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Question Number 108104 by bemath last updated on 14/Aug/20

Answered by bobhans last updated on 14/Aug/20

$$\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{★}$$$$\left(1\right)\mathrm{let}\mathrm{the}\mathrm{curve}\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)\mathrm{with}\mathrm{gradient}=\frac{\mathrm{dy}}{\mathrm{dx}}$$$$\mathrm{where}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{k}\left(\frac{\mathrm{y}-2}{\mathrm{x}-1}\right)\mathrm{or}\frac{\mathrm{dy}}{\mathrm{y}-2}=\mathrm{k}\frac{\mathrm{dx}}{\mathrm{x}-1}$$$$\int \frac{\mathrm{dy}}{\mathrm{y}-2}=\int \frac{\mathrm{k}\mathrm{dx}}{\mathrm{x}-1}\Rightarrow \ln (\mathrm{y}-2)=\mathrm{k}\ln (\mathrm{x}-1)+\mathrm{c}$$$$\mathrm{substitute}\mathrm{the}\mathrm{point}(2,5)\mathrm{\&}(9,8)$$$$\Rightarrow \{\begin{array}{l}(2,5)\to \ln \left(3\right)=\mathrm{k}.\ln \left(1\right)+\mathrm{c},\mathrm{c}=\ln \left(3\right)\\ (9,8)\to \ln \left(6\right)=\mathrm{k}.\ln \left(8\right)+\ln \left(3\right)\Rightarrow \mathrm{k}=\frac{1}{3}\end{array}$$$$\mathrm{therefore}\Rightarrow \ln (\mathrm{y}-2)=\frac{1}{3}\ln (\mathrm{x}-1)+\ln \left(3\right)$$$$\mathrm{y}-2=3.\sqrt[3]{\mathrm{x}-1}.$$

Commented by bemath last updated on 14/Aug/20

$$tqmrbob$$

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