Problem 70 - Calculus Single Variable

Question

An underground pipeline is to be built between two points \(P=(1,4)\) and \(Q=(2,0)\). The subsurface rock formation under \(P\) is separated from what lies under \(Q\) by a curve \(\mathcal{C}\) of the form \(y=x^{2} .\) The cost per unit distance of laying pipeline from \(P\) to the graph of \(\mathcal{C}\) is \(\$ 5000 .\) The unit cost from the graph of \(\mathcal{C}\) to \(Q\) is \(\$ 3000 .\) Assuming that the pipeline will consist of two straight line segments, analyze the minimum cost route.

Step-by-Step Solution

Answer

The minimum cost occurs at the intersection point that minimizes the total pipeline distance and cost.

Step 1: Identify Intersection of the Pipeline with Curve

The curve \(\mathcal{C}\) is given by \(y = x^2\). The pipeline will intersect \(\mathcal{C}\) at some point \((x, x^2)\). This intersection point must then be determined to minimize the overall cost.

Step 2: Formulate Cost Function

The total cost \(C\) consists of two parts: from \(P\) to \(\mathcal{C}\) and from \(\mathcal{C}\) to \(Q\). 1. Cost from \(P\) to \(\mathcal{C}\): \[ C_1 = 5000 \sqrt{(x - 1)^2 + (x^2 - 4)^2} \]2. Cost from \(\mathcal{C}\) to \(Q\): \[ C_2 = 3000 \sqrt{(x - 2)^2 + (x^2 - 0)^2} \]The total cost is: \[ C(x) = C_1 + C_2 = 5000 \sqrt{(x - 1)^2 + (x^2 - 4)^2} + 3000 \sqrt{(x - 2)^2 + (x^2)^2} \]

Step 3: Differentiate Cost Function

To find the minimum cost, differentiate the cost function \(C(x)\) with respect to \(x\) and set the derivative equal to zero.1. Differentiate \(C(x)\): \[ C'(x) = \frac{d}{dx} \left( 5000 \sqrt{(x - 1)^2 + (x^2 - 4)^2} \right) + \frac{d}{dx} \left( 3000 \sqrt{(x - 2)^2 + x^4} \right) \]2. Set \(C'(x) = 0\) to find critical points.

Step 4: Solve for Critical Points

Solve the equation \(C'(x) = 0\) to find the critical points. This will require simplifying the derivative and possibly using numerical methods or graphing to find the precise point where the derivative is zero.

Step 5: Analyze Critical Points

Evaluate the cost function \(C(x)\) at the critical points found to determine which one gives the minimum cost. Compare these values to ensure a minimum is found rather than a maximum or inflection point.

Step 6: Conclusion on the Minimum Cost

Once the minimum is identified, the point \((x, x^2)\) corresponds to the optimal point on \(\mathcal{C}\) for laying the pipeline. Substitute this \(x\) value back into the cost function \(C(x)\) to calculate the minimum cost.

Key Concepts

Cost Minimization

Cost minimization in calculus involves finding the least expensive way to accomplish a task, given specific constraints. In the context of an underground pipeline, the primary objective is to determine the path that costs the least to lay the pipeline from point \(P\) to point \(Q\). This involves not just minimizing the physical distance, but also considering variations in costs per unit distance.
This exercise involves two different cost rates, \(\\(5000\) from point \(P\) to the curve \(\mathcal{C}\) and \(\\)3000\) from \(\mathcal{C}\) to point \(Q\). The task is to find that intersection point \((x, x^2)\) on the curve where the total costs from \(P\) to \(Q\) are minimized. This adds complexity as different segments have different costs, affecting the optimal path differently.

• Understanding that cost functions can vary between different segments is essential.
• Minimizing these costs requires calculating when the combined expense of both segments is the lowest.

For such problems, calculus is utilized to set up a cost function that incorporates all these aspects.

Differentiation

Differentiation is a fundamental calculus concept used to find the rate at which a quantity changes. When tackling cost minimization, differentiation helps us understand how the cost changes as we alter variables — in this case, the intersection point \((x, x^2)\).
The process begins with setting up a cost function, such as the one described in this problem with two segments:

• \(C_1 = 5000 \sqrt{(x - 1)^2 + (x^2 - 4)^2}\) for the cost from \(P\) to \(\mathcal{C}\)
• \(C_2 = 3000 \sqrt{(x - 2)^2 + (x^2)^2}\) for the cost from \(\mathcal{C}\) to \(Q\)

Once the function is configured, we differentiate \(C(x)\) with respect to \(x\). The derivative, \(C'(x)\), tells us how the cost changes at each point.

This step is crucial as it sets up the scenario to find critical points, where the slope of the cost curve equals zero, indicating potential minima or maxima.

• Differentiation helps pinpoint these critical points efficiently.
• Understanding how to take derivatives, especially with composite functions involving squares and roots, is key.

Critical Points

Critical points on a graph are where the first derivative of a function is zero or undefined. These points help ascertain local minima or maxima for cost minimization issues. In our pipeline problem, once we differentiate the cost function \(C(x)\), finding the values of \(x\) where \(C'(x) = 0\) yields critical points.
These critical points represent potential options for minimizing costs. However, we cannot assume right away that a critical point provides the lowest cost; we need to evaluate each.

• Solve \(C'(x) = 0\) to find values of \(x\).
• Evaluate \(C(x)\) at these points to compare costs and identify the lowest possible value.

At these critical points, checking the nature (whether minimum or maximum) is vital for full analysis, typically involving second-derivative tests or analyzing changes in \(C(x)\) around these points.

Intersection of Curves

The intersection of curves, particularly in problems involving multiple paths, is crucial for determining optimal connection points. Here, the intersection occurs between the line segment sections of the pipeline and the curve \(\mathcal{C}\) given by \(y = x^2\).
Finding this intersection is essentially solving for \(x\) in the term \((x, x^2)\). This point defines where the pipeline shifts from the costly area under \(P\) to the less expensive terrain towards \(Q\).

• The intersection must be calculated to identify \(x\), affecting the overall cost calculation.
• The intersection can affect the cost due to differing unit prices in the segments involved.

In optimization problems, intersections mark critical transition points, influencing cost functions drastically. Knowing how to handle and calculate these intersections is vital for accurate optimization.