Here is a example 1.3 from analytical dynamics of Haim Baruh. a particle moves on a path on the xy plane defined by the curve y=3*x^2 , where x varies with the relation x= sin(a). find the radius of curvature of the path and unit vectors in the normal and tangential directions when a=pi/6. This is example with solution but I tried another way to find unit vectors.I have find the tangential vector true but normal vector was wrong but I dont know where I am wrong. Can you help me please. First I have tried to find tangential vector. r=sin(a)i + 3(sin(a))^2j as in the solution. Then I thought unit tangential unit vector must be in the direction of tangent to the path which is the same as derivative of position vector r. its derivative I found is cos(a) i + 6sin(a)cos(a)j. then I calculated derivative vector at a=pi/6 as sqrt(3)/2 i + 3/2 sqrt(3) j, which is the same as in the book. Then I have calculated the magnitude of derivative of position vector and then calculated its unit vector as the tangential vector. This was true result. But when calculating normal unit vector something is wrong. While calculating normal unit vector first I thought that it is orthogonal to the tangential unit vector and if I take derivative of tangential unit vector must be in the direction of normal unit vector. I thought derivative of tangential unit vector must be in the same direction with second derivative of position vector which is -sin(a)i + -6sin(a)cos(a)j. Then I have calculated second derivative vector at pi/6 as -1/2i + -2.59808 j. Then I calculate the magnitude as sqrt(7) and unit vector in the normal direction as (-1/2)/sqrt(7)i + -2.59808/sqrt(7) which is -0.18898i+ -0.98198 j. But solution for normal unit vector in the book is -0.9487i+0.3162j. I can't understand what is wrong for my solution.