# What is the instantaneous velocity of an object moving in accordance to # f(t)= (t^2,tcos(t-(5pi)/4)) # at # t=(pi)/3 #?

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To find the instantaneous velocity of an object moving according to ( f(t) = (t^2, t \cos(t - \frac{5\pi}{4})) ) at ( t = \frac{\pi}{3} ), we differentiate each component of the position function with respect to time ( t ) to get the velocity vector.

So, the velocity vector ( \mathbf{v} ) at ( t = \frac{\pi}{3} ) is:

[ \mathbf{v} = \left( \frac{d}{dt}(t^2), \frac{d}{dt}(t \cos(t - \frac{5\pi}{4})) \right) ]

Differentiating ( t^2 ) with respect to ( t ) gives ( 2t ).

Differentiating ( t \cos(t - \frac{5\pi}{4}) ) with respect to ( t ) involves the product rule and chain rule:

[ \frac{d}{dt}(t \cos(t - \frac{5\pi}{4})) = \cos(t - \frac{5\pi}{4}) - t \sin(t - \frac{5\pi}{4}) ]

Now, substitute ( t = \frac{\pi}{3} ) into each expression to find the velocity vector at ( t = \frac{\pi}{3} ).

So, ( \mathbf{v} = (2(\frac{\pi}{3}), \cos(\frac{\pi}{3} - \frac{5\pi}{4}) - \frac{\pi}{3} \sin(\frac{\pi}{3} - \frac{5\pi}{4})) ).

Now, simplify each component to get the velocity vector.

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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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