Problem 2 - 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. \(\sqrt{35} ;\) let \(f(x)=\sqrt{x}, a=36\), and \(x=35\)

Step-by-Step Solution

Answer

Using the approximation formula, \( \sqrt{35} \approx 5.9167 \), while a calculator gives \( \sqrt{35} \approx 5.9161 \). The approximation is quite close.

Step 1: Identify the function and its derivative

The given function is \( f(x) = \sqrt{x} \). To approximate using the formula, we first need the derivative of \( f(x) \), which is \( f'(x) = \frac{1}{2\sqrt{x}} \).

Step 2: Evaluate the function and its derivative at point a

We are given \( a = 36 \). Let's evaluate the function and the derivative at this point: - \( f(36) = \sqrt{36} = 6 \).- \( f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12} \).

Step 3: Set up the linear approximation formula

Substitute \( f(a) \), \( f'(a) \), \( a \), and \( x \) into the linear approximation formula: \[ f(x) \approx f(a) + f'(a)(x-a) \]\[ \sqrt{35} \approx 6 + \frac{1}{12}(35-36) \].

Step 4: Calculate the approximate value

Compute the approximation:- \( 35 - 36 = -1 \).- Substituting into the formula: \[ \sqrt{35} \approx 6 + \frac{1}{12}(-1) = 6 - \frac{1}{12} \].- Convert \( \frac{1}{12} \) to a decimal: \( 0.0833 \), thus, \[ \sqrt{35} \approx 6 - 0.0833 = 5.9167 \].

Step 5: Compare with calculator value

Using a calculator, the actual value of \( \sqrt{35} \) is approximately 5.9161. Compare this value with our approximation 5.9167 to assess the accuracy.

Key Concepts

Derivative

In mathematics, the derivative of a function represents the rate at which the function's value changes as its input changes. For the function given in this problem, \( f(x) = \sqrt{x} \), the derivative helps us understand how quickly the square root of a number changes as we vary the input around a point.
To find the derivative of \( f(x) = \sqrt{x} \), we use the power rule for derivatives. Since \( \sqrt{x} = x^{1/2} \), the power rule tells us that the derivative is \( f'(x) = \frac{d}{dx}[x^{1/2}] = \frac{1}{2}x^{-1/2} \).
After simplifying, this becomes \( f'(x) = \frac{1}{2\sqrt{x}} \). This derivative is essential because it allows us to predict how \( \sqrt{x} \) behaves near a specific point, which is crucial for approximations.

Function Evaluation

Function evaluation involves finding the exact value of the function at a particular point. In our problem, we evaluate the function \( f(x) = \sqrt{x} \) at the point \( a = 36 \).
When \( x = 36 \), we plug it into the function, resulting in \( f(36) = \sqrt{36} = 6 \). This gives us the value of the function at the chosen point.
Similarly, we also evaluate the derivative at \( a = 36 \). We previously found that \( f'(x) = \frac{1}{2\sqrt{x}} \). Substituting \( x = 36 \), we get \( f'(36) = \frac{1}{2\sqrt{36}} = \frac{1}{12} \).
These evaluations are used in the linear approximation formula to approximate the value of \( \sqrt{35} \).

Approximating Values

Approximating values is a way to estimate the function's output at a point where it isn't easily calculable.
Linear approximation involves using a tangent line to the curve of a function at a specific point as an estimate of the function itself. The linear approximation formula is given by:

• \( f(x) \approx f(a) + f'(a)(x-a) \)

For this problem, with \( f(x) = \sqrt{x} \), \( a = 36 \), and \( x = 35 \), we substitute into the formula:

• \( \sqrt{35} \approx 6 + \frac{1}{12}(35-36) \)

Calculate the values to get:

• \( 6 + \frac{1}{12} \cdot (-1) = 6 - 0.0833 \)
• So, \( \sqrt{35} \approx 5.9167 \)

This approximation is close to finding the square root directly, but it's much simpler to calculate, especially without a calculator.

Calculator Comparison

Once we have an approximation, it is helpful to compare it with the exact result from a calculator to assess accuracy.
Using a calculator, the actual value of \( \sqrt{35} \) is approximately 5.9161. When we compare this to our calculated estimation of 5.9167, we can see how close they are.
Small differences are expected in approximations, as they simplify calculations relying on tangent lines rather than the exact curve. Here:

• Estimated value: 5.9167
• Calculator value: 5.9161
• Difference: 0.0006

The close proximity of these numbers shows the accuracy and reliability of linear approximations for estimating difficult-to-calculate function values. It highlights the effectiveness of using derivatives in practical applications.