Question

Tangent Lines Show that the graphs of the two equations $$y=x\( and \)y=\frac{1}{x}$$ have tangent lines that are perpendicular to each other at their point of intersection.

Short answer

The proofs lies in the calculated slopes of the tangent lines at the intersection points being -1, confirming their perpendicularity. Hence, the tangent lines to the functions \(y=x\) and \(y=1/x\) at their intersection points (1,1) and (-1, -1) are indeed perpendicular to each other.

Step-by-step solution

Find the Intersection Point

To find the point where the graphs of the two equations intersect, set \(y=x\) and \(y=1/x\) equal to each other, giving \(x=1/x\). Multiplying each side by \(x\) leads to \(x^2=1\). Solving this quadratic equation, \(x = 1\) or \(x = -1\). The corresponding \(y\)-values are also \(1\) and \(-1\), respectively, as \(y = x\). Thus, the graphs intersect at points (1,1) and (-1,-1).

Calculate the Derivative (Slope)

The derivative of \(y = x\) is \(dy/dx = 1\), which represents the slope of the tangent line at each point along the graph of this function. For the equation \(y=1/x\), the derivative is \(dy/dx = -1/x^2\). Again, this represents the slope of the tangent line at each point along this graph.

Evaluate the slopes at the Intersection Points

Evaluate the slopes at the points of intersection: For \(y = x\), the slope is always \(1\). For \(y = 1/x\), the slope at \(x = 1\) is \(-1/(1)^2 = -1\), and at \(x = -1\) is \(-1/(-1)^2= -1\). The product of these slopes, in both cases, is \(-1*1 = -1\), confirming that the tangent lines at the intersection points are indeed perpendicular.

Key concepts

Tangent Line

A tangent line to a curve at a given point is a straight line that just touches the curve at that point. This concept is vital because the tangent line represents the direction in which the curve is heading at the point of tangency. In calculus, the equation of a tangent line can be found using the derivative of the function at the given point. The slope of this line is essentially the rate at which the function's value changes at that point.

When it comes to finding the tangent line at a point of intersection between two curves, it's essential to first determine the intersection point, and then find the derivative for each function separately. The slope of the tangent lines at this point will be critical in analyzing the relationship between the two curves at their intersection.

Derivative

The derivative of a function is the fundamental tool of calculus that measures how a function's output value changes as its input value changes. Technically speaking, it's the limit of the average rate of change of the function over an interval as the interval approaches zero. In more practical terms, it represents the slope of the tangent line to the function's graph at any point. For instance, the derivative of the function f(x) = x is a constant 1. This implies that for every unit increase in x, the function's output increases by one unit, indicating a constant slope and thus, a straight line as its graph.

Slope of a Tangent

The slope of a tangent line to a curve at a particular point is the instantaneous rate of change of the curve at that point and is numerically equal to the derivative of the function at that point. When working with curves such as y = x and y = 1/x, calculating the slope is essential to understand how steep the lines are at any given point. In the context of the exercise, the slopes at the intersection points are of specific interest because they will determine if the tangent lines are perpendicular. A key rule to remember is that two lines are perpendicular if the product of their slopes is -1, as seen with the slopes at points (1,1) and (-1,-1) in the given task.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation in a single variable x with the general form ax^2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. It can have two real solutions, one solution, or no real solutions, and these solutions are known as the roots of the equation. In our example exercise, the equation x^2 = 1 is a simple form of a quadratic equation and solving it yields the intersection points of the two given curves.

Intersection Point

The intersection point of two graphs is a point where the two curves cross each other. To find these points algebraically, you set the equations of the graphs equal to each other and solve for the variable. This is precisely what we did in the exercise by setting y = x equal to y = 1/x, resulting in the quadratic equation x^2 = 1. Once the intersection points are found, these points can be applied to additional analyses, such as evaluating the perpendicularity of tangent lines as in our example.