Problem 8 - Calculus for Biology and Medicine

Question

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. $$ \cos \left(\frac{\pi}{4}-0.01\right) $$

Step-by-Step Solution

Answer

The approximated value is 0.7000, and the calculator gives 0.7074.

Step 1: Identify the Function and Point of Tangency

The function we want to approximate is \( f(x) = \cos(x) \). We choose \( a = \frac{\pi}{4} \) because \( \frac{\pi}{4} \) is close to \( \frac{\pi}{4} - 0.01 \) and we know the exact value of \( f(a) \).

Step 2: Compute Values at the Point of Tangency

Compute \( f(a) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \). The derivative \( f^{\prime}(x) = -\sin(x) \), so \( f^{\prime}(a) = -\sin\left(\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2} \).

Step 3: Apply the Linear Approximation Formula

Using the formula \( f(x) \approx f(a) + f^{\prime}(a)(x-a) \), substitute \( a = \frac{\pi}{4} \) and \( x = \frac{\pi}{4} - 0.01 \). We get:\[\cos\left(\frac{\pi}{4} - 0.01\right) \approx \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} \times 0.01.\]This simplifies to:\[\approx \frac{\sqrt{2}}{2} - 0.01 \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \times (1 - 0.01).\]

Step 4: Calculate the Approximation

Calculate \( \frac{\sqrt{2}}{2} \approx 0.7071 \). Thus, the approximation is:\[0.7071 \times 0.99 \approx 0.7000.\]

Step 5: Use a Calculator for Comparison

Compute \( \cos\left(\frac{\pi}{4} - 0.01\right) \) using a calculator to get a more accurate value. Plug in the value directly to obtain:\[\cos(0.7853981633974483 - 0.01) \approx 0.7074.\]

Step 6: Compare and Analyze

Compare the approximation \( \approx 0.7000 \) with the calculator result \( \approx 0.7074 \). The linear approximation is close but slightly off by \( 0.0074 \). This shows the effectiveness of linear approximations for small changes around a known point.

Key Concepts

Trigonometric Functions

Trigonometric functions are the cornerstone of many mathematical applications, especially in geometry and calculus. Functions like sine, cosine, and tangent help describe relationships between the angles and sides of triangles. In calculus, they are valuable tools for predicting how things change. In our example, we focus on the cosine function, represented mathematically as \( \cos(x) \). When approximating values like \( \cos\left(\frac{\pi}{4}-0.01\right) \), trigonometric functions allow us to use known values and simplify our calculations.
We start with an angle like \( \frac{\pi}{4} \), which is well-known to have a cosine value of \( \frac{\sqrt{2}}{2} \). This helps us establish a baseline from which small changes are analyzed.

Derivative

A derivative represents the rate at which a function changes at any point. It's like a snapshot of a function's slope at a specific point. To find out how \( \cos(x) \) changes as \( x \) changes, we compute its derivative. The derivative of \( \cos(x) \) is \( f^{\prime}(x) = -\sin(x) \).
This derivative tells us how quickly \( \cos(x) \) changes in response to small changes in \( x \). In the problem at hand, evaluating the derivative at \( a = \frac{\pi}{4} \) gives us \( -\frac{\sqrt{2}}{2} \), a value that is crucial for using linear approximation methods.

• First, differentiate the function.
• Evaluate the derivative at the chosen point.

With this rate of change known, we use it to predict the function's value at slightly different points.

Approximation Error

Approximation error is the difference between an approximate value, derived from a method like linear approximation, and the actual value, often found using tools like calculators. In our exercise, we're finding \( \cos\left(\frac{\pi}{4} - 0.01\right) \). Using linear approximation gives us an estimated result of approximately 0.7000, while directly computing it with higher precision yields 0.7074.
This difference, known as the approximation error, is about 0.0074 in our example. Such an error arises because linear approximations are most accurate with infinitesimally small changes and become less precise with larger deviations.
Here are some key insights for understanding approximation error:

• Approximation error indicates the method's accuracy.
• Smaller errors suggest the approximation closely aligns with true values.
• The error grows with larger intervals from the known point.

Understanding and estimating these errors can guide decisions about when linear approximations are suitable or alternative methods need consideration.