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Question Number 176809 by cortano1 last updated on 27/Sep/22

   { ((x=cos^3 ∅)),((y=sin^3 ∅)) :} ⇒(d^2 y/dx^2 ) =?

$$\:\:\begin{cases}{\mathrm{x}=\mathrm{cos}\:^{\mathrm{3}} \emptyset}\\{\mathrm{y}=\mathrm{sin}\:^{\mathrm{3}} \emptyset}\end{cases}\:\Rightarrow\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:=? \\ $$

Answered by mr W last updated on 27/Sep/22

(dy/dφ)=−3 cos^2  φ sin φ  (dx/dφ)=3 sin^2  φ cos φ  ⇒(dy/dx)=((dy/dφ)/(dx/dφ))=((−3 cos^2  φ sin φ)/(3 sin^2  φ cos φ))=−(1/(tan φ))  (d^2 y/dx^2 )=(d/dx)((dy/dx))=(d/dφ)((dy/dx))×(1/(dx/dφ))        =(1/(sin^2  φ))×(1/(3 sin^2  φ cos φ))        =(1/(3 sin^4  φ cos φ))

$$\frac{{dy}}{{d}\phi}=−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\phi\:\mathrm{sin}\:\phi \\ $$$$\frac{{dx}}{{d}\phi}=\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\phi\:\mathrm{cos}\:\phi \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{\frac{{dy}}{{d}\phi}}{\frac{{dx}}{{d}\phi}}=\frac{−\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\phi\:\mathrm{sin}\:\phi}{\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\phi\:\mathrm{cos}\:\phi}=−\frac{\mathrm{1}}{\mathrm{tan}\:\phi} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{d}}{{d}\phi}\left(\frac{{dy}}{{dx}}\right)×\frac{\mathrm{1}}{\frac{{dx}}{{d}\phi}} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\phi}×\frac{\mathrm{1}}{\mathrm{3}\:\mathrm{sin}^{\mathrm{2}} \:\phi\:\mathrm{cos}\:\phi} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{3}\:\mathrm{sin}^{\mathrm{4}} \:\phi\:\mathrm{cos}\:\phi} \\ $$

Commented by Tawa11 last updated on 28/Sep/22

Great sirs

$$\mathrm{Great}\:\mathrm{sirs} \\ $$

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