The volume of a spherical hot air balloon expands as the air

The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon, in feet, is modeled by a twice-differentiable function r of time t, where t is measured in minutes. For 0

Solution

a)t=5.4

r(5)=30 ft

r\'(5)=2

by tangent line approximation

r(5.4)=r(5)+ r\'(5)*(5.4-5)

r(5.4)=30+ (2)*(0.4)

r(5.4)=30+0.8

r(5.4)=30.8 feet

estamate is less than the true value

because slope r\'(t) is decreasing at t =5

b)V=(4/3)r³

V\'(t)=4r²r\'(t)

V\'(5)=4*30²r\'(5)

V\'(5)=4*30²(2)

V\'(5)=7200 feet³/minute

c)right sum

_{[0 to 12]} r\'(t) dt (2*4)+(3*2)+(2*1.2)+(4*0.6)+(1*0.5)

_{[0 to 12]} r\'(t) dt 19.3

_{[0 to 12]} r\'(t) dt represents the change in radius of baloon in first 12 minutes in feet

d)right sum is less than _{[0 to 12]} r\'(t) dt because r\'(t) is decreasing