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

Answer 1

#v(pi/3)=1/3sqrt(4pi^2+9cos^2(pi/12)+pisin^2(pi/12)+6picos(pi/12)sin(pi/12))#

The equation #f(t)=(t^2;tcos(t-(5pi)/4))# gives you the object's coordinates with respect to time:
#x(t)=t^2# #y(t)=tcos(t-(5pi)/4)#
To find #v(t)# you need to find #v_x(t)# and #v_y(t)#
#v_x(t)=(dx(t))/dt=(dt^2)/dt=2t#
#v_y(t)=(d(tcos(t-(5pi)/4)))/dt=cos(t-(5pi)/4)-tsin(t-(5pi)/4)#
Now you need to replace #t# with #pi/3#
#v_x(pi/3)=(2pi)/3#
#v_y(pi/3)=cos(pi/3-(5pi)/4)-pi/3 cdot sin(pi/3-(5pi)/4)# #=cos((4pi-15pi)/12)-pi/3 cdot sin((4pi-15pi)/12)# #=cos((-11pi)/12)-pi/3 cdot sin((-11pi)/12)# #=cos(pi/12)+pi/3 cdot sin(pi/12)#
Knowing that #v^2=v_x^2+v_y^2# you find:
#v(pi/3)=sqrt(((2pi)/3)^2+(cos(pi/12)+pi/3 cdot sin(pi/12))^2)# #=sqrt((4pi^2)/9+cos^2(pi/12)+pi^2/9 cdot sin^2(pi/12)+(2pi)/3 cdot cos(pi/12)sin(pi/12))# #=1/3sqrt(4pi^2+9cos^2(pi/12)+pisin^2(pi/12)+6picos(pi/12)sin(pi/12))#
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Answer 2

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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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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