cssftw
Member
- Joined
- Jun 19, 2009
- Messages
- 207
- Gender
- Male
- HSC
- 2011
Hi guys, I'm having a bit of a problem with this question:
Q. A sphere is expanding so that its surface area is increasing at a rate of 0.04 cm^2 /second. Find the rate of increase of its volume at the instant when the radius is 10cm.
My solution: (please someone figure out where I went wrong)
A= surface area of sphere
dA/dt = 0.04 cm^2 / second --> this is constant
we are trying to find dV/dt
dV/dt = dA/dt * dV/dA
(here's where I think I might have gone wrong in):
V = (4/3)(pi)(r^3)
A = (4)(pi)(r^2)
therefore: 3V = (A)(r)
V = Ar/3 --> r=10, therefore V = 10A/3
therefore dV/dA = 10/3
dV/dt = dA/dt * dV/dA
dV/dt = 0.04 * 10/3
dv/dt = 2/15 cm^3 / second <--- final answer
However the answers the rate of change was 0.2 cm^3 / second.
So what exactly did I do wrong, and could someone please post a proper solution, and explain WHY I was wrong? Thanks for the help
Q. A sphere is expanding so that its surface area is increasing at a rate of 0.04 cm^2 /second. Find the rate of increase of its volume at the instant when the radius is 10cm.
My solution: (please someone figure out where I went wrong)
A= surface area of sphere
dA/dt = 0.04 cm^2 / second --> this is constant
we are trying to find dV/dt
dV/dt = dA/dt * dV/dA
(here's where I think I might have gone wrong in):
V = (4/3)(pi)(r^3)
A = (4)(pi)(r^2)
therefore: 3V = (A)(r)
V = Ar/3 --> r=10, therefore V = 10A/3
therefore dV/dA = 10/3
dV/dt = dA/dt * dV/dA
dV/dt = 0.04 * 10/3
dv/dt = 2/15 cm^3 / second <--- final answer
However the answers the rate of change was 0.2 cm^3 / second.
So what exactly did I do wrong, and could someone please post a proper solution, and explain WHY I was wrong? Thanks for the help