Test review please help

Question

Answer 1

(17)

Volume flow rate, dV/dt = 4 cubic feet per minute

Radius, r = 2 feet

Volume of the spherical balloon, V = (4/3)(πr³ )

Apply derivative each side with respect to t, where t is time in minute .

dV/dt = (4/3)3(πr²)dr/dt

4 = (4/3)3(π(2²)dr/dt

4 = 4(π(4)dr/dt

1 = (π(4)dr/dt

1/(4π) = dr/dt

dr/dt = 1/4π

Rate of change in radius ⇒ dr/dt = 1/4π feet per minute .

Answer 2

(18)

The rate of change of side of a square is da/dt = 2 feet per minute .

Side length of square is a = 7 feet .

Area of the square is A =  a² .

Apply derivative each side with respect to t, where t is time in minute .

dA/dt = ( 2a ) da/dt

dA/dt = ( 2*7 ) 2

dA/dt = 28 

Hence , the rate of change area of square is dA/dt = 28 square feet per minute .

Answer 3

(19)

The rate of change in radius of a circle is dr/dt = - 4 cm/min .

Radius r = 5 cm

Area of a circle is A = πr² .

Apply derivative each side with respect to t, where t is time in minute .

dA/dt = ( 2πr) dr/dt 

dA/dt = ( 2π(5))(-4)

dA/dt = - 40π 

Hence , the area of a circle decreasing at rate of 40π .