Linear Approximation (Need someone to check my work)

In summary, to find a good approximation to \sqrt{100.4}, we can use a linear approximation by setting x = 100.4 and y = \sqrt{100.4}. By using the point (x1 = 100, y1 = 10), we can find the equation y - 10 = \frac{1}{20}(100.4 - 100) and solve for y, which gives us y = 10.02. However, it is important to be careful with calculations involving decimals and to count the number of digits after the decimal point to avoid mistakes.
  • #1
shamieh
539
0
Use a linear approximation to find a good approximation to \(\displaystyle \sqrt{100.4}\)

\(\displaystyle x = 100.4\)
\(\displaystyle x1 = 100\)
\(\displaystyle y1 = 10\)

\(\displaystyle y - 10 = \frac{1}{20}(100.4 - 100) \)

\(\displaystyle y = 10.20 \)
 
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  • #2
shamieh said:
Use a linear approximation to find a good approximation to \(\displaystyle \sqrt{100.4}\)

\(\displaystyle x = 100.4\)
\(\displaystyle x1 = 100\)
\(\displaystyle y1 = 10\)

\(\displaystyle y - 10 = \frac{1}{20}(100.4 - 100) \)

\(\displaystyle y = 10.20 \)

Looks correct... except for a small calculation mistake.
Did you check what $10.20^2$ is?
That should immediately reveal the mistake.
 
  • #3
I like Serena said:
Looks correct... except for a small calculation mistake.
Did you check what $10.20^2$ is?
That should immediately reveal the mistake.

its 104.04 but I don't understand where I went wrong. Why can't I say 1/20 = .05 and then say .05 * .4 = .20, then finally 10 + .20 = 10.20 ?
 
  • #4
shamieh said:
its 104.04 but I don't understand where I went wrong. Why can't I say 1/20 = .05 and then say .05 * .4 = .20, then finally 10 + .20 = 10.20 ?

As you can see your fraction is off by a factor of 10.
Indeed .05 * .4 ≠ .20.
Instead .05 * .4 = .020.

The trick is to count the number of digits after the decimal point.
.05 has 2 digits, .4 has 1 digit, therefore their product (5 x 4 = 20) must have 2+1=3 digits after the decimal point (0.020).
 
  • #5
10.02 is the correct answer then correct?
 
  • #6
shamieh said:
10.02 is the correct answer then correct?

Let's see:
$$10.02^2 = (10 + 0.02)^2 = 10^2 + 2 \cdot 10 \cdot 0.02 + 0.02^2 = 100 + 0.4 + 0.0004 = 100.4004$$
Yep. I'd say that's the correct answer.
 

FAQ: Linear Approximation (Need someone to check my work)

What is linear approximation?

Linear approximation is a mathematical technique used to estimate the value of a function at a particular point by using the value of the function at a nearby point. It involves finding the equation of a line tangent to the function at the given point.

Why is linear approximation important?

Linear approximation is important because it allows us to approximate complex functions with simpler linear equations, making it easier to analyze and calculate values. It is also used in real-world applications such as engineering, physics, and economics.

How accurate is linear approximation?

The accuracy of linear approximation depends on how close the chosen point is to the point of interest. The closer the chosen point is, the more accurate the approximation will be. However, it is important to note that linear approximation is not exact and can introduce errors, especially for functions with high curvature or at points far away from the chosen point.

What are the limitations of linear approximation?

Linear approximation is only accurate for small intervals and cannot be used for functions with high curvature or sharp turns. It also assumes that the function is continuous and differentiable at the chosen point. Additionally, linear approximation may not be suitable for functions with multiple variables.

How is linear approximation different from linear interpolation?

Linear interpolation is a method of estimating values between known data points, while linear approximation is used to estimate the value of a function at a specific point. Linear interpolation involves finding the equation of a line between two known points and using it to find the value at a desired point, while linear approximation involves finding the tangent line at a chosen point and using it to estimate the function's value at that point.

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