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What is the instantaneous velocity of an object moving in accordance to # f(t)= (e^(t^2),2t-e^t) # at # t=-1 #?

Answer 1

Aurora Alvarado

Find the velocity vector by taking the derivatives of the #x# and #y# portions separately:

#f'(t)=(2te^(t^2),2-e^t)#

So at #t=-1# the velocity vector is:

#f'(-1)=(-2e^((-1)^2),2-e^-1)=(-2e,2-1/e)#

We could find the speed at this point by finding the magnitude of the vector:

#"speed"=sqrt((-2e)^2+(2-1/e)^2)#

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Answer 2

Scarlett Allison

To find the instantaneous velocity of the object at ( t = -1 ), we need to differentiate the function ( f(t) ) with respect to time ( t ) and then evaluate it at ( t = -1 ).

Given ( f(t) = (e^{t^2}, 2t - e^t) ), we differentiate each component separately.

( \frac{df}{dt} = (\frac{d}{dt}(e^{t^2}), \frac{d}{dt}(2t - e^t)) )

( = (2te^{t^2}, 2 - e^t) )

Now, substitute ( t = -1 ) into the derivative to find the instantaneous velocity at that point.

( \frac{df}{dt}(-1) = (2(-1)e^{(-1)^2}, 2 - e^{-1}) )

( = (-2e^{-1}, 2 - \frac{1}{e}) )

So, the instantaneous velocity of the object at ( t = -1 ) is ( (-2e^{-1}, 2 - \frac{1}{e}) ).

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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.