Question

If \(\mathrm{m}=\tan \theta\) is the slope of the tangent to the curve \(\mathrm{e}\) \(\mathrm{y}=1+\mathrm{x}^{2}\) than (a) \(|\tan \theta| \geq 1\) (b) \(|\tan \theta|<1\) (c) \(\tan \theta<1\) (d) \(|\tan \theta| \leq 1\)

Short answer

The correct option is (a) \(|\tan \theta| \geq 1\).

Step-by-step solution

Find the first derivative of the given equation

Given equation is: \[y = 1 + x^2\] To find the first derivative, differentiate with respect to x: \[\frac{dy}{dx} = \frac{d}{dx}(1+x^2)\] Using the power rule, we get: \[\frac{dy}{dx} = 0 + 2x\] So, the first derivative is: \[\frac{dy}{dx} = 2x\]

Compare the first derivative with the given slope

We have the slope `m` is equal to the tangent of the angle `θ`: \[m = \tan \theta\] Also, we found the first derivative as: \[\frac{dy}{dx} = 2x\] As the first derivative represents the slope of the tangent to the curve, therefore: \[ m = 2x \Rightarrow \tan\theta = 2x \] Now, we need to find the range of `|\tan θ|`.

Find the range of |\tan θ|

Recalling the inequality: \[ \tan\theta = 2x \] Therefore, \[ |\tan\theta| = |2x| \] As the function \(x^2\) is always non-negative, we can say that \[ 1 + x^2 \geq 1 \] So the minimum value of \(|2x|\) when \(x^2 \geq 1\), which is true when \(x \geq 1\) or \(x \leq -1\). Therefore, \[ |\tan\theta| \geq |\tan\theta| \] Comparing with the given options, we can conclude that the correct option is: (a) \(|\tan \theta| \geq 1\)

Key concepts

Differentiation and its Role in Calculus

Differentiation is a fundamental concept in calculus that allows us to find the rate at which a function is changing at any given point. It is essentially about finding the derivative of a function. The derivative, represented as \(\frac{dy}{dx}\), is symbolically a fraction whose numerator and denominator are very small changes in the variables \(y\) and \(x\), respectively.
When we differentiate a function, we apply rules, such as the power rule, to obtain this rate of change. In our exercise, we encountered the quadratic function \(y = 1 + x^2\), and by differentiating it using the power rule, we found that the derivative is \(2x\). This derivative, \(2x\), tells us how steep the curve \(y = 1 + x^2\) is at any point \(x\).
It’s like knowing how quickly or slowly a car is moving at a specific point in time. Differentiation helps us understand the ‘motion’ of the curve without plotting it. On the x-y plane, this derivative becomes crucial because it helps determine the slope of the tangent to the curve at a specific point. Without differentiation, understanding these instantaneous changes in functions would be much harder.

Understanding the Slope of a Tangent Line

The slope of a tangent line at a given point on a curve represents the inclination or steepness of the curve at that specific point. It's the very line that just barely "touches" the curve and does not cross it at that point. In the language of calculus, this slope is derived from the function's derivative.
In our example, the derivative \(\frac{dy}{dx} = 2x\) gives us the slope of the tangent line. This means, for any value \(x\), the slope is double that value. So if \(x = 1\), the slope is \(2 \times 1 = 2\). If \(x = -1\), the slope is \(2 \times -1 = -2\).
The concept becomes invaluable when analyzing the behavior of functions graphically. Tangents can tell us if and how a function is increasing or decreasing. They can also indicate turning points on graphs of functions, showing maximum or minimum points. From this small piece of information—the slope of the tangent—we can infer much about the nature of the broader function.

Exploring Quadratic Equations

Quadratic equations like \(y = 1 + x^2\) are a central topic in algebra and calculus. A quadratic equation is any equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). This form represents a parabola in the Cartesian plane, and each quadratic equation will graph as a curve that opens upwards or downwards.
When working with quadratic equations, we often seek their roots or solutions, which are the values of \(x\) where the equation equals zero. However, in calculus, we may more frequently be interested in how these parabolic shapes change or how steep their curves are, hence the differentiation we performed earlier.
For the equation \(y = 1 + x^2\), the graph is a parabola that opens upwards with its vertex at the point \((0, 1)\). A unique quality of such an equation is its symmetry about the line \(x = 0\). Understanding quadratic equations like this helps us explore the motion and behavior of parabolic curves, especially in physical systems or optimizing routines where curves represent performance metrics or cost functions.

The Relationship between Trigonometric Functions and Calculus

Trigonometric functions, such as \(\tan(\theta)\), \(\sin(\theta)\), and \(\cos(\theta)\), play a crucial role in calculus, particularly when dealing with periodic phenomena or solving certain types of differential equations. They describe relationships in right-angled triangles and can model wave-like patterns.
In our exercise, we specifically looked at the tangent function \(\tan(\theta)\), which is the ratio of the opposite side to the adjacent side in a right-angled triangle. In the realm of calculus, this tangent value is often associated with the slope of lines or surfaces, hence the direct link to differentiation. In fact, the slope \(m\) of a line can be given by \(\tan(\theta) = m\), which connects trigonometry directly to the slopes we calculate from derivative expressions.
The range of the tangent function, particularly its absolute value, was central to our exercise. Understanding how \(|\tan(\theta)|\) compares to values such as 1 helps us interpret and solve problems where angles and slopes are involved. Moreover, trigonometric identities and functions provide additional techniques for solving complex calculus problems, further linking these concepts in mathematical analysis.