BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 37RE
To determine

Tofind:how fast the distance between the boy and balloon increases at 3s later.

Expert Solution

Answer to Problem 37RE

It increases at 13ft/s .

Explanation of Solution

Given:

  dhdt=5ft/sec

  dxdt=15ft/sec

Concept used:

According to the Pythagoras theorem:

  s2=x2+h2 .

If d2ydx2>0. the concave will be open upward and local minima can be found.

If d2ydx2<0. the concave will open downward and local maxima can be found.

According to the Pythagoras theorem:

  s2=x2+h2 .

Calculation:

Let the figure of the question be:

  Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 4, Problem 37RE

According to the given data: dhdt=5ft/sec and dxdt=15ft/sec .

According to the Pythagoras theorem:

  s2=x2+h2 .

Differentiating with respect to t:

  2sdsdt=2xdxdt+2hdhdt .

  sdsdt=xdxdt+hdhdt .

  dsdt=1s(xdxdt+hdhdt) .

To find the length at 3 seconds:

  x=15fts3s=45ft .

  h=5fts.3s+45=60ft .

  s=452+602=5625=75ft .

Plug in known value for the moment in time:

  dsdt=1s(xdxdt+hdhdt)

  dsdt=175(45(15)+60(5)) .

  97575

  dsdt=13ft/s

Hence it increases at 13ft/s .

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