Question

Verify the following calculations:

$$\begin{gathered} \mathbf{a}\mathbf{)}\mathbf{}\sqrt{\mathbf{3}\mathbf{.}\mathbf{1415}\left(\pm 0.0011\right)\mathbf{=}}\mathbf{1}\mathbf{.}{\mathbf{7724}}_{\mathbf{3}}\left(\pm 0.{0003}_3\right)\mathbf{;} \\ \mathbf{b}\mathbf{)}\mathbf{}\mathbf{log}\left[3.1415\left(\pm 0.0011\right)\right]\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{4971}}_{\mathbf{4}}\left(\pm 0.{0001}_5\right)\mathbf{;} \\ \mathbf{c}\mathbf{)}\mathbf{}\mathbf{antilog}\left[3.1415\left(\pm 0.0011\right)\right]\mathbf{=}\mathbf{1}\mathbf{.}{\mathbf{1447}}_{\mathbf{0}}\left(\pm 0.{0003}_5\right)\mathbf{;} \\ \mathbf{d}\mathbf{)}\mathbf{}\mathbf{In}\left[3.1415\left(\pm 0.0011\right)\right]\mathbf{=}\mathbf{1}\mathbf{.}{\mathbf{1447}}_{\mathbf{0}}\left(\pm 0.{0003}_5\right)\mathbf{;} \\ \mathbf{e}\mathbf{)}\mathbf{}\mathbf{log}\left(\frac{\sqrt{0.104\left(\pm 0.006\right)}}{0.0511\left(\pm 0.0009\right)}\right)\mathbf{=}\mathbf{0}\mathbf{.}{\mathbf{80}}_{\mathbf{0}}\left(\pm 0.{01}_5\right) \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{a})\mathrm{y}={\mathrm{x}}^{1/2} \\ \%\mathrm{e}=1/2\left(100\times 0.0011/3.1415\right)=0.0175\% \\ 1.75\times {10}^{-4}\sqrt{3.1415}=3.1\times {10}^{-4} \end{gathered}$$

Step-by-step solution

Verification of solutions.

$$\begin{gathered} \mathrm{a})\mathrm{y}={\mathrm{x}}^{1/2} \\ \%\mathrm{e}=1/2\left(100\times 0.0011/3.1415\right)=0.0175\% \\ 1.75\times {10}^{-4}\sqrt{3.1415}=3.1\times {10}^{-4} \end{gathered}$$

Step 2:

$$\begin{gathered} \mathrm{b})\mathrm{y}=\log \mathrm{x} \\ \%{\mathrm{e}}_{\mathrm{y}}=0.43429\times \left(0.0011/3.1415\right)=1.52\times {10}^{-4} \end{gathered}$$

Step 3:

$$\begin{gathered} c)y=anti\log x={10}^x \\ {e}_y=y\times 2.3026e_x=\left({10}^{3.1415}\right)\left(2.3026\right)\left(0.0011\right)=3.51 \end{gathered}$$

Step 4:

$$\begin{gathered} \mathrm{d})\mathrm{y}=\mathrm{Inx} \\ {\mathrm{e}}_{\mathrm{y}}=0.0011/3.1415=3.5\times {10}^{-4} \end{gathered}$$

Step 5:

$$\begin{gathered} \mathrm{e})\mathrm{The}\mathrm{numerator}\mathrm{of}\log \mathrm{term}:\mathrm{y}={\mathrm{x}}^{1/2} \\ {\mathrm{e}}_{\mathrm{y}}=1/2\left(0.006/0.104\times 100\right)=2.88\% \\ \frac{0.3225\pm 2.88\%}{0.0511\pm 0.0009}=\frac{0.3225\pm 2.88\%}{0.0511\pm 1.76\%}=6.311\pm 3.3755=6.311\pm 0.213 \\ \mathrm{y}=\log \mathrm{x} \\ {\mathrm{e}}_{\mathrm{y}}=\mathrm{o}.42429{\mathrm{e}}_{\mathrm{x}}/\mathrm{x}=0.015 \end{gathered}$$