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Find the equation of the line tangent to the graph of tan \(x\) at $$ \text { (a) } x=0 \quad \text { (b) } x=\pi / 4 \quad \text { (c) } x=-\pi / 4 $$

Short Answer

Expert verified
(a) y = x, (b) y = 2x - \pi/2 + 1, (c) y = 2x + \pi/2 - 1.

Step by step solution

01

Understanding the Tangent Line Equation

The equation of the tangent line to a function at a given point is given by the formula: \[ y - f(a) = f'(a)(x - a) \]where \( f(a) \) is the function value at \( a \), and \( f'(a) \) is the derivative of the function at \( a \).
02

Finding the Derivative of tan(x)

The function given is \( f(x) = \tan(x) \). The derivative of \( \tan(x) \) is:\[ f'(x) = \sec^2(x) \]This derivative represents the slope of the tangent line at any given point \( x \).
03

Calculate Tangent Line at x = 0

For \( x = 0 \):- Function value: \( f(0) = \tan(0) = 0 \).- Derivative (slope): \( f'(0) = \sec^2(0) = 1 \).Using the tangent line formula: \[ y - 0 = 1(x - 0) \]The equation simplifies to:\[ y = x \]
04

Calculate Tangent Line at x = \( \pi/4 \)

For \( x = \pi/4 \):- Function value: \( f(\pi/4) = \tan(\pi/4) = 1 \).- Derivative (slope): \( f'(\pi/4) = \sec^2(\pi/4) = 2 \).Using the tangent line formula: \[ y - 1 = 2(x - \pi/4) \]The equation simplifies to:\[ y = 2x - \pi/2 + 1 \]
05

Calculate Tangent Line at x = \(-\pi/4\)

For \( x = -\pi/4 \):- Function value: \( f(-\pi/4) = \tan(-\pi/4) = -1 \).- Derivative (slope): \( f'(-\pi/4) = \sec^2(-\pi/4) = 2 \).Using the tangent line formula: \[ y + 1 = 2(x + \pi/4) \]The equation simplifies to:\[ y = 2x + \pi/2 - 1 \]

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus that measures how a function's output changes as its input changes. It tells us the rate of change or the slope of the function at a particular point.
The derivative is crucial for finding the tangent line to a curve, as it gives the slope of this line. For the function \( f(x) = \tan(x) \), the derivative is given by \( f'(x) = \sec^2(x) \), which means "the slope of the tangent" specified by \( \sec^2(x) \) changes with the value of \( x \).
  • This derivative can become quite large, as the secant function grows rapidly in some regions.
  • Knowing how to calculate derivatives of trigonometric functions is important since they often appear in various applications.
By knowing the derivative, we can determine how the function behaves around specific points and construct the equation of the tangent line.
Trigonometric Functions
Trigonometric functions are foundational in trigonometry, relating angles to side ratios in right-angled triangles.Common trigonometric functions include sine, cosine, and tangent, each with unique applications.
Tangent, in particular, is defined as the ratio of sine to cosine \( \tan(x) = \frac{\sin(x)}{\cos(x)} \).
  • It's periodic with a period of \( \pi \), meaning its values repeat every \( \pi \) units.
  • This property is important when looking at places where the tangent is undefined, such as \( x = \frac{\pi}{2} + k\pi \) for integer \( k \).
Understanding how tangent behaves helps locate the points where the derivative, \( \sec^2(x) \), which is \( \left( \frac{1}{\cos^2(x)} \right) \), becomes undefined or changes dramatically.In our problem, we focused on points like \( x = 0 \), \( x = \frac{\pi}{4} \), and \( x = -\frac{\pi}{4} \).At these points, the tangent has straightforward values, making calculations simpler.
Equation of the Line
The equation of the tangent line is essential for describing the line that just "touches" the curve at a specific point without crossing it.It shows a linear approximation of the curve around that point.
The standard formula for the tangent line at a point \( a \) is:\[ y - f(a) = f'(a)(x - a) \]where
  • \( f(a) \) is the value of the function at \( a \), and
  • \( f'(a) \) is the slope derived from the function's derivative at \( a \).
For example, at \( x = 0 \), the tangent line simplifies to \( y = x \).At \( x = \frac{\pi}{4} \), the line becomes \( y = 2x - \frac{\pi}{2} + 1 \).This formula forms the basis for finding how a line behaves at small intervals along the function, crucial for approximations and analyses.

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Most popular questions from this chapter

(a) Prove: $$ \begin{array}{l}{\frac{d^{2}}{d x^{2}}[c f(x)]=c \frac{d^{2}}{d x^{2}}[f(x)]} \\\ {\frac{d^{2}}{d x^{2}}[f(x)+g(x)]=\frac{d^{2}}{d x^{2}}[f(x)]+\frac{d^{2}}{d x^{2}}[g(x)]}\end{array} $$ (b) Do the results in part (a) generalize to \(n\) th derivatives? Justify your answer.

In the temperature range between \(0^{\circ} \mathrm{C}\) and \(700^{\circ} \mathrm{C}\) the re- sistance \(R[\mathrm{in} \text { ohms }(\Omega)]\) of a certain platinum resistance thermometer is given by \(R=10+0.04124 T-1.779 \times 10^{-5} T^{2}\) where \(T\) is the temperature in degrees Celsius. Where in the interval from \(0^{\circ} \mathrm{C}\) to \(700^{\circ} \mathrm{C}\) is the resistance of the ther- mometer most sensitive and least sensitive to temperature changes? [Hint: Consider the size of \(d R / d T\) in the interval \(0 \leq T \leq 700 .]\)

Given the following table of values, find the indicated derivatives in parts (a) and (b). $$ \begin{array}{|c|c|c|c|c|}\hline x & {f(x)} & {f^{\prime}(x)} & {g(x)} & {g^{\prime}(x)} \\ \hline 3 & {5} & {-2} & {5} & {7} \\ \hline 5 & {3} & {-1} & {12} & {4} \\ \hline\end{array} $$ (a) \(F^{\prime}(3),\) where \(F(x)=f(g(x))\) (b) \(G^{\prime}(3),\) where \(G(x)=g(f(x))\)

Let \(f(x)=5 \sqrt{x}\) and \(g(x)=4+\cos x\) (a) Find \((f \circ g)(x)\) and \((f \circ g)^{\prime}(x)\) (b) Find \((g \circ f)(x)\) and \((g \circ f)^{\prime}(x)\)

Suppose that \(L\) is the tangent line at \(x=x_{0}\) to the graph of the cubic equation \(y=a x^{3}+b x .\) Find the \(x\) -coordinate of the point where \(L\) intersects the graph a second time.

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