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

Question

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

Emily Ammerman · Accepted Answer

#abs(v)=2,047# 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#

Ezekiel Alexander · Answer

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

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"#

Aurora Alvarado · Answer

undefined To find the instantaneous velocity of an object moving according to the function $ f(t) = (\sin(2t)\cos(t), \tan(t) - \sec(t)) $ at $ t = \frac{13\pi}{12} $, we need to find the derivative of the position function with respect to time, and then evaluate it at $ t = \frac{13\pi}{12} $.

Let's find the derivative:

1. Find the derivative of $ f(t) $ with respect to $ t $ for each component.

$$ \frac{d}{dt}(\sin(2t)\cos(t)) = (2\cos(2t)\cos(t) - \sin(2t)\sin(t)) $$

$$ \frac{d}{dt}(\tan(t) - \sec(t)) = (\sec^2(t) - \sec(t)\tan(t)) $$

2. Evaluate the derivatives at $ t = \frac{13\pi}{12} $:

$$ \frac{d}{dt}(\sin(2t)\cos(t))\Bigg|_{t=\frac{13\pi}{12}} $$

$$ = (2\cos(2 \cdot \frac{13\pi}{12})\cos(\frac{13\pi}{12}) - \sin(2 \cdot \frac{13\pi}{12})\sin(\frac{13\pi}{12})) $$

$$ \frac{d}{dt}(\tan(t) - \sec(t))\Bigg|_{t=\frac{13\pi}{12}} $$

$$ = (\sec^2(\frac{13\pi}{12}) - \sec(\frac{13\pi}{12})\tan(\frac{13\pi}{12})) $$

These expressions give us the instantaneous velocity vector of the object at $ t = \frac{13\pi}{12} $.