EBK NUMERICAL METHODS FOR ENGINEERS
EBK NUMERICAL METHODS FOR ENGINEERS
7th Edition
ISBN: 8220100254147
Author: Chapra
Publisher: MCG
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Chapter 18, Problem 1P

Estimate the common logarithm of 10 using linear interpolation.

(a) Interpolate between log 8 = 0.9030900  and log  12 = 1.0791812 .

(b) Interpolate between log 9 = 0.9542425  and log  11 = 1.0413927 .

For each of the interpolations, compute the percent relative error based on the true value.

(a)

Expert Solution
Check Mark
To determine

To calculate: The common logarithmic of 10 by the linear interpolation between log8=0.9030900 and log12=1.0791812. Also, find the percentage relative error based on the true value.

Answer to Problem 1P

Solution:

The common logarithmic of 10 by the linear interpolation between log8 and log12 is 0.9911356 with relative percentage error of 0.88644%.

Explanation of Solution

Given Information:

The values,

log8=0.9030900log12=1.0791812

Formula used:

Linear interpolation formula:

f1(x)=f(x0)+f(x1)f(x0)x1x0(xx0)

And, formula for percentage relative error is,

εt=|True valueApproximatevalueTrue value|×100%

Calculation:

Consider the values, log8=0.9030900 and log12=1.0791812.

Here, x0=8 and x1=12. Therefore,

f(x0)=0.9030900f(x1)=1.0791812

Thus, the value of log 10 by the linear interpolation is,

f1(10)=0.9030900+1.07918120.9030900128(108)=0.9030900+0.17609124×2=0.9030900+0.0880456=0.9911356

Now, the true value of log10=1. Therefore, relative percentage error is,

εt=10.99113561×100%=0.0088644×100%=0.88644%

Hence, the value of log 10 by the linear interpolation is 0.9911356 with relative percentage error of 0.88644%.

(b)

Expert Solution
Check Mark
To determine

To calculate: The common logarithmic of 10 by the linear interpolation between log9=0.9542425 and log11=1.0413927. Also, find the percentage relative error based on the true value.

Answer to Problem 1P

Solution:

The common logarithmic of 10 by the linear interpolation between log9 and log11 is 0.9978176 with relative percentage error of 0.218%.

Explanation of Solution

Given Information:

The values,

log9=0.9542425log11=1.0413927

Formula used:

Linear interpolation formula:

f1(x)=f(x0)+f(x1)f(x0)x1x0(xx0)

And, formula for percentage relative error is,

εt=|True valueApproximatevalueTrue value|×100%

Calculation:

Consider the values, log9=0.9542425 and log11=1.0413927.

Here, x0=9 and x1=11. Therefore,

f(x0)=0.9542425f(x1)=1.0413927

Thus, the value of log 10 by the linear interpolation is,

f1(10)=0.9542425+1.04139270.9542425119(109)=0.9542425+0.08715022×1=0.9542425+0.0435751=0.9978176

Now, the true value of log10=1. Therefore, relative percentage error is,

εt=|10.99781761|×100%=0.0021824×100%=0.218%

Hence, the value of log 10 by the linear interpolation is 0.9978176 with relative percentage error of 0.218%.

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EBK NUMERICAL METHODS FOR ENGINEERS

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