Question Number 16499 by Tinkutara last updated on 23/Jun/17
$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{in}\:\mathrm{parabolic}\:\mathrm{path} \\ $$$${x}^{\mathrm{2}} \:=\:{y},\:\mathrm{with}\:\mathrm{constant}\:\mathrm{speed}\:{u}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{acceleration}\:\mathrm{of}\:\mathrm{the}\:\mathrm{particle}\:\mathrm{when}\:\mathrm{it} \\ $$$$\mathrm{crossess}\:\mathrm{origin}.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of} \\ $$$$\mathrm{curvature}\:\mathrm{at}\:\mathrm{origin}. \\ $$
Answered by ajfour last updated on 23/Jun/17
![x^2 =y ⇒ (dy/dx)=2x which is zero at the origin. And (d^2 y/dx^2 ) = 2 radius of curvature, r = ∣(([1+(dy/dx)^2 ]^(3/2) )/(d^2 y/dx^2 ))∣=(1/2) . x^2 =y 2x(dx/dt)=(dy/dt) =v_y (=0 at origin) ..(i) u^2 =((dx/dt))^2 +((dy/dt))^2 ....(ii) ⇒ ((dx/dt))^2 =(u^2 /(1+4x^2 )) ....(iii) so ((dx/dt))^2 = u^2 at the origin 2((dx/dt))^2 +2x(d^2 x/dt^2 )=(d^2 y/dt^2 ) .....(iv) means at origin, (d^2 y/dt^2 )=a_y =2u^2 ......(v) ⇒ ((dx/dt))^2 =(u^2 /(1+4x^2 )) 2((dx/dt))((d^2 x/dt^2 ))=− ((8u^2 x)/((1+4x^2 )^2 ))((dx/dt)) a_x = (d^2 x/dt^2 ) = −((4u^2 x)/((1+4x^2 )^2 )) , this is zero at the origin. acceleration a=(d^2 x/dt^2 )+(d^2 y/dt^2 ) at origin a = 0 +2u^2 [see (v)].](Q16510.png)
$${x}^{\mathrm{2}} ={y}\:\:\:\Rightarrow\:\:\frac{{dy}}{{dx}}=\mathrm{2}{x}\:\:\:\:{which}\:{is}\:{zero} \\ $$$$\:{at}\:{the}\:{origin}.\:{And}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:=\:\mathrm{2} \\ $$$${radius}\:{of}\:{curvature},\: \\ $$$$\:\:\:\:\:\:{r}\:=\:\mid\frac{\left[\mathrm{1}+\left({dy}/{dx}\right)^{\mathrm{2}} \right]^{\mathrm{3}/\mathrm{2}} }{{d}^{\mathrm{2}} {y}/{dx}^{\mathrm{2}} }\mid=\frac{\mathrm{1}}{\mathrm{2}}\:. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{2}} =\boldsymbol{{y}} \\ $$$$\:\mathrm{2}{x}\frac{{dx}}{{dt}}=\frac{{dy}}{{dt}}\:={v}_{{y}} \:\:\left(=\mathrm{0}\:{at}\:{origin}\right)\:..\left({i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{u}^{\mathrm{2}} =\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\left(\frac{{dy}}{{dt}}\right)^{\mathrm{2}} \:\:\:\:\:\:....\left({ii}\right) \\ $$$$\Rightarrow\:\:\:\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\:\:\:\:....\left({iii}\right) \\ $$$$\:\:\:{so}\:\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} \:=\:{u}^{\mathrm{2}} \:\:{at}\:{the}\:{origin} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{2}\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} +\mathrm{2}{x}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }=\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\:\:.....\left({iv}\right) \\ $$$${means}\:{at}\:{origin},\:\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }={a}_{{y}} =\mathrm{2}{u}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left({v}\right) \\ $$$$\Rightarrow\:\:\:\left(\frac{{dx}}{{dt}}\right)^{\mathrm{2}} =\frac{{u}^{\mathrm{2}} }{\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} }\: \\ $$$$\:\:\:\mathrm{2}\left(\frac{{dx}}{{dt}}\right)\left(\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\right)=−\:\frac{\mathrm{8}{u}^{\mathrm{2}} {x}}{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\left(\frac{{dx}}{{dt}}\right) \\ $$$${a}_{{x}} =\:\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }\:=\:−\frac{\mathrm{4}{u}^{\mathrm{2}} {x}}{\left(\mathrm{1}+\mathrm{4}{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\:,\:{this}\:{is}\:{zero} \\ $$$$\:{at}\:{the}\:{origin}. \\ $$$${acceleration}\:{a}=\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }\: \\ $$$${at}\:{origin}\:{a}\:=\:\mathrm{0}\:+\mathrm{2}{u}^{\mathrm{2}} \:\:\:\:\:\left[{see}\:\left({v}\right)\right]. \\ $$
Commented by Tinkutara last updated on 23/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$