Question

A gutter whose cross-section is an arc of a circle is to be made by bending into shape a strip of copper. If the width of the strip is \(a\), find the radius of the eross-section when the carrying capacity of the gutter is a maximum

Short answer

Answer: To find the optimal value of the radius (r) of the cross-section that maximizes the carrying capacity, we first need to express the area of the circle segment formed by bending the strip of copper in terms of r and a: Area(r) = 0.5 * a / (2r) * r² - 0.5 * r² * sin(a / (2r)). Then, find the derivative of the area with respect to r: Area'(r). Set this derivative equal to zero and solve for r. The obtained value of r represents the optimal radius of the cross-section that maximizes the carrying capacity of the gutter. This value can be found numerically or graphically.

Step-by-step solution

Draw and label the gutter

Draw the cross-section of the gutter, a circle segment with an arc length equal to the width of the copper strip (a). Mark the center angle as 2α, the radius as r, and the height of the segment as h.

Use the circumference formula

Since the width of the strip is the arc-length, we can write an equation for it by using the circumference formula as: a = 2rα

Express α in terms of r and a

From Step 2, isolate α. α = a / (2r)

Relate the segment height to the radius and angle

To find the relationship between the height, radius, and angle, we can use the following formula: h = r(1 - cos(α))

Express h in terms of r and a

Substitute α from Step 3 into the height formula from Step 4. h(r) = r(1 - cos(a / (2r)))

Find the area of the circle segment

The area of a circle segment is given by the formula: Area = 0.5 * α * r² - 0.5 * r² * sin(α)

Express the area in terms of r and a

Substitute our expression for α from Step 3 into the area formula: Area(r) = 0.5 * a / (2r) * r² - 0.5 * r² * sin(a / (2r))

Find the derivative with respect to r

To optimize the area, we need to find its derivative with respect to the radius: Area'(r) = \frac{d}{dr}(0.5 * a / (2r) * r² - 0.5 * r² * sin(a / (2r)))

Set the derivative to zero

Find the critical points by setting the derivative to zero: 0 = Area'(r)

Solve for r

Solve the equation 0 = Area'(r) to find the optimal value of r for the maximum carrying capacity. This can be done numerically or graphically.

Conclusion

Once the optimal value of r is found, it corresponds to the radius of the cross-section when the carrying capacity of the gutter is maximized.

Key concepts

Circle Segment

A circle segment is a part of a circle bounded by an arc and a chord. In the context of the gutter-shaped cross-section, the arc length corresponds to the width of the copper strip used to form the gutter. When you draw the segment, you see a sort of 'slice' from the circle, defined by its center angle and radius.

For understanding:

• Imagine slicing a pizza along its crust; this curved edge represents the arc of the segment.
• The straight line from one end of the slice to the other is the chord.
• The area enclosed between the chord and the arc represents the segment.

The segment relationship with the circle means that when you adjust one element like the radius or the angle, it changes the shape and size of the resulting segment. Understanding this relationship is crucial for solving problems on maximizing areas or volumes, such as those in optimization problems like the one described here.

Radius of Curvature

The radius of curvature in this context refers to the radius of the circle segment forming the cross-section of the gutter. This radius is a key component since a change in the radius will alter the curvature and consequently the shape and capacity of the gutter.

Key points about the radius of curvature:

• It determines the 'tightness' or 'openness' of the curve of the segment.
• Bigger radius means a flatter arc, leading potentially to a wider gutter.
• Smaller radius means a sharper curve, which could limit how much the gutter can carry.

In optimization problems, the task is often to find an optimal radius that maximizes a certain property, like the carrying capacity. This involves linking the radius to other known or controllable quantities of the problem, such as the width of the strip or the angle.

Derivative Calculation

Derivative calculation is key in identifying the rate at which a function changes at any point. In optimizing the gutter's carrying capacity, it's essential to calculate the derivative of the segment's area with respect to its radius in order to find maximum points.

Here's why derivatives are essential:

• They help determine where the function's rate of change is zero, which indicates potential maximum or minimum points.
• Understanding this concept proves valuable in identifying optimal conditions.
• In simple terms, we're finding where the 'slope' of our area function is flat, a common method in calculus optimization.

By setting the derivative of the area function to zero and solving for the radius, we identify the critical points. Only then can we verify if these points correspond to the maximum carrying capacity for the gutter.

Maximum Carrying Capacity

The goal is to determine the radius that maximizes the carrying capacity of the gutter. Maximum carrying capacity refers to the largest area or volume that can be efficiently utilized under given constraints.

To find the maximum carrying capacity:

• Follow the process of expressing the area of the segment in terms of the variable radius.
• After expressing, use derivative calculations to find the critical points.
• Solve for the radius by setting the derivative equal to zero.
• Check if these critical points provide a maximum value.

Ultimately, by calculating and optimizing within the constraints of the problem, you can adjust the radius to maximize what the gutter can carry. This step is crucial in engineering and design contexts, where material constraints and efficiency factors are always considered.