a. Write the equation of the line that represents the linear approximation to the following function at the given point a.
b. Use the linear approximation to estimate the given quantity.
c. Compute the percent error in the​ approximation, 100 times StartFraction StartAbsoluteValue approximation minus exact EndAbsoluteValue Over StartAbsoluteValue exact EndAbsoluteValue EndFraction100•
approximation−exact
exact​, where the exact value is given by a calculator.
​f(x)equals=cosine xcosx at aequals=StartFraction pi Over 2 EndFraction
π
2​, ​f(1.531.53​)

Let's go through each part of the problem step-by-step:

### Part (a): Write the equation of the linear approximation

Given:
- \( f(x) = \cos(x) \)
- Point \( a = \frac{\pi}{2} \)

1. **Find \( f(a) \):**
   \[
   f\left(\frac{\pi}{2}\right) = \cos\left(\frac{\pi}{2}\right) = 0
   \]

2. **Find \( f'(x) \):**
   The derivative of \( f(x) = \cos(x) \) is:
   \[
   f'(x) = -\sin(x)
   \]

3. **Evaluate \( f'(a) \):**
   \[
   f'\left(\frac{\pi}{2}\right) = -\sin\left(\frac{\pi}{2}\right) = -1
   \]

4. **Construct the linear approximation \( L(x) \):**
   The linear approximation (tangent line) at \( x = a \) is:
   \[
   L(x) = f(a) + f'(a)(x - a)
   \]
   Substituting the values we found:
   \[
   L(x) = 0 + (-1)(x - \frac{\pi}{2}) = -x + \frac{\pi}{2}
   \]
   So, the equation of the linear approximation is:
   \[
   L(x) = \frac{\pi}{2} - x
   \]

### Part (b): Use the linear approximation to estimate \( f(1.53) \)

Using our linear approximation \( L(x) = \frac{\pi}{2} - x \), we substitute \( x = 1.53 \):
\[
L(1.53) = \frac{\pi}{2} - 1.53
\]
Using \( \frac{\pi}{2} \approx 1.5708 \):
\[
L(1.53) \approx 1.5708 - 1.53 = 0.0408
\]

Thus, the estimated value of \( f(1.53) \) using the linear approximation is approximately \( 0.0408 \).

### Part (c): Compute the percent error

1. **Find the exact value of \( f(1.53) \):**
   \[
   f(1.53) = \cos(1.53) \approx 0.0423
   \]
   
2. **Calculate the percent error:**
   The formula for percent error is:
   \[
   \text{Percent Error} = 100 \times \frac{| \text{approximation} - \text{exact} |}{|\text{exact}|}
   \]
   Substituting the values:
   \[
   \text{Percent Error} = 100 \times \frac{|0.0408 - 0.0423|}{|0.0423|}
   \]
   \[
   = 100 \times \frac{0.0015}{0.0423} \approx 3.55\%
   \]

### Summary of Answers

1. **Linear Approximation Equation:** \( L(x) = \frac{\pi}{2} - x \)
2. **Estimated Value at \( x = 1.53 \):** \( L(1.53) \approx 0.0408 \)
3. **Percent Error:** Approximately \( 3.55\% \)

Would you like further clarification on any step?

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#### Expanding Questions

1. What is the process for finding the linear approximation of a function at any given point?
2. How does the choice of point \( a \) affect the accuracy of a linear approximation?
3. Why do we use the derivative to find the slope in a linear approximation?
4. How would the approximation change if we were to use a different function, such as \( \sin(x) \), at the same point?
5. How does the error percentage reflect the quality of an approximation?

---

**Tip:** Linear approximations work best when \( x \) is close to the point \( a \) because the tangent line more accurately represents the function’s behavior near that point.

a. Write the equation of the line that represents the linear approximation to the following function at the given point a. b. Use the linear approximation to estimate the given quantity. c. Compute the percent error in the approximation, 100 times StartFraction StartAbsoluteValue approximation minus exact EndAbsoluteValue Over StartAbsoluteValue exact EndAbsoluteValue EndFraction100 approximation−exact exact​, where the exact value is given by a calculator. f(x) equals cosine x at a equals StartFraction pi Over 2 EndFraction, f(1.53)

Explore linear approximation and percent error calculation for the function f(x) = cos(x) at the point π/2. This problem includes deriving the tangent line, estimating values, and calculating percent error to compare the approximation and exact result. Ideal for Grades 11-12 calculus students.

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