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Linear Approximation (Need someone to check my work)

shamieh

Active member
Sep 13, 2013
539
Use a linear approximation to find a good approximation to \(\displaystyle \sqrt{100.4}\)

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

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

\(\displaystyle y = 10.20 \)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.
 

shamieh

Active member
Sep 13, 2013
539
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 ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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).
 

shamieh

Active member
Sep 13, 2013
539
10.02 is the correct answer then correct?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.