Chapter 1.4, Problem 38E

To determine

To Find: and interpret the sensitivity of the height to the initial velocity when the initial velocity is $40\text{ft/sec}$ .

Answer to Problem 38E

The sensitivity of the height to the initial velocity when the initial velocity is $40\text{ft/sec}$ , is $\frac{80}{64}$ ft/sec.

Since the change is positive and greater than 1, the dosage $\Delta v$ will increase velocity by approximately $\frac{80}{64}\Delta v$ ft/sec.

Explanation of Solution

Given information:

The function H as a function of velocity v is given by $H=\frac{v^2}{64}\text{feet}$ .

Concept used:

Instantaneous rate of change of any function f at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

For the sensitivity of the height to the initial velocity when the initial velocity is $40\text{ft/sec}$ , find instantaneous rate of change of $H=\frac{v^2}{64}\text{feet}$ at $v=40\text{ft/sec}$ .

Let $a=40$ .

  $\begin{array}{c}{H}^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{H\left(a+h\right)-H\left(a\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{H\left(40+h\right)-H\left(40\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\frac{{\left(40+h\right)}^2}{64}-\frac{{\left(40\right)}^2}{64}}{h}\\ =\underset{h\to 0}{\lim }\frac{{\left(40\right)}^2+h^2+2\left(40\right)h-{\left(40\right)}^2}{64h}\\ =\underset{h\to 0}{\lim }\frac{h^2+2\left(40\right)h}{64h}\\ =\underset{h\to 0}{\lim }\frac{\left(h+2\left(40\right)\right)}{64}\\ =\frac{80}{64}\end{array}$

Hence the sensitivity of the height to the initial velocity when the initial velocity is $40\text{ft/sec}$ , is $\frac{80}{64}$ ft/sec.

Since the change is positive and greater than 1, the dosage $\Delta v$ will increase velocity by approximately $\frac{80}{64}\Delta v$ ft/sec.