Question

What is the instantaneous velocity of an object moving in accordance to `f(t)= (sin2tcost,tant-sect )` at `t=(13pi)/12`?

Answer 1

`abs(v)=2,047`

Explanation:

Let's derive the vectorial function `f(t)` and evaluate it in `t=13pi/12`.
we get
`vec(v)=f'(13pi/12)=(6cos^3(13pi/12)-4cos(13pi/12),1/sin(13pi/12))`
that is
`vec(v)=(-1,54;1,35)`
and its length is `abs(v)=2,047`

Answer 2

The instantaneous velocity `("speed", angle)` is `(-0.874, -0.718)`

Explanation:

Differentiate the equation, `y = tan(t) - sec(t)`, with respect to t:

`dy/dt = sec^2(t) - tan(t)sec(t)`

Use the product rule to differentiate the equation, `x = sin(2t)cos(t)`, with respect to t:

`dx/dt = 2cos(2t)cos(t) - sin(2t)sin(t)`

Solve the chain rule, `dy/dt = dy/dxdx/dt` for `dy/dx`

`dy/dx = (dy/dt)/(dx/dt)`

Substitute in the derivatives that we calculated:

`dy/dx = (sec^2(t) - tan(t)sec(t))/(2cos(2t)cos(t) - sin(2t)sin(t))`

The speed will be the above evaluated at `t = (13pi)/12`

`s = -0.874`

The angle with respect to horizontal is the inverse tangent of the speed: `theta = tan^-1(-0.874) = -0.718" radians"`