rate of change?

Question

As a spherical balloon is being inflated, its radius r (in cm) after t minutes is given by r = 3t^(1/3) for 0
a) Radius b) the volume V of the balloon c) the surface area S of the balloon

Answer 1

what is rate of change?? it is slope... since we have a function of t, we know if we differentiate the function with respect to t and solve the derivative at any t, we get rate of change at that value of t...

we know that r(t) = 3t^(1/3)

(a) radius
differentiate r with respect to t...
we get... r'(t) = 3(1/3)t^(-2/3)
now substitute t = 8...
r'(8) = 8^(-2/3) = 1/4 cm

(b) volume of the balloon
volume of a sphere = (4/3)*pi*r^3
r^3 = [3t^(1/3)]^3 = 27t
volume of the balloon = (4/3)*pi*27t = 36*pi*t

differentiate above fct with respect to t...
v'(t) = 36*pi
for any value of t, v'(t) remains same = 36*pi cm^3

(c) surface area
surface area of a sphere = 4*pi*r^2
r^2 = [3t^(1/3)]^2 = 9t^(2/3)
surface area of the balloon = 4*pi*9t^(2/3) = 36*pi*t^(2/3)

differentiate above fct with respect to t...
s'(t) = 36*pi*(2/3)*t^(-1/3)

solve for s'(8) = 24*pi*8^(-1/3) = 12*pi cm^2

Answer 2

As a spherical balloon is being inflated, its radius r (in cm) after t minutes is given by r = 3t^(1/3) for 0
First, we can find r = 3(8)^(1/3) = 6 cm

a) radius
dr/dt = t^(-2/3) = 8^(-2/3) = 1/4 cm/sec

b) the volume V of the balloon
dV/dt = d[(4/3)pi r^3]/dt = 4 pi r^2 dr/dt = 36 pi cm^3/sec

c) the surface area S of the balloon
dS/dt = d[4 pi r^2]/dt = 8 pi r dr/dt = 12 pi cm^2/sec

-------------------
raj,
You need to find the RATE of change wrt t. Also, r can be calculated from the given function r = 3t^(1/3).

Answer 3

dr/dt=(1/3)t^-2/3
when t=8
dr/dt=1/4 cmpm

2.V=(4/3)pir^3
dV/dt=dV/dr*dr/dt
=4pir^2*1/4
=pir^2

3.A=4pir^2
dA/dt=dA/dr*dr/dt
=8pi2(1/4)
=2pir

for 2 and 3 you need r