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

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

Commented by Acem last updated on 29/Oct/22

$$Isthelastlineimplyingthatthe{C}_f{doesn}^{\prime }t$$$$intersectwiththeaxis{y}^{\prime }y‘‘weknow{that}^{″}orthat$$$$pointofthetangent.So{what}^{\prime }smeantbythe$$$$‘‘andx\ne 0at{(0,5)}^{″}$$$$$$

Commented by cortano1 last updated on 30/Oct/22

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

Answered by DvMc last updated on 29/Oct/22

$$$$$$Weneedanlinealecuationf\left(x\right)of$$$$slopem=y^{\prime }andthatcontainsthe$$$$pointP(x,\frac{5}{x}-\frac{x}{5})$$$$$$$$y^{\prime }=-\frac{5}{x^2}-\frac{1}{5}$$$$$$$$then$$$$\frac{5}{x}-\frac{x}{5}=(-\frac{5}{x^2}-\frac{1}{5})x+b$$$$\frac{5}{x}-\frac{x}{5}=-\frac{5}{x}-\frac{x}{5}+b$$$$\frac{10}{x}=b$$$$Next$$$$f\left(x\right)=(-\frac{5}{k^2}-\frac{1}{5})x+\frac{10}{k}$$$$Wherekisthexoftheoriginalfunction$$$$inwhichthetanencyisneeded$$$$Theheightoftriangleis\frac{10}{k}$$$$anditsbaseisxwhenf\left(x\right)=0:$$$$0=(-\frac{5}{k^2}-\frac{1}{5})x+\frac{10}{k}$$$$x=\frac{10/k}{(25+k^2)/5k^2}=\frac{50k}{25+k^2}$$$$$$$$So,theareaoftriangletangentetoy$$$$inx=kis$$$$A=\frac{1}{2}\times \frac{10}{k}\times \frac{50k}{25+k^2}=\frac{250}{25+k^2}$$

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