Q2) Solve the wave equation u=u 00 " XX 11 subject to u(0,t) = u(2,t) =0, t>0 and u(x,0)=0, u )=0, u,(x,0): = sin (3лx), 0
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- 2. The position vector of a particle is given by r(t)= (2 cos t sin t)i +(cos^2 t - sin^2 t)j + (3t)k If the particle begins its motion at t = 0 and ends at t = pi, find the difference between the length of the path traveled and the distance between start position and end positionShow that the function Z = sin(wct)sin(wx) satisfies the wave equationThe motion of a point on the circumference of a rolling wheel of radius 3 feet is described by the vector function 7(t) = 3(10t – sin(10t))ỉ + 3(1 – cos(104))} Find the velocity vector of the point. v(t) = Find the acceleration vector of the point. a(t) Find the speed of the point. s(t) =
- for wave equation, seperation of vairables u(x,t)=X=(x)T(t)Which of the following is most suitable solution for wave equation = c2? %3D at2 ax2 .(a) y = (AeP* + Be-P*)(Ce pt + De ept) (b) y = (Acos px + Bsin px)(Ccos cpt + Dsin cpt) (c) y = (Ax + B)(Ct + D) (d) y = (Aepx + Be-px)(Cep*t + De-ep*r) O a O b O c O dThe motion of a point on the circumference of a rolling wheel of radius 3 feet is described by the vector function 7(t) = 3(22t – sin(22t))i + 3(1 – cos(22t))} Find the velocity vector of the point. v(t) = (66 – 66 cos(22t) )i + 66 sin( 22t)jv Find the acceleration vector of the point. a(t) = Find the speed of the point. s(t) = 66y cos( 22t) + sin( 22t) x
- What are the coordinates of A' after a reflection over the x-axis? 8particle moves along a path with velocity (t) = sin(t) i + t³j + etk. ts) Find its acceleration. Suppose the particle's initial position is P(2,1,-1). Find the particle'sWrite parametric equations for a(t) at (1,1) a(t) = <1, t, 1/(e+t^3)>
- Caculate both Parts a) y''-2y'-3y=x+1 b) y''+4y=sin3x6. (a) Find the directional derivative of w = x²y² at the point (1, -3) in the direc- 5T 5T tion of the unit vector u = cos i+sinj. (b) What is the maximum value of the directional derivative of w = ²y at (1,-3) and it what direction is it attained? Q Search ASketch the curve whose vector equation is Solution r(t) = 6 cos(t) i + 6 sin(t) j + 3tk. The parametric equations for this curve are X = I y = 6 sin(t), z = Since x² + y² = + 36. sin²(t) = The point (x, y, z) lies directly above the point (x, y, 0), which moves counterclockwise around the circle x² + y2 = in the xy-plane. (The projection of the curve onto the xy-plane has vector equation r(t) = (6 cos(t), 6 sin(t), 0). See this example.) Since z = 3t, the curve spirals upward around the cylinder as t increases. The curve, shown in the figure below, is called a helix. ZA (6, 0, 0) (0, 6, 37) I the curve must lie on the circular cylinder x² + y² =