Question

Sector of a hyperbola Let \(H\) be the right branch of the hyperbola \(x^{2}-y^{2}=1\) and let \(\ell\) be the line \(y=m(x-2)\) that passes through the point (2,0) with slope \(m,\) where \(-\infty < m < \infty\). Let \(R\) be the region in the first quadrant bounded by \(H\) and \(\ell\) (see figure). Let \(A(m)\) be the area of \(R .\) Note that for some values of \(m\) \(A(m)\) is not defined. a. Find the \(x\) -coordinates of the intersection points between \(H\) and \(\ell\) as functions of \(m ;\) call them \(u(m)\) and \(v(m),\) where \(v(m) > u(m) > 1 .\) For what values of \(m\) are there two intersection points? b. Evaluate \(\lim _{m \rightarrow 1^{+}} u(m)\) and \(\lim _{m \rightarrow 1^{+}} v(m)\) c. Evaluate \(\lim _{m \rightarrow \infty} u(m)\) and \(\lim _{m \rightarrow \infty} v(m)\) d. Evaluate and interpret \(\lim _{m \rightarrow \infty} A(m)\)

Short answer

Based on the step-by-step solution, a short answer would be as follows: To find the area of the region enclosed by the hyperbola and the line within the first quadrant, we first found the intersection points between them, calculated the limits of the x-coordinates of the intersection points (u(m) and v(m)) as m approaches 1 and infinity, and then found the limit of the area (A(m)) as m approaches infinity. The limit of A(m) was found to be 0, meaning that the enclosed area becomes infinitely thin as the line becomes parallel to the asymptote of the hyperbola.

Step-by-step solution

Solve for intersection points

To find the \(x\)-coordinates of the intersection points, we must first find the intersection points \((x, y)\). To do this, we can set the equations equal to each other: \(m(x-2) = x^2 - y^2\) and \(y = mx - 2m\). Now substituting the second equation into the first, we have \(m(x-2) = x^2 - (mx - 2m)^2\)

Find u(m) and v(m)

To find \(u(m)\) and \(v(m)\), solve for \(x\) in the above equation and express it as a function of \(m\): \(m(x-2) = x^2 - m^2x^2 + 4m^2x - 4m^3 \Rightarrow (m^2 + 1)x^2 - (4m^2-2m)x + 4m^3 = 0\) Let \(f(x) = (m^2+1)x^2 -(4m^2 - 2m)x + 4m^3\). Now, we need to find the values of \(x\) for which \(f(x) = 0\). These two x-coordinates should be distinct and greater than 1. The quadratic equation for \(x\) is given by: \(x = \frac{-(-4m^2+2m) \pm \sqrt{(-4m^2 + 2m)^2 - 4(m^2+1)(4m^3)}}{2(m^2+1)}\) Now let's calculate the discriminant: \(D = (-4m^2 + 2m)^2 - 4(m^2+1)(4m^3) = 16m^4 - 16m^3 + 4m^2 - 16m^2 - 16m^3 +16 = 16m^4 - 32m^3 - 12m^2 + 16\) For the equation to have two distinct real roots, the discriminant must be positive, which means that there are two intersection points: \(D > 0 \Rightarrow 16m^4-32m^3-12m^2+16 > 0\) The inequality holds true for values of \(m\) greater than zero (excluding \(m = 1\)). Now we can define \(u(m)\) and \(v(m)\): \(u(m) = \frac{2m(2m - 1) - \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)}\) \(v(m) = \frac{2m(2m - 1) + \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)}\) Next, we must evaluate the limits of \(u(m)\) and \(v(m)\).

Calculate limits of u(m) and v(m) as m → 1+

By applying L'Hopital's rule: \(\lim_{m \rightarrow 1^+} u(m) = \lim_{m \rightarrow 1^+} \frac{2m(2m-1) - \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)} = 1\) \(\lim_{m \rightarrow 1^+} v(m) = \lim_{m \rightarrow 1^+} \frac{2m(2m-1) + \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)} = 1\)

Calculate limits of u(m) and v(m) as m → ∞

Using the limit rules for rational functions, we can determine the limits of \(u(m)\) and \(v(m)\) as \(m\) approaches infinity: \(\lim_{m \rightarrow \infty} u(m) = \lim_{m \rightarrow \infty} \frac{2m(2m-1) - \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)} = 0\) \(\lim_{m \rightarrow \infty} v(m) = \lim_{m \rightarrow \infty} \frac{2m(2m-1) + \sqrt{16m^4-32m^3-12m^2+16}}{2(m^2+1)} = \infty\)

Calculate the limit of A(m) as m → ∞

To find the limit of A(m) as \(m\) approaches infinity, we can use integration by finding the integral of the line minus the integral of the hyperbola within the interval [u(m), v(m)]: \(A(m) = \int_{u(m)}^{v(m)} (m(x-2) - (x^2 - y^2)) dx\) Now we can compute the limit as \(m\) approaches infinity: \(\lim_{m \rightarrow \infty} A(m) = \lim_{m \rightarrow \infty} \int_{0}^{\infty} (m(x-2) - (x^2 - y^2)) dx = 0\) Thus, as \(m\) approaches infinity, the area of the region enclosed by the hyperbola and the line ultimately becomes zero. This can be interpreted as the line becoming parallel to the asymptote of the hyperbola, and the enclosed region becoming infinitely thin.

Key concepts

Limits of Functions

When we discuss the limits of functions, we're talking about what happens to the values of the function as the input approaches a certain point. It's a fundamental concept in calculus that helps us understand the behavior of functions near specific points, even if that function is not defined directly at that point. In our exercise, evaluating the limits of functions is crucial for finding the intersection points as the slope of the line, denoted by m, approaches particular values.

One could see the limit as a way to forecast the value of a function just before it reaches a certain point, similar to predicting the path of a ball as it rolls towards an edge. This concept is notably essential when we cannot simply plug a value into a function to get an answer, due to either discontinuity or because the value tends towards infinity, as in the case with our hyperbola intersection problem.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation typically written in the form ax^2 + bx + c = 0. These equations are fundamental in algebra and appear in various scientific contexts, including our hyperbola exercise. The solutions to these equations, where they intersect the x-axis, are called the roots.

In the context of our problem, finding the x-coordinates of the intersection points between a hyperbola and a line leads to solving a quadratic equation. The process involves substituting one function into another and then rearranging the terms to yield a quadratic equation that depends on the slope m. Notably, these equations can have two real solutions, one solution, or no real solutions, which impacts the number of intersection points.

Discriminant of a Quadratic

The discriminant of a quadratic equation provides critical information about the nature of its roots. Represented by the symbol D, the discriminant is the part of the quadratic formula under the square root: D = b^2 - 4ac.

When D is positive, there are two distinct real roots. If it's zero, there's exactly one real root (a repeated root). And if it's negative, there are no real roots, meaning the quadratic function does not intersect the x-axis. In our exercise, we use the discriminant to determine the conditions under which the line and the hyperbola intersect within the first quadrant, which is essential for finding the function A(m), representing the area of the region in question.

Area Under a Curve

The concept of area under a curve is essential in mathematics, particularly within the realm of integral calculus. It involves calculating the size of the region bounded by a curve, the x-axis, and given vertical boundaries.

Applying this to our example of the hyperbola, we determine the area A(m) by integrating the difference between the functions that define our bounds over a specific interval. In simpler terms, we're essentially adding up an infinite number of infinitesimally thin rectangles under the curve to calculate the total area. The results from such an analysis can provide a wealth of information, including, for instance, the physical space that a certain process or object occupies, as symbolized by the curve.