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Question Number 179458 by cortano1 last updated on 29/Oct/22

Find the area of the triangle in the first quadrant between the x and y axis and the tangent to the relationship curve y=(5/x)−(x/5) and x dont equal 0 at (0,5)

$$ \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{quadrant}\:\mathrm{between}\:\mathrm{the}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$$$\mathrm{axis}\:\mathrm{and}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{relationship}\:\mathrm{curve}\:\mathrm{y}=\frac{\mathrm{5}}{\mathrm{x}}−\frac{\mathrm{x}}{\mathrm{5}}\:\mathrm{and}\:\mathrm{x}\:\mathrm{dont} \\ $$$$\mathrm{equal}\:\mathrm{0}\:\mathrm{at}\:\left(\mathrm{0},\mathrm{5}\right) \\ $$

Commented by Acem last updated on 29/Oct/22

Is the last line implying that the C_f doesn′t intersect with the axis y′y “we know that” or that point of the tangent. So what′s meant by the “and x ≠ 0 at (0, 5)”

$${Is}\:{the}\:{last}\:{line}\:{implying}\:{that}\:{the}\:{C}_{{f}} \:{doesn}'{t} \\ $$$$\:{intersect}\:{with}\:{the}\:{axis}\:{y}'{y}\:``{we}\:{know}\:{that}''\:{or}\:{that} \\ $$$$\:{point}\:{of}\:{the}\:{tangent}.\:{So}\:{what}'{s}\:{meant}\:{by}\:{the} \\ $$$$\:``{and}\:{x}\:\neq\:\mathrm{0}\:{at}\:\left(\mathrm{0},\:\mathrm{5}\right)'' \\ $$$$\: \\ $$

Commented by cortano1 last updated on 30/Oct/22

in this book like that

$$\mathrm{in}\:\mathrm{this}\:\mathrm{book}\:\mathrm{like}\:\mathrm{that} \\ $$

Answered by DvMc last updated on 29/Oct/22

We need an lineal ecuation f(x) of slope m = y′ and that contains the point P(x,(5/x)−(x/5)) y′=−(5/x^2 )−(1/5) then (5/x)−(x/5)=(−(5/x^2 )−(1/5))x+b (5/x)−(x/5)=−(5/x)−(x/5)+b ((10)/x)=b Next f(x)=(−(5/k^2 )−(1/5))x+((10)/k) Where k is the x of the original function in which the tanency is needed The height of triangle is ((10)/k) and its base is x when f(x)=0: 0=(−(5/k^2 )−(1/5))x+((10)/k) x=((10/k)/((25+k^2 )/5k^2 ))=((50k)/(25+k^2 )) So, the area of triangle tangente to y in x=k is A=(1/2)×((10)/k)×((50k)/(25+k^2 ))=((250)/(25+k^2 ))

$$ \\ $$$${We}\:{need}\:{an}\:{lineal}\:{ecuation}\:{f}\left({x}\right)\:{of}\: \\ $$$${slope}\:{m}\:=\:{y}'\:{and}\:{that}\:{contains}\:{the}\: \\ $$$${point}\:{P}\left({x},\frac{\mathrm{5}}{{x}}−\frac{{x}}{\mathrm{5}}\right) \\ $$$$\: \\ $$$${y}'=−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}}\:\:\: \\ $$$$\: \\ $$$${then} \\ $$$$\frac{\mathrm{5}}{{x}}−\frac{{x}}{\mathrm{5}}=\left(−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}}\right){x}+{b} \\ $$$$\frac{\mathrm{5}}{{x}}−\frac{{x}}{\mathrm{5}}=−\frac{\mathrm{5}}{{x}}−\frac{{x}}{\mathrm{5}}+{b}\:\:\:\:\:\:\:\:\: \\ $$$$\frac{\mathrm{10}}{{x}}={b} \\ $$$$\:{Next} \\ $$$${f}\left({x}\right)=\left(−\frac{\mathrm{5}}{{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}}\right){x}+\frac{\mathrm{10}}{{k}}\: \\ $$$${Where}\:{k}\:{is}\:{the}\:{x}\:{of}\:{the}\:{original}\:{function} \\ $$$${in}\:{which}\:{the}\:{tanency}\:{is}\:{needed} \\ $$$${The}\:{height}\:{of}\:{triangle}\:{is}\:\frac{\mathrm{10}}{{k}} \\ $$$${and}\:{its}\:{base}\:{is}\:{x}\:{when}\:{f}\left({x}\right)=\mathrm{0}: \\ $$$$\mathrm{0}=\left(−\frac{\mathrm{5}}{{k}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{5}}\right){x}+\frac{\mathrm{10}}{{k}}\: \\ $$$${x}=\frac{\mathrm{10}/{k}}{\left(\mathrm{25}+{k}^{\mathrm{2}} \right)/\mathrm{5}{k}^{\mathrm{2}} }=\frac{\mathrm{50}{k}}{\mathrm{25}+{k}^{\mathrm{2}} } \\ $$$$\: \\ $$$${So},\:{the}\:{area}\:{of}\:{triangle}\:{tangente}\:{to}\:{y} \\ $$$${in}\:{x}={k}\:{is} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{10}}{{k}}×\frac{\mathrm{50}{k}}{\mathrm{25}+{k}^{\mathrm{2}} }=\frac{\mathrm{250}}{\mathrm{25}+{k}^{\mathrm{2}} }\:\:\: \\ $$

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