pleaseee help!!. Calculus!!!...

Question

A large sphere of material is placed in a dissolving liquid.

(a) When the radius is 1 m, it is observed to be changing at 7 cm/hour. At that instant, what is the rate at which the sphere's volume is being dissolved with respect to time?

 

___ cm³/hr

(b) Suppose, instead, that the liquid is known to dissolve the material at a rate of 1 m³/hour, at what rate will its radius r change with respect to time when r = 1 meter? when r = 1/2 meter?

r = 1			m/hr
r = 1/2			m/hr

(c) What happens to the rate of change of the radius in part (b) as r → 0⁺? r → ∞?

r → 0+
r → ∞

Answer 1

a) v= 4pir^3/3= 4pi/3 m^3

dr/dt= 7cm/hr= .07 cm/hr

r=1 m=100cm

dv/dt= 4r^2pidr/dt= 4(100^2)pi(7)

=280,000pi cm^3/hr= .28pi m^3/hr

b) dv/dt= 1 m^3/hr= 4pir^2dr/dt= 4pi(1)dr/dt

dr/dt=.25/pi m/hr= 25/pi cm/hr

Answer 2

Here are the 3 keys to the solution

V=(4/3)πR³, (1)

dV/dt=4πR²dR/dt (2)

Also :use R=(3V/4π)^1/3 (3) when you'd need to express things as a function volume V from results you'd get as functions of R (like e.g in (2). In opposite case (switching from V-,dependence to R-dependence) use (1)

Best,

Dr.A.B.