Chapter 2.4, Problem 28E

To determine

To find: The speed of the climbing of rocket after $10\sec$ of lift-off.

Answer to Problem 28E

The speed of the climbing of rocket after $10\sec$ of lift-off is $60\text{ft}/\sec$.

Explanation of Solution

Given information:

At $t\sec$ after the lift-off, the height of a rocket is $3t^2\text{ft}$.

Calculation:

The speed of the climb of rocket after $t\sec$ is the instantaneous rate of change of height function with respect to time.

The formula for the instantaneous rate of change of height with respect to time at $t$ seconds is equal to $\underset{h\to 0}{\lim }\frac{f\left(t+h\right)-f\left(t\right)}{h}$.

The formula for the instantaneous rate of change of $y=3t^2\text{ft}$ at $t=10\sec$ is equal to $\underset{h\to 0}{\lim }\frac{f\left(10+h\right)-f\left(10\right)}{h}$.

Substitute $10+h$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=3t^2\\ f\left(10+h\right)=3{\left(10+h\right)}^2\\ f\left(10+h\right)=3\left(100+h^2+20h\right)\\ f\left(10+h\right)=300+3h^2+60h\end{array}$

Substitute $10$ for $t$ in the function.

  $\begin{array}{c}f\left(t\right)=3t^2\\ f\left(10\right)=3{\left(10\right)}^2\\ f\left(10\right)=3\left(100\right)\\ f\left(10\right)=300\end{array}$

Substitute $300+3h^2+60h$ for $f\left(10+h\right)$ and $300$ for $f\left(10\right)$ in the formula.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(10+h\right)-f\left(10\right)}{h}=\underset{h\to 0}{\lim }\frac{300+3h^2+60h-300}{h}\\ =\underset{h\to 0}{\lim }\frac{3h^2+60h}{h}\\ =\underset{h\to 0}{\lim }\frac{h\left(3h+60\right)}{h}\\ =\underset{h\to 0}{\lim }\left(3h+60\right)\end{array}$

Further simplify,

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(10+h\right)-f\left(10\right)}{h}=3\left(0\right)+60\\ =60\end{array}$

Therefore, the speed of the climbing of rocket after $10\sec$ of lift-off is $60\text{ft}/\sec$.