

Since water weighs 62.4 lb/ft\(^3\text{,}\) the approximate weight of a representative slice is

This is also the approximate force the pump must exert to move the slice.

Because the slice is located at a depth of \(x\) feet below the top of the crock, the slice being moved by the pump must move \(x\) feet to get to the level of the basement floor, and then, as stated in the problem description, another 9 feet to reach the drain at ground level. Hence, the total distance a representative slice travels is

Finally, the work to move a representative slice is given by

We sum the work required to move slices throughout the tank (from \(x = 0\) to \(x = 4\)), let \(\Delta x \to 0\text{,}\) and hence

When evaluated using appropriate technology, the integral shows that the total work is \(W = 3463.2 \pi\) foot-pounds.

1. Note that slices of water at constant depth are all cylinders and all have the same radius. Also, remember to convert mass to weight when computing force.

2. Recall that the top half of a circle of radius 10 centered at the origin has equation \(y = \sqrt{100-x^2}\text{.}\)

3. The equation of the line that determines the right side of the front face of the tank is \(y=\frac{5}{2} - \frac{5}{8}x\text{.}\) Note that a slice of water at constant depth is a rectangular slab whose width is \(2y\text{.}\)

1. \begin{equation*} W = \int_{0}^{1} 9.8 \cdot 4000\pi \cdot x \, dx = 19600 \pi \, \text{newton-meters}\text{.} \end{equation*}
2. \begin{equation*} W = \int_{3}^{8} 62.4 \pi (100-x^2)(x+5) \, dx \approx 673593 \, \text{foot-pounds}\text{.} \end{equation*}
3. \begin{equation*} W = \int_{1}^{3} 62.4 (50 - \frac{25}{2}x) x \, dx = 5720 \, \text{foot-pounds}\text{.} \end{equation*}

1. A typical slice is a thin cylinder of radius 2 and thickness \(\Delta x\text{.}\) The volume of such a cylinder is \(V_{\text{slice}} = \pi (2)^2 \Delta x\text{,}\) and its weight is thus \(F_{\text{slice}} = 1000 \cdot 9.8 \cdot V_{\text{slice}} = 4000 \pi \Delta x\text{.}\) Finally, the distance such a slice travels to the top of the tank is \(d_{\text{slice}} = x\text{.}\) If we only pump out the top 1 m of water, it follows that the total work is

   \begin{equation*} W = \int_{0}^{1} 9.8 \cdot 4000\pi \cdot x \, dx = 19600 \pi \, \text{newton-meters}\text{.} \end{equation*}

2. Viewing the tank as a solid of revolution, the tank is generated by the function \(R(x) = \sqrt{100-x^2}\text{,}\) and a typical slice has volume \(V_{\text{slice}} = \pi (\sqrt{100-x^2})^2 \Delta x\text{.}\) The force to move such a slice is its weight, \(F_{\text{slice}} = 62.4 \cdot V_{\text{slice}} = 62.4 \pi (100-x^2) \Delta x\text{.}\) Each slice has to move \(x\) feet to clear the top of the tank, and then an additional 5 feet up to the truck, so the distance each slice moves is \(d_{\text{slice}} = x + 5\text{.}\) Finally, since the tank is full to a depth of 7 feet, and we wish to pump out the top 5 feet, \(x\) ranges from \(x=3\) (the top of the water) to \(x=8\) (the point at which there are 2 feet remaining in the tank). Hence, the total work done is

   \begin{equation*} W = \int_{3}^{8} 62.4 \pi (100-x^2)(x+5) \, dx \approx 673593 \, \text{foot-pounds}\text{.} \end{equation*}

3. The line that bounds the right edge of the triangular face of the tank lying in the first quadrant is \(y=\frac{5}{2} - \frac{5}{8}x\text{.}\) If we take a slice of water that lies at depth \(x\text{,}\) note that the slice is rectangular with thickness \(\Delta x\text{,}\) length 10 (the tank is 10 feet long), and width \(2(\frac{5}{2} - \frac{5}{8}x)\text{.}\) Thus, the volume of a typical slice is

   \begin{equation*} V_{\text{slice}} = 10 \cdot 2(\frac{5}{2} - \frac{5}{8}x) \cdot \Delta x = (50 - \frac{25}{2}x) \Delta x\text{,} \end{equation*}

   and the force that has to be exerted to move such a slice is

   \begin{equation*} F_{\text{slice}} = 62.4 \cdot V_{\text{slice}} = 62.4 (50 - \frac{25}{2}x) \Delta x\text{.} \end{equation*}

   The distance a typical slice moves when pumped to the top of the tank is \(d_{\text{slice}} = x\text{,}\) and since the tank is full to a depth of 3 feet (within 1 foot of the top) and we are going to empty the tank until 1 foot of water remains in it, we see that we need to integrate from \(x=1\) to \(x=3\text{,}\) so

   \begin{equation*} W = \int_{1}^{3} 62.4 (50 - \frac{25}{2}x) x \, dx = 5720 \, \text{foot-pounds}\text{.} \end{equation*}

First, we sketch a picture of the dam, as shown in Figure6.74. Note that, as in problems involving the work to pump out a tank, we let the positive \(x\)-axis point down.

