Have Trouble on Problem 5 Write the process of finding the length as a matrix op

Question

Have Trouble on Problem 5

Write the process of finding the length as a matrix operation. e) Find v as a linear combination of the original basis. f) Verify the length you previously calculated is correct using the Pythagorean theorem. The motion of a particle is given by the position vector: r(t) = At cos(t)i + Bt sin(t)j + C squareroot t^2 + 1k a) The particle is constrained to move on a surface in 3D space, what is the equation of the surface? (Eliminate the parameter t) b) What is the velocity of the particle at t = 0? c) What is the first non-zero time where r and v are perpendicular? Set A = 1, B = 2, and C = 3. d) Is this problem easier in a different coordinate system? What if A = B? The motion of a particle is described in cylindrical coordinates as: r = e^-4z

Explanation / Answer

Assuming that we are to solve question No. 5 only ,I will solve the four parts of the question.

a). The given equation in spherical cordinates is

r(t) =At cos(t) i+Bt sin (t) j+C(1+t2)0.5 k

equating it it general representation of a vector as xi+yj+zk

we get x =At cost ,y=Bt sint and z2= (1+t2)

t.cost =x/A ,t sint =y/B and 1+t2=z2/C2

sqauring both sides of first two equations and equating their sum to thir equation we get

t2.cos2t =x2/A2 ,t2 sin2t =y2/B2

x2/A2+y2/B2=z2/C2-1

or, -x2/A2-y2/B2+z2/C2=1   (equation of the required surface)

b). Velocity of the particle at any time t is given by dr/dt where r is the position vector in any time t

thus V=dr(t)/dt =d/dt(At cos(t) i+Bt sin (t) j+C(1+t2)0.5 k)

=d/dt(At cos(t) i) +d/dt (Bt sin (t) j +d/dt(C(1+t2)0.5 k)

=A(cost-tsint)i+B(sint+tcost)j+Ct*1/(1+t2)1/2 k

Thus the velcoity v(t) is given by A(cost-tsint)i+B(sint+tcost)j+Ct*1/(1+t2)1/2 k

At t=0,

v(0) =A(1-0)i+B(0)j+0k

thus the velocity of the particle at t =0 is A in the x axis direction given by Ai

c). When r and v are orthogonal then their dot product will be equal to 0.

r.v=0

Finding the dot product of r and v we get

r.v ={A(cost-tsint)i+B(sint+tcost)j+Ct*1/(1+t2)1/2 k} .{At cos(t) i+Bt sin (t) j+C(1+t2)0.5 k}

=>A2t(cos2t-tsint cost) +B2t.(sin2t +tcost.sint)+C2t=0

t* {A2(cos2t-tsint cost)+B2(sin2t +tcost.sint)+C2 =0

Keeping A=1, B=2 and C=3 and solving for t we get

t~25 seconds (we ommit t=0 since non zero second is required)

d). The spherical ocrdinate system (the system in which the equation is provided) is one of the easiest frame to solve the equation as it involves orientation which is in one variable.

If A=B then it will be a hyperboloid in sheet 2.

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