Answer:
[tex]\lim_{t \to \infty} v(t) =vT[/tex]
Explanation:
Using distributive propierty:
[tex]v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}[/tex]
So:
[tex]\lim_{t\to \infty} vT-e^{-gt}[/tex]
The limit of the sum of two functions is equal to the sum of their limits, therefore:
[tex]\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}[/tex]
The limit of a constant function is the constant, hence:
[tex]\lim_{t\to \infty} vT=vT[/tex]
Now, let's solve the other limit:
[tex]\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}[/tex]
The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:
[tex]\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)[/tex]
[tex]-g(\infty)=-\infty[/tex]
Therefore:
[tex]e^{(-\infty)} =0[/tex]
Finally:
[tex]\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT[/tex]
The velocity function for the drop of water will be [tex]Lim_{t- > oo}\ v(t)=vT[/tex]
The velocity of an object is the rate of change of its position with respect to a frame of reference
Using distributive propierty:
[tex]v(t0=vT(1-\dfrac{e^{-gt}}{vt})=vT-ew^{-gt}[/tex]
So:
[tex]lim_{t- > oo}\ vT-e^{-gt}[/tex]
The limit of the sum of two functions is equal to the sum of their limits, therefore:
[tex]lim_{t- > oo}\ vT-e^{-gt}=lim_{t- > oo}vT-lim_{t- > oo}\ e^{-gt}[/tex]
The limit of a constant function is the constant, hence:
[tex]lim_{t- > oo}\ vT =vT[/tex]
Now, let's solve the other limit:
[tex]lim_{t- > oo}\ e^{-gt}=e^{lim_{t- > oo\ -gt}[/tex]
The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:
[tex]-g(oo)=-oo[/tex]
Therefore:
[tex]e^{(-oo)}=0[/tex]
Finally:
[tex]Lim_{t- > oo}\ v(t)=vT[/tex]
Hence the velocity function for the drop of water will be [tex]Lim_{t- > oo}\ v(t)=vT[/tex]
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