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Chapter 3: Problem 36

Graph the curves over the given intervals, together with their tangents at the given values of \(x\). Label each curve and tangent with its equation. $$\begin{aligned} &y=\tan x, \quad-\pi / 2 < x < \pi / 2\\\ &x=-\pi / 3,0, \pi / 3 \end{aligned}$$

Short Answer

Expert verified
Graph the curve \( y=\tan x \) and its tangents at \( x=-\pi/3, 0, \pi/3 \): 1. At \(-\pi/3\), \( y = 4x + (4\pi/3 - \sqrt{3}) \). 2. At \(0\), \( y = x \). 3. At \(\pi/3\), \( y = 4x - (4\pi/3) + \sqrt{3} \).

Step by step solution

01

Understand the problem

We are tasked with graphing the curve of \( y = \tan x \) over the interval \( -\pi/2 < x < \pi/2 \). Additionally, we need to find the equations of tangents at given points \( x = -\pi/3, 0, \pi/3 \) and graph these tangents alongside the curve.
02

Graph the curve \( y = \tan x \)

Recall that the tangent function \( y = \tan x \) has vertical asymptotes at \( x = -\pi/2 \) and \( x = \pi/2 \), where the function is undefined. It increases from negative infinity to positive infinity crossing the origin \((0,0)\). Plot the curve between these asymptotes on the Cartesian plane.
03

Calculate the derivative

To find the tangent lines, we need the derivative of \( y = \tan x \), which gives us the slope of the tangent line at any point. The derivative is \( \frac{dy}{dx} = \sec^2 x \).
04

Find the slope at each given point

Calculate the slope of the tangent lines at the points \( x = -\pi/3, 0, \pi/3 \). - At \( x = -\pi/3 \), slope = \( \sec^2(-\pi/3) = 4 \).- At \( x = 0 \), slope = \( \sec^2(0) = 1 \).- At \( x = \pi/3 \), slope = \( \sec^2(\pi/3) = 4 \).
05

Write the equation of tangent lines

Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equations of the tangent lines, where \( m \) is the slope and \( (x_1, y_1) \) are the coordinates on the curve.- At \( x = -\pi/3 \): \( y - (-\sqrt{3}) = 4(x + \pi/3) \).- At \( x = 0 \): \( y - 0 = 1(x - 0) \), simplified to \( y = x \).- At \( x = \pi/3 \): \( y - \sqrt{3} = 4(x - \pi/3) \).
06

Simplify the equations of tangent lines

Simplify the tangent lines' equations:- At \( x = -\pi/3 \): \( y = 4x + (4\pi/3 - \sqrt{3}) \).- At \( x = 0 \): \( y = x \).- At \( x = \pi/3 \): \( y = 4x - (4\pi/3) + \sqrt{3} \).
07

Graph the tangent lines

Plot the tangent lines using their simplified equations on the same graph as the tangent curve \( y = \tan x \). Ensure the tangents touch the curve at their respective points and label each line with its equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are mathematical functions that relate angles of a right-angled triangle to the lengths of its sides. They are fundamental in studying periodic phenomena, like waves and circular motion. Here, we focus on the tangent function, represented as \( y = \tan x \). This function is unique because it results in a repeating pattern over intervals, with important concepts such as:
  • The tangent function has vertical asymptotes, points where the function starts to rise upwards or downwards to infinity, found at \( x = -\pi/2 \) and \( x = \pi/2 \).
  • It is defined as the ratio of sine to cosine: \( \tan x = \frac{\sin x}{\cos x} \). Whenever \( \cos x = 0 \), the tangent function becomes undefined, leading to the vertical asymptotes.
  • The function crosses the x-axis at the origin \((0, 0)\) when both the sine and cosine are zero and one, respectively, resulting in the tangent being zero.
Tangent functions are crucial in calculus and engineering for analyzing oscillations and waves.
Graphing
Graphing a function involves plotting its values on a coordinate plane to visualize its behavior. For the tangent function, it's essential to remember characteristics like periodicity and asymptotes.
  • Since \( y = \tan x \) has vertical asymptotes at \( x = -\pi/2 \) and \( x = \pi/2 \), the graph approaches these lines but never touches them.
  • It spans continuously from negative to positive infinity, climbing steeply as it nears an asymptote.
  • The plotted points and sections of the graph between the asymptotes help to clearly visualize the function's behavior over the specified interval \( -\pi/2 < x < \pi/2 \).
By understanding these features, students can effectively graph the tangent function between its critical points and ensure accurate representation.
Derivatives of Functions
Finding the derivative of a function helps determine the slope of the tangent line at any point on the curve. For trigonometric functions like \( y = \tan x \), this is essential for calculus applications like finding tangent lines.
  • The derivative of \( y = \tan x \) is \( \frac{dy}{dx} = \sec^2 x \), where secant squared is \( \frac{1}{\cos^2 x} \).
  • This derivative formula tells us how sharply or gently the tangent function changes at any given point.
  • By evaluating \( \sec^2 x \) at specific points, such as \( x = -\pi/3, 0, \pi/3 \), we can find the slopes of the tangent lines at these points, which are 4, 1, and 4 respectively.
These calculations allow for a precise determination of tangent descriptors, integrating smoothly with graphing efforts.
Calculus Applications
Calculus provides tools to model real-world scenarios and understand changes in functional behavior. One of its widespread applications is determining tangent lines to curves, which reveal instantaneous rates of change.
  • By using the point-slope form \( y - y_1 = m(x - x_1) \), tangent equations at specific \( x \)-values can be easily derived, where \( m \) is the slope, and \((x_1, y_1)\) is a point on the curve.
  • For the curve \( y = \tan x \), we applied this to find tangents at \( x = -\pi/3, 0, \pi/3 \), resulting in equations: \( y = 4x + (4\pi/3 - \sqrt{3}) \), \( y = x \), and \( y = 4x - (4\pi/3) + \sqrt{3} \) respectively.
  • These lines were then graphed alongside the curve, illustrating the tangent's concept geometrically and algebraically.
Understanding how derivatives aid in creating tangent lines not only enhances mathematical understanding but also equips students for more complex analytical tasks.

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Most popular questions from this chapter

Find \(d y / d t\) $$y=4 \sin (\sqrt{1+\sqrt{t}})$$

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

Two ships are steaming straight away from a point \(O\) along routes that make a \(120^{\circ}\) angle. Ship \(A\) moves at 14 knots (nautical miles per hour; a nautical mile is \(1852 \mathrm{m}\) ). Ship \(B\) moves at 21 knots. How fast are the ships moving apart when \(O A=5\) and \(O B=3\) nautical miles?

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$2 y^{2}+(x y)^{1 / 3}=x^{2}+2, \quad P(1,1)$$

a. Find equations for the tangents to the curves \(y=\sin 2 x\) and \(y=-\sin (x / 2)\) at the origin. Is there anything special about how the tangents are related? Give reasons for your answer. b. Can anything be said about the tangents to the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) at the origin \((m \text { a constant } \neq 0) ?\) Give reasons for your answer. c. For a given \(m,\) what are the largest values the slopes of the curves \(y=\sin m x\) and \(y=-\sin (x / m)\) can ever have? Give reasons for your answer. d. The function \(y=\sin x\) completes one period on the interval \([0,2 \pi],\) the function \(y=\sin 2 x\) completes two periods, the function \(y=\sin (x / 2)\) completes half a period, and so on. Is there any relation between the number of periods \(y=\sin m x\) completes on \([0,2 \pi]\) and the slope of the curve \(y=\sin m x\) at the origin? Give reasons for your answer.
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