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 sinθ
a = 2 cm
A = $\frac{1}{2}$ ( 2 )b sinθ
b = 3 cm
A = $\frac{1}{2}$ ( 2 )( 3 ) sinθ
A = 3 sinθ
$\frac{dA}{dt}$ = 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 sinθ
a = 2 cm
A = $\frac{1}{2}$ ( 2 )b sinθ
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 sinθ
$\frac{dA}{dt}$ = [ $\frac{1}{2}$ a ]( b )( sinθ )+[ b ]( $\frac{1}{2}$ a )( sinθ )+[ sinθ ][ θ ] ( $\frac{1}{2}$ a )( b )
$\frac{dA}{dt}$ = [ $\frac{1}{2}$ a ]( b sinθ )+[ b ]( $\frac{1}{2}$ a sinθ )+[ sinθ ][ θ ] ( $\frac{1}{2}$ ab )
$\frac{dA}{dt}$ = ( $\frac{1}{2}$ da/dt )( b sinθ )+( db/dt )( $\frac{1}{2}$ a sinθ )+( cosθ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ 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 sinθ )+( db/dt )( $\frac{1}{2}$ a sinθ )+( cosθ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab )
$\frac{dA}{dt}$ = ( $\frac{5}{4}$ )( b sinθ )+( db/dt )( $\frac{1}{2}$ a sinθ )+( cosθ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab )
$\frac{db}{dt}$ = 1.5 cm per minute = $\frac{3}{2}$
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ +( $\frac{3}{2}$ )( $\frac{1}{2}$ a sinθ )+( cosθ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab )
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ + $\frac{3}{4}$ a sinθ +( cosθ )( $\frac{dθ}{dt}$ )( $\frac{1}{2}$ ab )
$\frac{dθ}{dt}$ = 0.2 radians per minute = $\frac{1}{5}$
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ + $\frac{3}{4}$ a sinθ +( cosθ )( $\frac{1}{5}$ )( $\frac{1}{2}$ ab )
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ + $\frac{3}{4}$ a sinθ + $\frac{1}{10}$ ab cosθ
When a = 2 cm
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ + $\frac{3}{4}$ ( 2 ) sinθ + $\frac{1}{10}$ ( 2 )b cosθ
$\frac{dA}{dt}$ = $\frac{5}{4}$ b sinθ + $\frac{3}{2}$ sinθ + $\frac{1}{5}$ b cosθ
When b = 3 cm
$\frac{dA}{dt}$ = $\frac{5}{4}$ ( 3 ) sinθ + $\frac{3}{2}$ sinθ + $\frac{1}{5}$ ( 3 ) cosθ
$\frac{dA}{dt}$ = $\frac{15}{4}$ ( sinθ ) + $\frac{3}{2}$ ( sinθ ) + $\frac{3}{5}$ ( cosθ )
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.