Question #97731
The magnitude of the velocity of a particle which moves along the curve x=2 sin ??¡3t,y=2 cos ??¡3t and z=8t at any time t>0 is ?
1
Expert's answer
2019-11-11T10:39:35-0500

Curve given is

x=2sin(a3t)x=2sin(a^{3t})

y=2cos(a3t)y=2cos(a^{3t})

z=8tz=8t

vx=dx/dt=2.cos(a3t).a3tlna.d(3t)/dtv_x=dx/dt=2.cos(a^{3t}).a^{3t}\ln a.d(3t)/dt

vx=2.cos(a3t).a3t.lna.3v_x​=2.cos(a^{3t}).a^{3t}.\ln a.3

vx=6.cos(a3t).a3t.lnav_x=6.cos(a^ {3t} ).a^{3t}.\ln a -------(i)

vy=dy/dt=2.sin(a3t).a3tlna.d(3t)/dtv_y=dy/dt=-2.sin(a^{3t}).a^{3t}\ln a.d(3t)/dt

vy=2.sin(a3t).a3t.lna.3v_y=-2.sin(a^{3t}).a^{3t}.\ln a.3

vy=6.sin(a3t).a3t.lnav_y=-6.sin(a^{3t}).a^{3t}.\ln a ---------(ii)

vz=dz/dt=8v_z= dz/dt= 8 ------(iii)

v=vxi^+vyj^+vzk^v = v_x \hat{i}+v_y\hat{j}+v_z\hat{k}

v=vx2+vy2+vz2|v| = \sqrt{v_x^2 + v_y^2+v_z^2}

v=(6.cos(a3t).a3t.lna)2+(6.sin(a3t).a3t.lna)2+82|v|=\sqrt{(6.cos(a^ {3t} ).a^{3t}.\ln a)^2+(6.sin(a^ {3t} ).a^{3t}.\ln a)^2+ 8^2}

v=(6.a3t.lna)2(sin(a3t)2+cos(a3t)2)+82|v|=\sqrt{(6.a^{3t}.\ln a)^2(sin(a^{3t})^2+cos(a^{3t})^2)+8^2}

v=36.a6t(lna)2+64|v| =\sqrt{36.a^{6t}(\ln a)^2+64} (Answer)


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