The radius of air bubble is increasing at 0.8cm/s. At what rate the volume is increasing when radius is 2cm?

2 Answers
Ratnaker Mehta
Feb 25, 2018

# 12.8pi (cm)^3/sec#.

Explanation:

Let, #v=#the volume, and #r=#the radius of the bubble.

#:. v=4/3pir^3#.

#:. d/dt(v)=d/dt(4/3pir^3), i.e., #

# (dv)/dt=4/3pi*d/dt(r^3)=4/3pi*d/(dr)(r^3)*(dr)/dt..."[Chain Rule]"#

#:. (dv)/dt=4/3pi*3r^2*(dr)/dt=4pir^2*(dr)/dt#.

Here, #r=2cm, and #

#(dr)/dt="the rate of increasing of the radius"=0.8(cm)/sec#,

#:."The Rate at which the Volume "v" increases="(dv)/dt,#

#=4pi(2)^2(0.8)#,

#=12.8pi (cm)^3/sec#.

Parabola
Feb 25, 2018

#(dv)/dt=(12.8picm^3)/s#

Explanation:

There are two functions we consider:

#r(t)# radius in respect to time

#v(t)# volume in respect to time

Now, the speed in which radius increases in respect to time is #r'(t)# or #(dr)/dt#

Similarly, the speed in which volume increases in respect to time is #v'(t)# or #(dv)/dt#

We know that:

#(0.8cm)/s=(dr)/dt#

Air bubble we are talking about here is [most likely] a sphere.

The volume of a sphere is:

#v=4/3pir^3#

Let's apply #d/dt# to both sides.

#d/dt(v)=d/dt(4/3pir^3)# Bring the constants outside the derivative

#(dv)/dt=4/3pid/dt(r^3)# DO NOT use the power rule alone, for we are dealing with #t#.

Use the chain rule:

If #f(x)=g(h(x))#, then #f'(x)=g'(h(x))*h'(x)#

We have:

#(dv)/dt=4/3pi*[(dr)/dt3r^2]#

We know that #(0.8cm)/s=(dr)/dt#, so we can replace #(dr)/dt# with #(0.8cm)/s#

#(dv)/dt=4/cancel3pi*cancel3[(0.8cm)/s*r^2]#

Since we want the rate of change when the radius equals two centimeter, we can replace #r# with #2cm#

#(dv)/dt=4pi*[(0.8cm)/s*(2cm)^2]#

#(dv)/dt=4pi*[(0.8cm)/s*4cm^2]#

#(dv)/dt=4pi*[(0.8cm*4cm^2)/s]#

#(dv)/dt=4pi*[(3.2cm^3)/s]#

#(dv)/dt=(4pi*3.2cm^3)/s#

#(dv)/dt=(12.8picm^3)/s#

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