Exemplo 5 de Precalculus, 12.4 Derivatives

Finding the Instantaneous Rate of Change

Problema e solução em inglês, como no livro original.

Using the function above, $s (t) = −16 t^{2} + 64 t + 6,$ what is the instantaneous velocity of the ball at 1 second and 3 seconds into its flight?

Resolva no papel primeiro. Depois abra a solução um passo de cada vez e pare assim que conseguir terminar sozinho.

Solução
The velocity at $t = 1$ and $t = 3$ is the instantaneous rate of change of distance per time, or velocity. Notice that the initial height is 6 feet. To find the instantaneous velocity, we find the derivative and evaluate it at $t = 1$ and $t = 3 :$
$\begin{array}{l} f^{'} (a) = \lim_{h \to 0} \frac{f (a + h) - f (a)}{h} \\ = \lim_{h \to 0} \frac{- 16 {(t + h)}^{2} + 64 (t + h) + 6 - (- 16 t^{2} + 64 t + 6)}{h} & Substitute s (t + h) and s (t) . \\ = \lim_{h \to 0} \frac{- 16 t^{2} - 32 h t - h^{2} + 64 t + 64 h + 6 + 16 t^{2} - 64 t - 6}{h} & Distribute . \\ = \lim_{h \to 0} \frac{- 32 h t - h^{2} + 64 h}{h} & Simplify . \\ = \lim_{h \to 0} \frac{h (- 32 t - h + 64)}{h} & Factor the numerator . \\ = \lim_{h \to 0} - 32 t - h + 64 & Cancel out the common factor h . \\ s^{'} (t) = - 32 t + 64 & Evaluate the limit by letting h = 0. \end{array}$
For any value of $t$ , $s^{'} (t)$ tells us the velocity at that value of $t .$

Evaluate $t = 1$ and $t = 3.$

$\begin{array}{l} s^{'} (1) = −32 (1) + 64 = 32 \\ s^{'} (3) = −32 (3) + 64 = −32 \end{array}$
The velocity of the ball after 1 second is 32 feet per second, as it is on the way up.

The velocity of the ball after 3 seconds is $−32$ feet per second, as it is on the way down.

Sua vez

O Try It do livro logo depois deste exemplo: mesma ideia, outros números. Só a resposta é dada.

The position of the ball is given by $s (t) = −16 t^{2} + 64 t + 6.$ What is its velocity 2 seconds into flight?

Como foi?