What is the instantaneous velocity of an object moving in accordance to # f(t)= (t-e^t,e^(2t)) # at # t=2 #?
Instantaneous velocity is given by the vector:
# vec(v) = << 1-e^2, 2e^4 >> # , or,#(1-e^2) hat(i)+2e^4hat(j)# , or,#( (1-e^2),(2e^4) )#
We have:
And so the instantaneous velocity is given by the vector:
If we want the instantaneous speed , it is given by:
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To find the instantaneous velocity of an object moving according to the function ( f(t) = (t - e^t, e^{2t}) ) at ( t = 2 ), we need to find the derivative of the function with respect to time ( t ), which represents the velocity vector. Then, we substitute ( t = 2 ) into the derivative to obtain the instantaneous velocity at that specific time.
The derivative of ( f(t) ) with respect to ( t ) is given by:
[ f'(t) = \left(\frac{d}{dt}(t - e^t), \frac{d}{dt}(e^{2t})\right) ]
[ f'(t) = \left(1 - e^t, 2e^{2t}\right) ]
Substituting ( t = 2 ) into ( f'(t) ) gives:
[ f'(2) = \left(1 - e^2, 2e^{4}\right) ]
So, the instantaneous velocity of the object at ( t = 2 ) is ( (1 - e^2, 2e^{4}) ).
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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.
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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