Linear Approximation and Applications

[hsiwin3]

1) Notice that 15.9 is close to a real number whose square root can be computed easily.

(a) Use this observation and the linear approximation to estimate (15.9)^1/2. (Give your answer correct to at least four decimal places.)
i got 3.9875 for this one and it was right

(b) Find the error using a calculator. (Give your answer correct to at least six decimal places.)
but how do i do this one? i did sqrt(16)-sqrt(15.9) but that was incorrect.


any help would be appreciated

 

[soroban]

Hello, hsiwin3!

1) Notice that 15.9 is close to a real number whose square root can be computed easily.

(a) Use this observation and the linear approximation to estimate \(\displaystyle (15.9)^{\frac{1}{2}}\)
. . (Give your answer correct to at least four decimal places.)

i got 3.9875 for this one and it was right

I agree . . .

(b) Find the error using a calculator.
. . (Give your answer correct to at least six decimal places.)

but how do i do this one? .i did \(\displaystyle \sqrt{16} - \sqrt{15.9}\), but that was incorrect.

They want the difference between your answer and the actual answer.

Your answer is: .\(\displaystyle 3.9875\)

The actual answer is: .\(\displaystyle \sqrt{15.9} \:=\:3.987480407\)


\(\displaystyle \text{And: }\;3.9875 - 3.987480407 \;=\;0.000019593\;\text{ is the error they want.}\)

 

[hsiwin3]

Thank you but that answer in incorrect also.