Pressure is constant at a fixed depth, so we consider a slice of water at constant depth on the face, as shown in the figure. The area of this slice is approximately the area of the rectangle pictured. Since the width of that rectangle depends on the variable \(x\text{,}\) we find a formula for the line that represents one side of the dam. It is straightforward to find that \(y = 45 - \frac{3}{5}x\text{.}\) Hence, the approximate area of a representative slice is

At any point on this slice, the depth is approximately constant, so the pressure can be considered constant. Because the water rises to within 5 feet of the top of the dam, the depth of any point on the representative slice is approximately \((x-5)\text{.}\) Now, since pressure is given by \(P = 62.4d\text{,}\) we have that at any point on the slice

Knowing both the pressure and area, we can find the force the water exerts on the slice. Using \(F = PA\text{,}\) it follows that

Finally, we use a definite integral to sum the forces over the appropriate range of \(x\)-values. Since the water rises to within 5 feet of the top of the dam, we start at \(x = 5\) and take slices all the way to the bottom of the dam, where \(x = 30\text{.}\) Hence,

Using technology to evaluate the integral, we find \(F = 848 640\) pounds.

1. Put the center of the top of the dam at the origin and let the positive \(x\)-axis point down.

2. With the positive \(x\)-axis pointing down, the equation for the edge of the semi-circular dam is \(y = \sqrt{900-x^2}\text{.}\)

3. The equation for the line that bounds the right side of the trough is \(y = -\frac{5}{8}x + 2.5 \text{.}\)

1. \(F = \int_{x = 0}^{x=50} (6240 x) dx = 7~800~000 \text{ pounds } \text{.}\)

2. \(F = \int_{x=4}^{x=30} 124.8 (x - 4)\sqrt{900 - x^2} dx = 800~244 \text{ pounds } \text{.}\)

3. \(F = \int_{x=1}^{x=4} 62.4 (x - 1)(5 - 1.25x) dx = 351 \text{ pounds } \text{.}\)

1. We let the positive \(x\)-axis point down and let \(x\) be the distance from the top of the dam. The area of a typical horizontal slice (in square feet) is

   \begin{equation*} A_{\text{ slice } } = 100 \Delta x\text{.} \end{equation*}

   The pressure (in pounds per square foot) at depth \(x\) is given by \(P_{\text{ slice } } = 62.5 x\text{.}\) So the force on a typical slice (in pounds) is

   \begin{equation*} F_{\text{ slice } } = (62.4x)(100 \Delta x) = 6240 x \Delta x\text{.} \end{equation*}

   So the force of the water against the dam is

   \begin{equation*} F = \int_{x = 0}^{x=50} (6240 x) dx = 7~800~000 \text{ pounds }\text{.} \end{equation*}

2. We let the positive \(x\)-axis point down and let \(x\) be the distance from the top of the dam. We place the top of the dam along the \(x\)-axis with the center at the origin. So the equation for the semicircle is \(y = \sqrt{30^2 - x^2} = \sqrt{900 - x^2}\text{.}\) The area of a typical horizontal slice (in square feet) is

   \begin{equation*} A_{\text{ slice } } = 2y \Delta x = 2 \sqrt{900 - x^2} \Delta x\text{.} \end{equation*}

   At a given value of \(x\text{,}\) the depth of the water is \((x - 4)\text{.}\) So the pressure (in pounds per square foot) at \(x\) is given by \(P_{\text{ slice } } = 62.4 (x - 4)\text{.}\) So the force on a typical slice (in pounds) is

   \begin{equation*} F_{\text{ slice } } = 62.4(x - 4) \left(2 \sqrt{900 - x^2} \Delta x \right) = 124.8 (x - 4)\sqrt{900 - x^2} \Delta x\text{.} \end{equation*}

   So the force of the water against the dam is

   \begin{equation*} F = \int_{x=4}^{x=30} 124.8 (x - 4)\sqrt{900 - x^2} dx = 800~244 \text{ pounds }\text{.} \end{equation*}

3. We let the positive \(x\)-axis point down and let \(x\) be the distance from the top of the dam. We place the top of the trough along the \(x\)-axis with the center at the origin. We need the equation for the straight line from the point \((0, 2.5)\) to the point \((4, 0)\text{.}\) The equation for this line is \(y = -\frac{5}{8}x + 2.5 = 2.5 - 0.625x\text{.}\) The area of a typical horizontal slice (in square feet) is

   \begin{equation*} A_{\text{ slice } } = 2y \Delta x = 2(2.5 - 0.625x) \Delta x = (5 - 1.25x) \Delta x\text{.} \end{equation*}

   At a given value of \(x\text{,}\) the depth of the water is \((x - 1)\text{.}\) So the pressure (in pounds per square foot) at \(x\) is given by \(P_{\text{ slice } } = 62.4 (x - 1)\text{.}\) So the force on a typical slice (in pounds) is

   \begin{equation*} F_{\text{ slice } } = 62.4(x - 1) \left( 5 - 1.25x \Delta x \right) = 62.4 (x - 1)(5 - 1.25x) \Delta x\text{.} \end{equation*}

   So the force of the water against the trough is

   \begin{equation*} F = \int_{x=1}^{x=4} 62.4 (x - 1)(5 - 1.25x) dx = 351 \text{ pounds }\text{.} \end{equation*}