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Question Number 59733 by otchereabdullai@gmail.com last updated on 14/May/19

$$\mathrm{If}\mathrm{y}=\sqrt{\mathrm{x}},\mathrm{find}\mathrm{approximate}\mathrm{value}\mathrm{of}$$$$\sqrt{101}$$

Commented by prakash jain last updated on 14/May/19

$${\left(101\right)}^{1/2}={100}^{1/2}{(1+\frac{1}{100})}^{1/2}$$$$=10(1+\frac{1}{200}+\frac{\frac{1}{2}(\frac{1}{2}-1)}{{100}^2}+...)$$$$\mathrm{ignore}\frac{1}{{100}^2}\mathrm{and}\mathrm{smaller}\mathrm{terms}$$$$\approx 10.05$$

Commented by otchereabdullai@gmail.com last updated on 14/May/19

$$\mathrm{thanks}\mathrm{sir}$$

Answered by tanmay last updated on 14/May/19

$$y=\sqrt{x}$$$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$$$$\frac{dy}{dx}\approx \frac{△y}{△x}$$$$△y=\frac{dy}{dx}\times △x$$$$nowx=100y=10$$$$x+△x=101y+△y=?$$$$△x=1$$$$△y=\frac{dy}{dx}\times △x$$$$△y=\frac{1}{2\sqrt{x}}\times △x$$$$△y=\frac{1}{2\times \sqrt{100}}=0.05$$$$soy+△y=10+0.05=10.05$$$$so\sqrt{101}$$$$=10.05$$

Commented by otchereabdullai@gmail.com last updated on 14/May/19

$$\mathrm{fantastic}\mathrm{solution}\mathrm{prof}$$

Answered by MJS last updated on 14/May/19

$${\mathrm{Heron}}^{\prime }\mathrm{s}\mathrm{Algorithm}\mathrm{for}\sqrt[n]{a}:$$$$x_{i+1}=\frac{(n-1)x_i^n+a}{{nx}_i^{n-1}}$$$${\mathrm{we}}^{\prime }\mathrm{re}\mathrm{free}\mathrm{to}\mathrm{choose}\mathrm{any}{x}_1>0$$$$\mathrm{for}n=2:$$$$x_{i+1}=\frac{x_i^2+a}{2x_i}$$$$a=101;x_1=10$$$$x_2=\frac{{10}^2+101}{2\times 10}=\frac{201}{20}=10.050000$$$$x_3=\frac{{\left(\frac{201}{20}\right)}^2+101}{2\times \frac{201}{20}}=\frac{80801}{8040}\approx 10.049876$$

Commented by otchereabdullai@gmail.com last updated on 14/May/19

$$\mathrm{fantastic}\mathrm{solution}\mathrm{prof}$$

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