Chapter 4.6, Problem 22E

To determine

To calculate: Distance when cycle $3sec$ later

Answer to Problem 22E

The distance between the bicycle and balloon increasing $3sec$ later is $11ft/sec$

Explanation of Solution

Given information:

Balloon vertically rising at $1ft/sec$ .

When Balloon $65ft$ above the ground at the time bicycle moving $17ft/sec$ .

Calculation:

  ${\left.\frac{ds}{dt}\right|}_{t=3}$

Where $t=0$ is the moment when the balloon is $65\text{ft}$ above the ground.

Using the Pythagorean Theorem

  $x^2(t)+y^2(t)=s^2(t)$

By differentiation

  $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2s\frac{ds}{dt}$

Simplifying (dividing by 2),

  $x\frac{dx}{dt}+y\frac{dy}{dt}=s\frac{ds}{dt}$   $\left(\ast \right)$

Given

  $\begin{array}{c}\frac{dx}{dt}=17\text{ft}/\text{sec}\\ \frac{dy}{dt}=1\text{ft}/\text{sec}\end{array}$

The formula is

  $\text{ Position }=\text{ Velocity }\times \text{ Time }$ if velocity is constant (like here).

So, at $t=3$, the positions of the bicycle and the balloon are:

Bicycle:

  $\begin{array}{c}x(3)=17\times 3\\ =51\end{array}$

Balloon:

  $\begin{array}{c}y(3)=65+(1\times 3)\\ =68\end{array}$

The distance $s$ , at $t=3$ is

  $\begin{array}{c}s(3)=\sqrt{{51}^2+{68}^2}\\ =85\end{array}$

Substituting all values into the equation $(*)$ ,

  $51\cdot 17+68\cdot 1={\left.85\cdot \frac{ds}{dt}\right|}_{t=3}$

Follows

  $\begin{array}{c}{\left.\frac{ds}{dt}\right|}_{t=3}=\frac{867+68}{85}\\ =11\text{ft}/\text{sec}\\ \end{array}$