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Chapter 3: Problem 19

$$ \begin{array}{l} \text { C } \text { . Find the equation of the tangent line to } y=\cos x \text { at }\\\ x=1 \end{array} $$

Short Answer

Expert verified
The tangent line at \( x = 1 \) is \( y = -\sin(1)x + \sin(1) + \cos(1) \).\)

Step by step solution

01

Find the Derivative

To find the tangent line to the curve at a specific point, we first need to find the derivative of the function. The given function is \( y = \cos x \). The derivative of \( \cos x \) is \( y' = -\sin x \).
02

Evaluate the Derivative at the Given Point

Next, evaluate the derivative at the point \( x = 1 \) to find the slope of the tangent line. Substitute \( x = 1 \) into the derivative: \( y'(1) = -\sin(1) \). Keep in mind that \( \sin(1) \) is a numeric value that can be found using a calculator.
03

Calculate the Value of the Function at the Given Point

We need to find the y-coordinate of the point on the curve where the tangent touches. Substitute \( x = 1 \) into the original function: \( y = \cos(1) \). This will give us the y-coordinate of the point \((1, \cos(1))\).
04

Apply Point-Slope Formula

Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point on the curve and \(m\) is the slope. Here, \( m = -\sin(1), x_1 = 1, y_1 = \cos(1) \). The equation becomes: \( y - \cos(1) = -\sin(1)(x - 1) \).
05

Simplify the Equation

Now simplify the equation from Step 4: \( y - \cos(1) = -\sin(1)x + \sin(1) \). Then rearrange it in the standard form: \( y = -\sin(1)x + \sin(1) + \cos(1) \).

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Key Concepts

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

Derivative
When dealing with calculus and functions, the derivative is a powerful tool. A derivative provides us with the slope of the function at any given point. It essentially tells us how the function changes at a specific moment. For the function given in the exercise, which is \( y = \cos x \), the derivative can be calculated using standard rules for differentiation. The derivative of \( \cos x \) is \( y' = -\sin x \). This result is based on the differentiation rules of trigonometric functions. Once you have the derivative, you can determine how the function is changing at a specific point, which is crucial for finding the equation of a tangent line.
Trigonometric Functions
Trigonometric functions like sine and cosine are fundamental in calculus. These functions arise from the geometric interpretation of a unit circle. In the function \( y = \cos x \), \( \cos x \) represents the cosine of an angle \( x \), which corresponds to the x-coordinate of a point on a unit circle. Trigonometric functions are periodic and oscillate between -1 and 1, making them unique and interesting to work with. When we differentiate \( \cos x \), we use the property that the derivative results in another trigonometric function, \(-\sin x\). Understanding these functions and their derivatives allows for deeper problem-solving strategies in calculus.
Calculus Problem-Solving
Tackling calculus problems often involves systematic steps. In this particular problem, we are tasked with finding the equation of a tangent line. The steps include:
  • Finding the derivative of the function to determine the general slope at any point.
  • Substituting a specific point into the derivative to find the slope at that point.
  • Calculating the corresponding function value to get the y-coordinate of the tangent point.
  • Applying the point-slope formula to derive the tangent line equation.
  • Simplifying the equation for clarity and accuracy.
These steps guide us through understanding the function behavior and applying mathematical frameworks to real calculations.
Point-Slope Form
The point-slope form is a fundamental element in geometry and algebra for determining linear equations. It uses the format \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is a specific point on the line, and \(m\) is the slope. In this exercise, once we calculated the slope \(-\sin(1)\) at \(x = 1\) and found the function value \(y = \cos(1)\), we could apply this form. The equation from point-slope form directly facilitates writing the equation of a tangent line. Rearranging the terms or simplifying allows obtaining a standard line equation that's easier to understand and use.

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Most popular questions from this chapter

Given that \(f(1)=2, f^{\prime}(1)=-1, g(1)=0 \quad\) and \(g^{\prime}(1)=1\), find \(F^{\prime}(1)\) where \(F(x)=f(x) \cos g(x)\).

Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ g(x)=x^{2} \cos x \text { at } a=\pi / 2,[0, \pi] $$

Suppose \(f\) is a function satisfying \(f(3)=8\) and \(f^{\prime}(3.05)=\frac{1}{4} .\) Use this information to approximate \(f(3.05) .\)

Find the linear approximation to the given functions at the specified points. Plot the function and its linear approximation over the indicated interval. $$ G(x)=x+\sin 2 x, \text { at } a=\pi / 2,[0, \pi] $$

If \(y=x^{2}-3\), find the values of \(\Delta y\) and \(d y\) in each case. (a) \(x=2\) and \(d x=\Delta x=0.5\) (b) \(x=3\) and \(d x=\Delta x=-0.12\)
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