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Question Number 138702 by peter frank last updated on 16/Apr/21

Answered by mr W last updated on 17/Apr/21

y=sinh x+k cosh x    =(√(k^2 −1)) ((1/( (√(k^2 −1))))sinh x+(k/( (√(k^2 −1)))) cosh x)    =(√(k^2 −1)) (sinh α sinh x+cosh α cosh x)    =(√(k^2 −1)) cosh (x+α)    ≥(√(k^2 −1))  i.e. y_(min) =(√(k^2 −1)) when x+α=0, i.e.  x=−α=−coth^(−1)   k=−(1/2)ln ((k+1)/(k−1))=(1/2)ln ((k−1)/(k+1))

$${y}=\mathrm{sinh}\:{x}+{k}\:\mathrm{cosh}\:{x} \\ $$$$\:\:=\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:\left(\frac{\mathrm{1}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\mathrm{sinh}\:{x}+\frac{{k}}{\:\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}}\:\mathrm{cosh}\:{x}\right) \\ $$$$\:\:=\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:\left(\mathrm{sinh}\:\alpha\:\mathrm{sinh}\:{x}+\mathrm{cosh}\:\alpha\:\mathrm{cosh}\:{x}\right) \\ $$$$\:\:=\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{cosh}\:\left({x}+\alpha\right) \\ $$$$\:\:\geqslant\sqrt{{k}^{\mathrm{2}} −\mathrm{1}} \\ $$$${i}.{e}.\:{y}_{{min}} =\sqrt{{k}^{\mathrm{2}} −\mathrm{1}}\:{when}\:{x}+\alpha=\mathrm{0},\:{i}.{e}. \\ $$$${x}=−\alpha=−\mathrm{coth}^{−\mathrm{1}} \:\:{k}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{k}+\mathrm{1}}{{k}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$

Commented by peter frank last updated on 17/Apr/21

thank you

$${thank}\:{you} \\ $$

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