Question-200035

Question Number 200035 by ajfour last updated on 12/Nov/23

Commented by ajfour last updated on 12/Nov/23

Find equation of parabola having same curvature as sin x, at shown point

$${Find}\:{equation}\:{of}\:{parabola}\:{having}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:{same}\:{curvature}\:{as}\:\mathrm{sin}\:{x}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{at}\:{shown}\:{point} \\ $$

Commented by Matica last updated on 13/Nov/23

$${Use}\:{approxitation}\:{methods}: \\ $$$$−{Maclurine}\:{serie} \\ $$$$−{Pade}\:{approximation} \\ $$

Answered by mr W last updated on 12/Nov/23

y=sin x at x=(π/2): y′=cos x∣_(x=(π/2)) =0 y′′=−sin x∣_(x=(π/2)) =−1 R=∣(((1+0)^(3/2) )/(−1))∣=1 y=−a(x−(π/2))^2 +1 at x=(π/2): y′=−2a(x−(π/2))∣_(x=(π/2)) =0 y′′=−2a R=∣(((1+0)^(3/2) )/(−2a))∣=(1/(2a))=1 ⇒a=(1/2) ⇒the parabola is y=−(1/2)(x−(π/2))^2 +1

$${y}=\mathrm{sin}\:{x}\:{at}\:{x}=\frac{\pi}{\mathrm{2}}: \\ $$$${y}'=\mathrm{cos}\:{x}\mid_{{x}=\frac{\pi}{\mathrm{2}}} =\mathrm{0} \\ $$$${y}''=−\mathrm{sin}\:{x}\mid_{{x}=\frac{\pi}{\mathrm{2}}} =−\mathrm{1} \\ $$$${R}=\mid\frac{\left(\mathrm{1}+\mathrm{0}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{−\mathrm{1}}\mid=\mathrm{1} \\ $$$${y}=−{a}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}\:{at}\:{x}=\frac{\pi}{\mathrm{2}}: \\ $$$${y}'=−\mathrm{2}{a}\left({x}−\frac{\pi}{\mathrm{2}}\right)\mid_{{x}=\frac{\pi}{\mathrm{2}}} =\mathrm{0} \\ $$$${y}''=−\mathrm{2}{a} \\ $$$${R}=\mid\frac{\left(\mathrm{1}+\mathrm{0}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{−\mathrm{2}{a}}\mid=\frac{\mathrm{1}}{\mathrm{2}{a}}=\mathrm{1}\:\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\Rightarrow{the}\:{parabola}\:{is}\:{y}=−\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1} \\ $$

Commented by ajfour last updated on 13/Nov/23