Answer:
a) the radius of the balloon increases at a rate of 5.42 in/s
b) the surface area of the balloon increases at a rate of 3 in²/s
Step-by-step explanation:
a) since the volume of a sphere V is
V= 4/3*π*R³
where R= radius , then the rate of change of the volume is
V' = dV/dR= 4*π*R²
using the chain rule
dV/dt = dV/dR*dR/dt
thus
k = 4*π*R² * dR/dt
dR/dt = k/(4*π*R²)
replacing values
dR/dt = k/(4*π*R²) = (33 in³/s) /(4*π*(22 in)²] = 5.42 in/s
then the radius of the balloon increases at a rate of 5.42 in/s
b) since the surface area is
S=4*π*R²
then
S' = dS/dR= 8*π*R
and
dS/dt = dS/dR*dR/dt = 8*π*R * k/(4*π*R²) = 2*k/R
replacing values
dS/dt = 2*k/R = 2*(33 in³/s)/( 22 in) = 3 in²/s
then the surface area of the balloon increases at a rate of 3 in²/s
Noon to 2:00 pm is 2 hours
3 degrees to -1 degree is a total of 4 degrees.
The temperature changed 4 degrees in 2 hours.
4/2 = 2
The temperature dropped 2 degrees per hour.
The slope is -3/2
Use the two ordered pairs (-2,0) and (0,-3) and do y2-y1/x2-x1
-3-0/0+2= -3/2