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Question-108104




Question Number 108104 by bemath last updated on 14/Aug/20
Answered by bobhans last updated on 14/Aug/20
     ((BobHans)/★)  (1) let the curve y = f(x) with gradient = (dy/dx)  where (dy/dx) = k (((y−2)/(x−1))) or (dy/(y−2)) = k (dx/(x−1))  ∫ (dy/(y−2)) = ∫ ((k dx)/(x−1)) ⇒ln (y−2)= k ln (x−1) + c  substitute the point (2,5) & (9,8)  ⇒ { (((2,5)→ln (3)= k. ln (1) + c , c = ln (3))),(((9,8)→ln (6) = k.ln (8)+ ln (3)⇒k=(1/3))) :}  therefore ⇒ln (y−2)=(1/3)ln (x−1)+ln (3)  y−2 = 3 .((x−1))^(1/3)  .
$$\:\:\:\:\:\frac{\mathbb{B}\mathrm{ob}\mathbb{H}\mathrm{ans}}{\bigstar} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{let}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{y}\:=\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{with}\:\mathrm{gradient}\:=\:\frac{\mathrm{dy}}{\mathrm{dx}} \\ $$$$\mathrm{where}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{k}\:\left(\frac{\mathrm{y}−\mathrm{2}}{\mathrm{x}−\mathrm{1}}\right)\:\mathrm{or}\:\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{2}}\:=\:\mathrm{k}\:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{1}} \\ $$$$\int\:\frac{\mathrm{dy}}{\mathrm{y}−\mathrm{2}}\:=\:\int\:\frac{\mathrm{k}\:\mathrm{dx}}{\mathrm{x}−\mathrm{1}}\:\Rightarrow\mathrm{ln}\:\left(\mathrm{y}−\mathrm{2}\right)=\:\mathrm{k}\:\mathrm{ln}\:\left(\mathrm{x}−\mathrm{1}\right)\:+\:\mathrm{c} \\ $$$$\mathrm{substitute}\:\mathrm{the}\:\mathrm{point}\:\left(\mathrm{2},\mathrm{5}\right)\:\&\:\left(\mathrm{9},\mathrm{8}\right) \\ $$$$\Rightarrow\begin{cases}{\left(\mathrm{2},\mathrm{5}\right)\rightarrow\mathrm{ln}\:\left(\mathrm{3}\right)=\:\mathrm{k}.\:\mathrm{ln}\:\left(\mathrm{1}\right)\:+\:\mathrm{c}\:,\:\mathrm{c}\:=\:\mathrm{ln}\:\left(\mathrm{3}\right)}\\{\left(\mathrm{9},\mathrm{8}\right)\rightarrow\mathrm{ln}\:\left(\mathrm{6}\right)\:=\:\mathrm{k}.\mathrm{ln}\:\left(\mathrm{8}\right)+\:\mathrm{ln}\:\left(\mathrm{3}\right)\Rightarrow\mathrm{k}=\frac{\mathrm{1}}{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{therefore}\:\Rightarrow\mathrm{ln}\:\left(\mathrm{y}−\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\left(\mathrm{x}−\mathrm{1}\right)+\mathrm{ln}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{y}−\mathrm{2}\:=\:\mathrm{3}\:.\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{1}}\:. \\ $$
Commented by bemath last updated on 14/Aug/20
tq mr bob
$${tq}\:{mr}\:{bob} \\ $$

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