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

Answer 1

Please read below.

We have:

#x=e^sqrtt# #y=1/t+2#

We manipulate the equations a bit.

#=>ln(x)=sqrtt#
#=>(ln(x))^2=t#
#=>y-2=1/t#
#=>1/(y-2)=t#

We now have:

#(ln(x))^2=1/(y-2)#
#=>1/(ln(x))^2=y-2#
#=>(ln(x))^-2+2=y#
#=>d/dx[(ln(x))^-2+2]=d/dx[y]#
#=>d/dx[(ln(x))^-2]+d/dx[2]=dy/dx#

Power rule:

#d/dx[x^n]=nx^(n-1)#
#d/dx[ln(x)]=1/x#

Chain rule:

#d/dx[f(g(x))]=f'(g(x))*g'(x)#
#=>-2(ln(x))^(-2-1)*d/dx[ln(x)]+0*2*x^(0-1)=dy/dx#
#=>-2(ln(x))^(-3)*1/x+0=dy/dx#
#=>-2/(ln(x))^(3)*1/x=dy/dx#
#=>-2/(x(ln(x))^(3))=dy/dx#
Substitute by using the fact that #e^sqrtt=x#
#=>-2/(e^sqrtt(ln(e^sqrtt))^(3))=dy/dx#
#=>-2/(e^sqrtt(t^(1/2))^(3))=dy/dx#
#=>-2/(e^sqrtt(t^(3/2)))=dy/dx#
Replace #t# with #1#.
#=>-2/(e^sqrt1(1^(3/2)))=f'(1)#
#=>-2/(e)=f'(1)#
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Answer 2

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

Given ( f(t) = (e^{\sqrt{t}}, \frac{1}{t} + 2) ), the velocity vector is ( \frac{df}{dt} = \left(\frac{d}{dt} e^{\sqrt{t}}, \frac{d}{dt} \left(\frac{1}{t} + 2\right)\right) ).

Differentiating each component separately, we get:

( \frac{d}{dt} e^{\sqrt{t}} = \frac{1}{2\sqrt{t}}e^{\sqrt{t}} )

( \frac{d}{dt} \left(\frac{1}{t} + 2\right) = -\frac{1}{t^2} )

At ( t = 1 ), we have:

( \frac{d}{dt} e^{\sqrt{1}} = \frac{1}{2}e )

( \frac{d}{dt} \left(\frac{1}{1} + 2\right) = -1 )

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

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Answer 3

The instantaneous velocity of an object moving according to the parametric function ( f(t) = (e^{\sqrt{t}}, \frac{1}{t} + 2) ) at ( t = 1 ) can be found by taking the derivative of each component of the function with respect to time ( t ), and then evaluating them at ( t = 1 ).

The derivative of ( e^{\sqrt{t}} ) with respect to ( t ) is ( \frac{d}{dt}(e^{\sqrt{t}}) = \frac{1}{2\sqrt{t}} e^{\sqrt{t}} ).

The derivative of ( \frac{1}{t} + 2 ) with respect to ( t ) is ( \frac{d}{dt}(\frac{1}{t} + 2) = -\frac{1}{t^2} ).

Evaluating these derivatives at ( t = 1 ), we get:

[ \frac{1}{2\sqrt{1}} e^{\sqrt{1}} = \frac{1}{2} e ]

[ -\frac{1}{1^2} = -1 ]

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

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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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