Problem 47 - Basic Technical Mathematics with Calculus

Question

A cam is designed such that one face of it is described as being the area between the curves \(y=x^{3}-2 x^{2}-x+2\) and \(y=x^{2}-1\) (units in \(\mathrm{cm}\) ). Show that this description does not uniquely describe the face of the cam. Find the area of the face of the cam, if a complete description requires that \(x \leq 1\)

Step-by-Step Solution

Answer

The area of the face of the cam is 18 cm², and it is not uniquely described without the constraint \(x \leq 1\).

Step 1: Identify Intersection Points

To find the area between the curves, we begin by determining where the two curves intersect. We set the equations equal to each other: \[x^{3} - 2x^{2} - x + 2 = x^{2} - 1.\]Rearranging gives:\[x^{3} - 3x^{2} - x + 3 = 0.\]We solve this equation to find the intersection points.

Step 2: Solve the Cubic Equation

We factor the cubic equation as \((x-1)(x^{2}-2x-3)=0\). The quadratic can be factored further:\[(x-1)(x-1)(x+3)=0.\]This gives us the solutions: \(x=1\) (a repeated root) and \(x=-3\). Thus, the curves intersect at \(x=-3\) and \(x=1\).

Step 3: Set Integration Limits

Because the condition is \(x \leq 1\), we need to integrate from \(-3\) to \(1\). The area between the curves is represented by:\[\int_{-3}^{1} ((x^{2} - 1) - (x^{3} - 2x^{2} - x + 2))\, dx.\]

Step 4: Simplify the Integrand

Simplify the expression inside the integral:\[x^{2} - 1 - (x^{3} - 2x^{2} - x + 2) = -x^{3} + 3x^{2} + x - 3.\]Thus, the integral becomes:\[\int_{-3}^{1} (-x^{3} + 3x^{2} + x - 3) \, dx.\]

Step 5: Integrate the Function

Calculate the definite integral:\[\int (-x^{3} + 3x^{2} + x - 3)\, dx = -\frac{x^{4}}{4} + x^{3} + \frac{x^{2}}{2} - 3x.\]Evaluate this antiderivative from \(-3\) to \(1\).

Step 6: Substitute the Limits and Compute the Area

Substitute the limits into the antiderivative:For \(x = 1\):\[-\frac{1}{4} + 1 + \frac{1}{2} - 3 = -\frac{9}{4}.\]For \(x = -3\):\[-\frac{81}{4} - 27 + \frac{9}{2} + 9 = -\frac{81}{4}.\]The area is:\[-\frac{9}{4} - \left(-\frac{81}{4}\right) = \frac{72}{4} = 18.\]

Step 7: Conclude Description Uniqueness

Since the integration is performed over a specific interval based on the condition \(x \leq 1\), the shape of the cam will differ if other intervals without such constraints were considered. Thus, the cam's shape is not uniquely described by the intersection points alone without the additional constraint on \(x\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, allowing us to find areas, volumes, and other quantities under curves. This process is often described as the reverse of differentiation. To integrate a function means to find the accumulation of quantities, especially when the quantities change as represented by the function's curve.
To find the area between two curves, the integral subtracts one curve from another. For example, if you want to find the area between curves given by functions, you would set up an integral of the top function minus the bottom one:

• Select the limits of integration, often defined by where the curves intersect.
• Subtract the lower function from the upper function inside the integrand.
• Carry out the integration over the defined limits to find the area.

In the exercise, integrating between \(-3\) and \(1\) involves calculating the integral of \(-x^3 + 3x^2 + x - 3\), which requires handling the differentials and boundaries carefully.
Remember, each part of the function contributes proportionally to the total area, so accurate integration is key.

Area Between Curves

The area between two curves is the total space encapsulated by two functions on a graph. This space is found by integrating the difference between the functions over a specific interval bounded by points of intersection.
To calculate the area:

• Identify points of intersection by setting the functions equal and solving for x-values. This will give the bounds for your integration.
• Set up an integral using these bounds, subtracting the lower function from the upper one.
• Evaluate the integral to find the area.

In our example, the curves \(y=x^3-2x^2-x+2\) and \(y=x^2-1\) are considered from \(-3\) to \(1\). The intersection points determine the slice over which we measure the difference in their heights.
The integral of their differences gives the full measurement of the area between, ensuring all space between the curves is accounted for by our calculated bound.

Cubic Equations

Cubic equations are polynomials of degree three, taking the form \(ax^3 + bx^2 + cx + d = 0\). Solving cubic equations often involves finding the roots, which can be done through various methods including factoring, synthetic division, or graphical interpretation.
In this exercise, the cubic equation was obtained by equalizing the curves and simplifying:

• First, simplify the equation to standard form.
• Identify possible integer roots, often by trying values that simplify to zero.
• Factor these roots out step by step.

This specific equation factored into \((x-1)(x^2-2x-3)=0\), with additional factoring of \(x^2-2x-3\) yielding the roots \(x=1\) (a repeated root) and \(x=-3\).
Balanced handling of these results ensures understanding of how intersection points like these directly affect integration limits and, subsequently, how areas are accurately calculated between curves.