Chapter 5.6, Problem 22E

To determine

To find: the distance between the bicycle and balloon increasing 3 seconds later.

Answer to Problem 22E

The distance between the bicycle and balloon is $11\text{ feet/sec}$ .

Explanation of Solution

Given information:

From the given diagram in the question we can see that,

  $y=65\text{ feet}$ .

  $\frac{dy}{dt}=1\text{ feet/sec}$ .

  $\frac{dx}{dt}=17\text{ feet/sec}$

All variables are differentiable functions of t .

Calculation :

We have to calculate the distance between the bicycle and balloon increasing 3 seconds later.

First, we have to find the value of x .

x travels $17\text{ feet/sec}$ .

Therefore,

  $\begin{array}{l}x=\left(17\times 3\right)\text{feet}\\ x=51\text{ feet}\end{array}$ .

We have to find y will be after 3 seconds.

y travels 1 feet/sec. so add 3 feet to 65 feet.

  $\begin{array}{l}y=\left(3+6\right)\text{ feet}\\ y=68\text{ feet}\end{array}$

From Pythagoras theorem,

  $\begin{array}{l}{\text{S}}^2=x^2+y^2\\ \text{S}=\sqrt{x^2+y^2}\end{array}$

Putting the value of $x=51$ and $y=68$ in above equation,

  $\begin{array}{l}\text{S}=\sqrt{{\left(51\right)}^2+{\left(68\right)}^2}\\ \text{S}=\sqrt{2601+4624}\\ \text{S}=\sqrt{7225}\\ \text{S}=85\text{ feet}\end{array}$

Again,

From Pythagoras theorem,

  ${\text{S}}^2=x^2+y^2$

Differentiate with respect to t .

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

Putting the value of $\text{S}=85$ , $x=51$ , $\frac{dx}{dt}=17$ , $\frac{dy}{dt}=1$ and $y=68$ in above equation,

  $\begin{array}{l}2s\frac{ds}{dt}=2\left(51\right)\left(17\right)+2\left(68\right)\left(1\right)\\ 170\frac{ds}{dt}=1734+136\\ 170\frac{ds}{dt}=1870\\ \frac{ds}{dt}=\frac{1870}{170}\\ \frac{ds}{dt}=11\text{ feet/sec}\text{.}\end{array}$

Hence, the distance between the bicycle and balloon is $11\text{ feet/sec}$ .