Question

In a later chapter we will be able to show, under certain assumptions, that the velocity v(t) of a falling raindrop at time t is v(t) = vT(1 − e−gt/vT) where g is the acceleration due to gravity and vT is the terminal velocity of the raindrop. (a) Find lim t→[infinity] v(t).

Answer 1

Answer:

$\lim_{t \to \infty} v(t) =vT$

Explanation:

Using distributive propierty:

$v(t)=vT(1-\frac{e^{-gt} }{vT} )=vT-e^{-gt}$

So:

$\lim_{t\to \infty} vT-e^{-gt}$

The limit of the sum of two functions is equal to the sum of their limits, therefore:

$\lim_{t\to \infty} vT-e^{-gt} = \lim_{t\to \infty} vT - \lim_{t\to \infty} e^{-gt}$

The limit of a constant function is the constant, hence:

$\lim_{t\to \infty} vT=vT$

Now, let's solve the other limit:

$\lim_{t\to \infty} e^{-gt}=e^{ \lim_{t \to \infty} -gt}$

The limit of a constant times a function is equal to the product of the constant and the limit of the function, so:

$\lim_{t \to \infty} -gt}=-g\lim_{t \to \infty} t}=-g(\infty)$

$-g(\infty)=-\infty$

Therefore:

$e^{(-\infty)} =0$

Finally:

$\lim_{t\to \infty} vT-e^{-gt}=vT-0=vT$