Answer to Question #97731 in Algebra for Idahosa

Question

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 ?

Answer 1

Curve given is

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

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

$z=8t$

$v_x=dx/dt=2.cos(a^{3t}).a^{3t}\ln a.d(3t)/dt$

$v_x​=2.cos(a^{3t}).a^{3t}.\ln a.3$

$v_x=6.cos(a^ {3t} ).a^{3t}.\ln a$ -------(i)

$v_y=dy/dt=-2.sin(a^{3t}).a^{3t}\ln a.d(3t)/dt$

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

$v_y=-6.sin(a^{3t}).a^{3t}.\ln a$ ---------(ii)

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

$v = v_x \hat{i}+v_y\hat{j}+v_z\hat{k}$

$|v| = \sqrt{v_x^2 + v_y^2+v_z^2}$

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

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

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