Chapter 2: Problem 26
A function \(f\) and an \(x\) -value \(a\) are given. Approximate the equation of the tangent line to the graph of \(f\) at \(x=a\) by numerically approximating \(f^{\prime}(a),\) using \(h=0.1 .\) $$f(x)=\cos x, x=0$$
Short Answer
Expert verified
The equation of the tangent line is \( y = 1 - 0.05x \).
Step by step solution
01
Understand the Problem
We are given the function \( f(x) = \cos x \) and the point \( a = 0 \). We need to find the equation of the tangent line at \( x = a \) by approximating the derivative \( f'(a) \) using a small interval \( h = 0.1 \).
02
Use the Difference Quotient
The derivative \( f'(a) \) at \( x = a \) can be approximated using the difference quotient:\[f'(a) \approx \frac{f(a + h) - f(a)}{h}\]Substitute \( f(x) = \cos x \), \( a = 0 \), and \( h = 0.1 \) into the formula.
03
Evaluate the Function at Relevant Points
Calculate \( f(0) = \cos(0) = 1 \) and \( f(0.1) = \cos(0.1) \). We use these in the difference quotient. Using a calculator, find \( f(0.1) \approx 0.995 \).
04
Calculate the Difference Quotient
Substitute the values of \( f(0) \) and \( f(0.1) \) into the difference quotient to approximate the derivative:\[f'(0) \approx \frac{0.995 - 1}{0.1} = \frac{-0.005}{0.1} = -0.05\]
05
Write the Equation of the Tangent Line
The equation of the tangent line at \( x = a \) is given by the formula: \( y = f(a) + f'(a)(x - a) \). Substitute \( a = 0 \), \( f(a) = 1 \), and \( f'(a) \approx -0.05 \) into the formula:\[ y = 1 - 0.05(x - 0) \]which simplifies to \( y = 1 - 0.05x \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Numerical Derivative
Numerical derivatives are useful when it's difficult to find the derivative algebraically or when dealing with complicated functions. Instead of using calculus rules, a numerical derivative approximates the slope of a function at a certain point by evaluating how much the function changes over a tiny interval. In the exercise, the derivative of the function \( f(x) = \cos x \) at \( x = 0 \) is approximated using a numerical derivative. The chosen interval is \( h = 0.1 \), a small value, which helps in capturing the local behavior of the function closely.
To achieve this, we calculate the value of the function "slightly ahead" (\( a + h \)), and "exactly at" the point \( a \). By dividing the change in the function value from \( f(a+h) \) to \( f(a) \) by the interval \( h \), we can effectively estimate the derivative \( f'(a) \). A very small \( h \) can give a closer approximation but might be limited by numerical precision, which is why \( h = 0.1 \) is a reasonable choice here.
To achieve this, we calculate the value of the function "slightly ahead" (\( a + h \)), and "exactly at" the point \( a \). By dividing the change in the function value from \( f(a+h) \) to \( f(a) \) by the interval \( h \), we can effectively estimate the derivative \( f'(a) \). A very small \( h \) can give a closer approximation but might be limited by numerical precision, which is why \( h = 0.1 \) is a reasonable choice here.
Difference Quotient
The difference quotient is a fundamental tool in calculus used to compute derivatives. It is defined as:
\[ f'(a) \approx \frac{f(a + h) - f(a)}{h} \] This method provides an approximation for the instantaneous rate of change of a function at a particular point \( a \). In the given exercise, \( f(x) = \cos x \) is considered, with \( a = 0 \) and \( h = 0.1 \).
The approximation centers around the value \( a \) and involves two function evaluations: \( f(a) \) and \( f(a + h) \). By calculating these values, the difference quotient approximates \( f'(a) \), which gives the slope of the tangent line at the point. This slope is essential for constructing the tangent line equation, which represents the linear approximation of \( f(x) \) at that particular location.
\[ f'(a) \approx \frac{f(a + h) - f(a)}{h} \] This method provides an approximation for the instantaneous rate of change of a function at a particular point \( a \). In the given exercise, \( f(x) = \cos x \) is considered, with \( a = 0 \) and \( h = 0.1 \).
The approximation centers around the value \( a \) and involves two function evaluations: \( f(a) \) and \( f(a + h) \). By calculating these values, the difference quotient approximates \( f'(a) \), which gives the slope of the tangent line at the point. This slope is essential for constructing the tangent line equation, which represents the linear approximation of \( f(x) \) at that particular location.
Trigonometric Functions
Trigonometric functions, such as sine, cosine, and tangent, play a significant role in various fields of science and engineering. In this exercise, the function \( f(x) = \cos x \) is used. The cosine function oscillates between -1 and 1, creating a wave-like pattern. It is periodic with a period of \(2\pi\).
When working with trigonometric functions, it is often necessary to compute their values at various points. In this particular problem, evaluating \( \cos(0) = 1 \) and \( \cos(0.1) \approx 0.995 \) helps to form the difference quotient. Calculators or software are usually used to evaluate trigonometric functions at non-standard angles, ensuring precise computation without algebraic complexities. Understanding these functions' behavior is crucial when dealing with derivatives, as their rates of change vary at different points on the curve.
When working with trigonometric functions, it is often necessary to compute their values at various points. In this particular problem, evaluating \( \cos(0) = 1 \) and \( \cos(0.1) \approx 0.995 \) helps to form the difference quotient. Calculators or software are usually used to evaluate trigonometric functions at non-standard angles, ensuring precise computation without algebraic complexities. Understanding these functions' behavior is crucial when dealing with derivatives, as their rates of change vary at different points on the curve.
Calculus
Calculus is the mathematical study of change and motion, often used to find rates of change, such as velocity or slope, and areas under curves. Calculus is broadly divided into two fields: differential calculus and integral calculus.
In this exercise, we focus on differential calculus, which deals with the concept of a derivative — essentially a measure of how a function's output value changes as its input changes. Here, by applying the difference quotient, we seek to approximate the derivative of \( f(x) = \cos x \) at \( x = 0 \).
The aim is to find the tangent line's equation to the graph at this point, making calculus a practical tool for solving real-world problems involving rates, optimization, and dynamic changes. By understanding and applying calculus concepts such as derivatives and tangent lines, students can better navigate the complexities of understanding how functions behave and change.
In this exercise, we focus on differential calculus, which deals with the concept of a derivative — essentially a measure of how a function's output value changes as its input changes. Here, by applying the difference quotient, we seek to approximate the derivative of \( f(x) = \cos x \) at \( x = 0 \).
The aim is to find the tangent line's equation to the graph at this point, making calculus a practical tool for solving real-world problems involving rates, optimization, and dynamic changes. By understanding and applying calculus concepts such as derivatives and tangent lines, students can better navigate the complexities of understanding how functions behave and change.
