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1.A Particle moves in a curve y=A LOG(secx/a) such that tangent to the curve rotates uniformly > prove that the resultant acceleration of the particle varies as the square of the radius of curvature.2.A Particle is moving in a parabola p2 = ar with uniform angular velocity about the focus . show that its normal accleration at any point is proportional to the radius of curvature of its path at that point.

3 years ago
Y=album(sec(x)/a)Putting a =1Y=ln(sec(x))Slope of this curve,Tan(θ)=dy/dx=tan(x) ;equation(1)θ=xAngular velocity of tangent i.e. ω=dθ/dt=dx/dt=2Hence dx/dt=2By equation (1) dy/dt =tan(x) dx/dtHence velocity along y direction is 2 tan(x)So acceleration along y,= 2 sec(x) sec(x)At x=π/4, acceleration=2 √2 √2 =4
3 years ago
Y=ln(sec(x)/a),Putting a =1,Y=ln(sec(x)),Slope of this curve,Tan(θ)=dy/dx=tan(x)........equation(1),θ=x,Angular velocity of tangent i.e. ω=dθ/dt=dx/dt=2,Hence dx/dt=2,By equation (1),dy/dt =tan(x) dx/dt,Hence velocity along y direction=2tan(x),So acceleration along y,= 2 sec(x) sec(x),At x=π/4, acceleration=2 √2 √2 =4
3 years ago
Y=ln(sec(x)/a);Putting a =1;Y=ln(sec(x));Slope of this curve;Tan(θ)=dy/dx=tan(x)........equation(1);θ=x;Angular velocity of tangent i.e. ω=dθ/dt=dx/dt=2;Hence dx/dt=2;By equation (1);dy/dt =tan(x) dx/dt;Hence velocity along y direction=2tan(x);So acceleration along y;= 2 sec(x) sec(x);At x=π/4; Acceleration=2 √2 √2 =4