What is the speed for the motion of a particle given by the parametric equations x=2sin(t^3) and y=3cos(t^3)?

Answer 1

the qn has given u the speed of the particle (pte) only tt it is in vector form.
hence any vector comprises of its x & y components. (& z, if in 3D.)

given: x = 2sin(t^3), y = 3cos(t^3), where t = time passed
thus, speed, say v = square root of [ (x^2) + (y^2) ]

**recall [sin(x)]^2 + [cos(x)]^2 =1

therefore, v = sqrt [ (x^2) + (y^2) ]
= sqrt { [2sin(t^3)]^2 + [3cos(t^3)]^2 }
= sqrt { 4[sin(t^3)]^2 + 9(1 - [sin(t^3)]^2) } **
= sqrt { 9 - 5[sin(t^3)]^2 }

sub in ur value(s) of "t" & u will obtain ur respective speed at that moment of time. (or velocity, depending on direction)

Answer 2

The velocity vector is x´= 2 cos t^3 *3 t^2 = 6t^2cost^3
y´= -9t^2 sin t^3 and the speed is the modulos=
= sqrt( t^4(36cos^2 +81sin^2) = t^2sqrt(36+45sin^2t^3)