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Question: How do you estimate the quantity using linear approximation and find the error using a calculator of...

How do you estimate the quantity using linear approximation and find the error using a calculator of 195−198\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}} ?

Explanation

Solution

For answering this question we will assume a function f(x)=1xf\left( x \right)=\dfrac{1}{\sqrt{x}} and find the slopef′(x){{f}^{'}}\left( x \right) of the following curve. We will use the equation of the curve and find the approximate value and compare with the calculated value.

Complete step by step solution:
Now considering from the question we have been asked to estimate the quantity using linear approximation and find the error using a calculator of 195−198\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}} .
We will assume a function f(x)=1xf\left( x \right)=\dfrac{1}{\sqrt{x}} with slope ddxf(x)=−12x−32\dfrac{d}{dx}f\left( x \right)=\dfrac{-1}{2}{{x}^{\dfrac{-3}{2}}} .
Let us consider x=100x=100 then we will have f(100)=110f\left( 100 \right)=\dfrac{1}{10} and slope
ddxf(100)=−12(100)−32 ⇒ddxf(100)=−12(10)−3 \begin{aligned} & \dfrac{d}{dx}f\left( 100 \right)=\dfrac{-1}{2}{{\left( 100 \right)}^{\dfrac{-3}{2}}} \\\ & \Rightarrow \dfrac{d}{dx}f\left( 100 \right)=\dfrac{-1}{2}{{\left( 10 \right)}^{-3}} \\\ \end{aligned} .
Now we will try to get the equation of the curve. The equation of the curve will be given as
(y−f(x1))=f′(x1)(x−x1) ⇒(y−110)=−12(10)−3(x−100) ⇒y=−12000x+320 \begin{aligned} & \left( y-f\left( {{x}_{1}} \right) \right)={{f}^{'}}\left( {{x}_{1}} \right)\left( x-{{x}_{1}} \right) \\\ & \Rightarrow \left( y-\dfrac{1}{10} \right)=\dfrac{-1}{2}{{\left( 10 \right)}^{-3}}\left( x-100 \right) \\\ & \Rightarrow y=\dfrac{-1}{2000}x+\dfrac{3}{20} \\\ \end{aligned} .
Now we need to get the value of
f(95)−f(98)=−12000(95)+320−(−12000(98)+320) ⇒−12000(95)+320+12000(98)−320=12000(98−95) ⇒32000=0.00150 \begin{aligned} & f\left( 95 \right)-f\left( 98 \right)=\dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20}-\left( \dfrac{-1}{2000}\left( 98 \right)+\dfrac{3}{20} \right) \\\ & \Rightarrow \dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20}+\dfrac{1}{2000}\left( 98 \right)-\dfrac{3}{20}=\dfrac{1}{2000}\left( 98-95 \right) \\\ & \Rightarrow \dfrac{3}{2000}=0.00150 \\\ \end{aligned}.
Hence we can say that the estimated value is 0.001500.00150 and the calculator value is given as 0.001580.00158 .
The error percentage will be %error=\dfrac{0.00150-0.00158}{0.00158}\Rightarrow 5% .
Hence we can say that the percentage error is 55% .
Therefore we can conclude that the estimated value of the given quantity 195−198\dfrac{1}{\sqrt{95}}-\dfrac{1}{\sqrt{98}} using linear approximation is 0.001500.00150 and the value found with calculator is 0.001580.00158 and the error is 55%.

Note: While answering this question we should be sure with our concept that we apply and the calculations we make. This is a question related to errors and approximations chapter. We can find the value of 195\dfrac{1}{\sqrt{95}} and 198\dfrac{1}{\sqrt{98}} as given
195=−12000(95)+320 ⇒195=−952000+320 ⇒195=−0.0475+0.15 ⇒195=−0.1025 \begin{aligned} & \dfrac{1}{\sqrt{95}}=\dfrac{-1}{2000}\left( 95 \right)+\dfrac{3}{20} \\\ & \Rightarrow \dfrac{1}{\sqrt{95}}=\dfrac{-95}{2000}+\dfrac{3}{20} \\\ & \Rightarrow \dfrac{1}{\sqrt{95}}=-0.0475+0.15 \\\ & \Rightarrow \dfrac{1}{\sqrt{95}}=-0.1025 \\\ \end{aligned}
and similarly
198=−12000(98)+320 ⇒198=−982000+320 ⇒198=−0.049+0.15 ⇒198=−0.101 \begin{aligned} & \dfrac{1}{\sqrt{98}}=\dfrac{-1}{2000}\left( 98 \right)+\dfrac{3}{20} \\\ & \Rightarrow \dfrac{1}{\sqrt{98}}=\dfrac{-98}{2000}+\dfrac{3}{20} \\\ & \Rightarrow \dfrac{1}{\sqrt{98}}=-0.049+0.15 \\\ & \Rightarrow \dfrac{1}{\sqrt{98}}=-0.101 \\\ \end{aligned}.