Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.9 - Related Rates - 3.9 Exercises - Page 249: 8

Answer

8a) The area of the triangle is increasing by 0.3 cm² per minute 8b) The area of the triangle is increasing by 3.2 cm² per minute. 8c) The area of the triangle is increasing by 4.8 cm² per minute.

Work Step by Step

a) A = $\frac{1}{2}$ ab sin⁡θ a = 2 cm A = $\frac{1}{2}$ ( 2 )b sin⁡θ b = 3 cm A = $\frac{1}{2}$ ( 2 )( 3 ) sin⁡θ A = 3 sin⁡θ $\frac{dA}{dt}$ = 3 cos θ $\frac{dθ}{dt}$ θ = $\frac{π}{3}$ $\frac{dA}{dt}$ = 3 $\frac{1}{2}$ $\frac{dθ}{dt}$ $\frac{dθ}{dt}$ = 0.2 radians per minute, how fast is the area increasing ( $\frac{dA}{dt}$ ) when θ = $\frac{π}{3}$ $\frac{dA}{dt}$ = ( $\frac{3}{2}$ )( 0.2 ) = ( 1.5 )( 0.2 ) $\frac{dA}{dt}$ = 0.3 cm² 8a) The area of the triangle is increasing by 0.3 cm² per minute A = $\frac{1}{2}$ ab sin⁡θ a = 2 cm A = $\frac{1}{2}$ ( 2 )b sin⁡θ A = b sin θ $\frac{dA}{dt}$ = [ b ]( sin θ ) + [ sin θ ] [ θ ]( 2b ) [differentiate](term for multiplying) $\frac{dA}{dt}$ = ( $\frac{db}{dt}$ )( sin θ ) + ( cos θ )( $\frac{dθ}{dt}$ )( 2b ) $\frac{db}{dt}$ = 1.5 cm per minute $\frac{dA}{dt}$ = ( 1.5 )( sin θ ) + ( cos θ )( $\frac{dθ}{dt}$ )( 2b ) $\frac{dθ}{dt}$ = 0.2 radians per minute $\frac{dA}{dt}$ = ( 1.5 )( sin θ ) + ( cos θ )( 0.2 )( 2b ) When b = 3 cm $\frac{dA}{dt}$ = ( 1.5 )( sin θ ) + ( cos θ )( 0.2 )( 2 * 3 ) When θ = $\frac{π}{3}$ $\frac{dA}{dt}$ = ( 1.5 )( sin $\frac{π}{3}$ ) + ( cos $\frac{π}{3}$ )( 0.2 )( 2 * 3 ) $\frac{dA}{dt}$ = ( 1.5 )( $\frac{\sqrt 3}{2}$ ) + ( $\frac{1}{2}$ )( 0.2 )( 2 * 3 ) $\frac{dA}{dt}$ = ( $\frac{3}{2}$ )( $\frac{\sqrt 3}{2}$ ) + ( $\frac{1}{2}$ )( 0.2 )( 2 * 3 ) $\frac{dA}{dt}$ = $\frac{3\sqrt 3}{4}$ + ( $\frac{1}{2}$ )( $\frac{1}{5}$ )( 6 ) $\frac{dA}{dt}$ = $\frac{3\sqrt 3}{4}$ + $\frac{3}{5}$ = 0.75 $\sqrt 3$ + 0.6 ≈ 1.9 cm² per minute 8b) The area of the triangle is increasing by 1.9 cm² per minute. c) A = $\frac{1}{2}$ ab sin⁡θ $\frac{dA}{dt}$ = [ $\frac{1}{2}$ a ]( b )( sin⁡θ )+[ b ]( $\frac{1}{2}$ a )( sin⁡θ )+[ sin⁡θ ][ θ ] ( $\frac{1}{2}$ a )( b ) $\frac{dA}{dt}$ = [ $\frac{1}{2}$ a ]( b sin⁡θ )+[ b ]( $\frac{1}{2}$ a sin⁡θ )+[ sin⁡θ ][ θ ] ( $\frac{1}{2}$ ab ) $\frac{dA}{dt}$ = ( $\frac{1}{2}$ da/dt )( b sin⁡θ )+( db/dt )( $\frac{1}{2}$ a sin⁡θ )+( cos⁡θ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab ) $\frac{da}{dt}$ = 2.5 cm per minute = $\frac{5}{2}$ $\frac{5}{2}$ $\frac{dA}{dt}$ = ( $\frac{1}{2}$ * $\frac{5}{2}$ )( b sin⁡θ )+( db/dt )( $\frac{1}{2}$ a sin⁡θ )+( cos⁡θ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab ) $\frac{dA}{dt}$ = ( $\frac{5}{4}$ )( b sin⁡θ )+( db/dt )( $\frac{1}{2}$ a sin⁡θ )+( cos⁡θ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab ) $\frac{db}{dt}$ = 1.5 cm per minute = $\frac{3}{2}$ $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ +( $\frac{3}{2}$ )( $\frac{1}{2}$ a sin⁡θ )+( cos⁡θ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab ) $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ + $\frac{3}{4}$ a sin⁡θ +( cos⁡θ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab ) $\frac{dθ}{dt}$ = 0.2 radians per minute = $\frac{1}{5}$ $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ + $\frac{3}{4}$ a sin⁡θ +( cos⁡θ )( $\frac{1}{5}$ )( $\frac{1}{2}$ ab ) $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ + $\frac{3}{4}$ a sin⁡θ + $\frac{1}{10}$ ab cos⁡θ When a = 2 cm $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ + $\frac{3}{4}$ ( 2 ) sin⁡θ + $\frac{1}{10}$ ( 2 )b cos⁡θ $\frac{dA}{dt}$ = $\frac{5}{4}$ b sin⁡θ + $\frac{3}{2}$ sin⁡θ + $\frac{1}{5}$ b cos⁡θ When b = 3 cm $\frac{dA}{dt}$ = $\frac{5}{4}$ ( 3 ) sin⁡θ + $\frac{3}{2}$ sin⁡θ + $\frac{1}{5}$ ( 3 ) cos⁡θ $\frac{dA}{dt}$ = $\frac{15}{4}$ ( sin⁡θ ) + $\frac{3}{2}$ ( sin⁡θ ) + $\frac{3}{5}$ ( cos⁡θ ) When θ = $\frac{π}{3}$ $\frac{dA}{dt}$ = $\frac{15}{4}$ ( sin⁡ $\frac{π}{3}$ ) + $\frac{3}{2}$ ( sin⁡ $\frac{π}{3}$ ) + $\frac{3}{5}$ ( cos⁡ $\frac{π}{3}$ ) $\frac{dA}{dt}$ = $\frac{15}{4}$ ( $\frac{\sqrt 3}{2}$ ) + $\frac{3}{2}$ ( $\frac{\sqrt 3}{2}$ ) + $\frac{3}{5}$ ( $\frac{1}{2}$ ) $\frac{3\sqrt 3}{4}$ $\frac{dA}{dt}$ = $\frac{15\sqrt 3}{8}$ + $\frac{3\sqrt 3}{4}$ + $\frac{3}{10}$ $\frac{dA}{dt}$ = 1.875$\sqrt 3$ + 0.75$\sqrt 3$ + 0.3 $\frac{dA}{dt}$ = 2.625$\sqrt 3$ + 0.3 ≈ 4.8 cm² per minute. 8c) The area of the triangle is increasing by 4.8 cm² per minute.
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