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Question Number 188083 by Kalebwizeman last updated on 25/Feb/23

determine the value of b for which y=((−x)/3) +b meets the graph of y^2 =x^3 orthogonally

$${determine}\:{the}\:{value}\:{of}\:{b}\:{for}\:{which}\: \\ $$$$\:\:\:{y}=\frac{−{x}}{\mathrm{3}}\:+{b}\:\:{meets}\:{the}\:{graph}\:{of} \\ $$$$\:{y}^{\mathrm{2}} ={x}^{\mathrm{3}} \:\:{orthogonally} \\ $$

Commented by a.lgnaoui last updated on 27/Feb/23

tan θ=(y/x)=(dy/dx) y=x(dy/dx) 3=((b−(x/3)+(c/3))/x) [(1)](meme tengente) b−(x_0 /3) =(√x_0 ^3 ) (2)(intersect) (1)⇒((10)/3)x_0 =b+(c/3) x_0 =((10)/3)b+((10c)/9) b=((10)/3)x_0 −(c/3) ⇒ b−(x_0 /3)=3x_0 −(c/3)=f(x_0 ) ou 3x_0 +(b−((10x_0 )/3))=tangente 1)− b−(x_0 /3)+(c/3)=3x_0 2 )− 3x_0 −(c/3)=(√x_0 ^3 ) x^3 −9x^2 +2cx−((c/3))^2 =0 (a completer...) A suivre...................

$$\mathrm{tan}\:\theta=\frac{{y}}{{x}}=\frac{{dy}}{{dx}}\:\:\:{y}={x}\frac{{dy}}{{dx}} \\ $$$$\:\mathrm{3}=\frac{{b}−\frac{{x}}{\mathrm{3}}+\frac{{c}}{\mathrm{3}}}{{x}}\:\:\:\:\:\left[\left(\mathrm{1}\right)\right]\left({meme}\:{tengente}\right) \\ $$$$\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}\:\:=\sqrt{{x}_{\mathrm{0}} ^{\mathrm{3}} }\:\:\:\:\:\:\:\left(\mathrm{2}\right)\left({intersect}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\frac{\mathrm{10}}{\mathrm{3}}{x}_{\mathrm{0}} ={b}+\frac{{c}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:{x}_{\mathrm{0}} =\frac{\mathrm{10}}{\mathrm{3}}{b}+\frac{\mathrm{10}{c}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{b}=\frac{\mathrm{10}}{\mathrm{3}}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}\:\:\:\:\:\Rightarrow \\ $$$$\:\:\:\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}=\mathrm{3}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}={f}\left({x}_{\mathrm{0}} \right) \\ $$$${ou}\:\:\:\:\mathrm{3}{x}_{\mathrm{0}} +\left({b}−\frac{\mathrm{10}{x}_{\mathrm{0}} }{\mathrm{3}}\right)={tangente} \\ $$$$\left.\:\mathrm{1}\right)−\:\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}+\frac{{c}}{\mathrm{3}}=\mathrm{3}{x}_{\mathrm{0}} \:\:\: \\ $$$$\left.\mathrm{2}\:\right)−\:\:\:\mathrm{3}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}=\sqrt{{x}_{\mathrm{0}} ^{\mathrm{3}} }\: \\ $$$${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}{cx}−\left(\frac{{c}}{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}\:{completer}...\right) \\ $$$${A}\:\:{suivre}................... \\ $$

Commented by a.lgnaoui last updated on 26/Feb/23

Commented by Kalebwizeman last updated on 26/Feb/23

thank you what is now the value of b?

$${thank}\:{you} \\ $$$${what}\:{is}\:{now}\:{the}\:{value}\:{of}\:{b}? \\ $$

Answered by mr W last updated on 27/Feb/23

at intersection point: 2y(dy/dx)=3x^2 (dy/dx)=3 ⇒2y×3=3x^2 ⇒y=(x^2 /2)>0 on the other side y^2 =x^3 ((x^2 /2))^2 =x^3 ⇒x=4 ⇒y=(4^2 /2)=8 8=−(4/3)+b ⇒b=(4/3)+8=((28)/3) ✓

$${at}\:{intersection}\:{point}: \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{2}{y}×\mathrm{3}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}>\mathrm{0} \\ $$$${on}\:{the}\:{other}\:{side} \\ $$$${y}^{\mathrm{2}} ={x}^{\mathrm{3}} \\ $$$$\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{8} \\ $$$$\mathrm{8}=−\frac{\mathrm{4}}{\mathrm{3}}+{b} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{8}=\frac{\mathrm{28}}{\mathrm{3}}\:\checkmark \\ $$

Commented by mr W last updated on 27/Feb/23

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