Problem 23 - Calculus Early Transcendentals

Question

Use linear approximations to estimate the following quantities. Choose a value of a that produces a small error. \(\sqrt{146}\)

Step-by-Step Solution

Answer

Question: Use linear approximation to estimate the value of √146. Answer: Using linear approximation with a chosen value of a=144, we estimate that √146 is about 12.083333.

Step 1: Choose a suitable value for a

We need to find an \(a\) value for which the square root is easy to compute and that is close to the number whose root we're estimating (\(146\)). The closest square number to \(146\) is \(144\), so we'll choose \(a = 144\) because \(\sqrt{144} = 12\) is easy to compute.

Step 2: Define the function f(x) and find its derivative

Define the function \(f(x) = \sqrt{x}\). Next, we need to find its derivative, \(f'(x)\). Using the power rule, we have: $$f'(x) = \frac{d}{dx}x^{\frac{1}{2}} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}.$$

Step 3: Evaluate the derivative at a

Now, we need to evaluate the derivative \(f'(a)\) at \(a = 144\). Plugging in the value of \(a\), we get: $$f'(144) = \frac{1}{2\sqrt{144}} = \frac{1}{2\cdot 12} = \frac{1}{24}.$$

Step 4: Apply the Linear Approximation Formula

The linear approximation formula is as follows: $$f(x) \approx f(a) + f'(a)(x-a).$$ Now, we need to plug in the values we found in Step 1 and Step 3, as well as our target value \(x = 146\), to estimate \(\sqrt{146}\): $$\sqrt{146} \approx f(144) + f'(144)(146-144) \Rightarrow \sqrt{146} \approx 12 + \frac{1}{24}(2).$$

Step 5: Compute the final estimate

Finally, we can compute the estimate for \(\sqrt{146}\) using the values from the previous steps: $$\sqrt{146} \approx 12 + \frac{1}{24}(2) = 12 + \frac{1}{12} = 12 + 0.083333 = 12.083333$$ Using linear approximation with a chosen value of \(a=144\), we estimate that \(\sqrt{146}\) is about \(12.083333\).

Key Concepts

Derivative

When we talk about derivatives in mathematics, we're essentially discussing how a function changes. Imagine you're on a long road trip, and you want to know how fast you're going at various points. The speed at any specific moment is like the derivative of your journey's distance function – it tells you how your distance changes over time.
In our exercise, the function in question is the square root function, \(f(x) = \sqrt{x}\). We find the derivative of this using rules of calculus, illustrating how this function increases as \(x\) changes.

• The derivative \(f'(x)\) helps estimate the rate at which \(\sqrt{x}\) changes near the point \(x = 144\).
• For the square root function, \(f'(x) = \frac{1}{2\sqrt{x}}\), computed using differentiation techniques from calculus.

This derivative helps us with linear approximation by giving us insight into the behavior of the function near our chosen point.

Power Rule

The power rule is crucial in calculus for finding the derivative of functions in the form \(x^n\). It's like having a superpower that makes the process quick and easy! You simply multiply by the exponent and reduce the exponent by one.
For example, if you have \(x^3\), its derivative using the power rule is \(3x^2\). It's straightforward and very handy.

• For our exercise, we applied the power rule to \(f(x) = \sqrt{x} = x^{\frac{1}{2}}\).
• Using the power rule, the derivative becomes \(f'(x) = \frac{1}{2}x^{-\frac{1}{2}}\).

This rule leads us directly to understanding how the function behaves locally and is an essential tool in differential calculus.

Square Root Estimation

Linear approximation is a great way to estimate functions like square roots when exact computation is challenging. In our exercise, we estimated \(\sqrt{146}\) using a linear approximation around a value easily manageable, \(a = 144\), where the square root is exactly 12.
Here's a simple way to think about square root estimation:

• Choose a nearby 'nice' number you know the square root of, in this case, 144.
• Use the linear approximation formula: \(f(x) \approx f(a) + f'(a)(x-a)\).
• Calculate the approximated value of \(\sqrt{146}\) giving a result that is close to the real value.

This method helps when dealing with non-perfect squares and gives a quick, yet reliable estimate of square roots.

Differential Calculus

Differential calculus is the branch that studies how things change; it's like the world's knowledge about motion and growth wrapped up in mathematics. It provides tools to understand rates of change, slopes of curves, and so much more.
In our exercise, we harnessed differential calculus to perform linear approximation, allowing us to estimate function values easily.

• Differential calculus involves concepts such as derivatives which are key to understanding how functions behave.
• Its applications range from physics to economics, anytime you need to analyze change.
• Through concepts like the power rule and linear approximations, we see how differential calculus simplifies complex problems.

It's a powerful branch of mathematics that serves as a foundation for many scientific analyses and problem-solving techniques.