What is the instantaneous velocity of an object moving in accordance to # f(t)= (sin^2t,cos3t) # at # t=pi/12 #?

Answer 1

The instantaneous velocity is #=(0.5, -2.12)#

#(sint)'=cost#
#(cost)'=-sint#

The velocity is the derivative of the position

#f(t)=(sin^2t,cos3t)#
#v(t)=f'(t)=(2sintcost, -3sin3t)#
Therefore, when #t=1/12pi#
#v(1/12pi)=(2sin(1/12pi)cos(1/12pi), -3sin(3/12pi))#
#=(0.5, -2.12)#
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Answer 2

To find the instantaneous velocity of an object moving according to ( f(t) = (\sin^2 t, \cos 3t) ) at ( t = \frac{\pi}{12} ), we need to find the derivative of the function ( f(t) ) with respect to ( t ). Then, we can evaluate this derivative at ( t = \frac{\pi}{12} ) to find the instantaneous velocity vector.

The derivative of ( f(t) ) with respect to ( t ) is given by:

[ f'(t) = \left( \frac{d}{dt} (\sin^2 t), \frac{d}{dt} (\cos 3t) \right) ]

Using the chain rule and the derivative of sine and cosine functions, we get:

[ f'(t) = (2\sin t \cdot \cos t, -3\sin 3t) ]

Now, we evaluate this derivative at ( t = \frac{\pi}{12} ):

[ f'\left(\frac{\pi}{12}\right) = \left(2\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right), -3\sin \left(\frac{3\pi}{12}\right)\right) ]

[ f'\left(\frac{\pi}{12}\right) = \left(2\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right), -3\sin \left(\frac{\pi}{4}\right)\right) ]

[ f'\left(\frac{\pi}{12}\right) = \left(2\sin \left(\frac{\pi}{12}\right) \cdot \cos \left(\frac{\pi}{12}\right), -3\sin \left(\frac{\pi}{4}\right)\right) ]

Using trigonometric identities:

[ \sin\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4}, \quad \cos\left(\frac{\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} ]

[ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} ]

Substitute these values into the expression for ( f'\left(\frac{\pi}{12}\right) ):

[ f'\left(\frac{\pi}{12}\right) = \left(2 \cdot \frac{\sqrt{6} - \sqrt{2}}{4} \cdot \frac{\sqrt{6} + \sqrt{2}}{4}, -3 \cdot \frac{\sqrt{2}}{2}\right) ]

[ f'\left(\frac{\pi}{12}\right) = \left(\frac{1}{2}, -\frac{3\sqrt{2}}{2}\right) ]

Therefore, the instantaneous velocity of the object at ( t = \frac{\pi}{12} ) is ( \left(\frac{1}{2}, -\frac{3\sqrt{2}}{2}\right) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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