quick questions! (1 Viewer)

[shkspeare]

"The volume (V) and the surface area (S) of a sphere of radius r are given by V = (4/3)pir^3 and S = 4pir^2 respectively

a) Show that dV/dr = S and dV/dt=s (dr/dt) [DONE]

b) A spherical ball, of radius 24mm, is immersed in an acid bath and its volume decreases at a rate equal to three times its surface area at thet time. Find the time taken for

(i) the radius to be reduced to one eighth of its original size.
(ii) the volume to be reduced to one-eigth of its original size.

 

[Estel]

If you are having that much trouble with rates perhaps you should talk to your teacher about it...

Anyhow...
a)
dV/dr = (4/3)(pi)r^2.3
= 4(pi)r^2
= S
dV/dt = dV/dr.dr/dt
= S.dr/dt

EDIT: was incorrect.
I'll try again some other time if nobody does it.

 

[shkspeare]

he's been away...

nehooz...the answer is 7 and 4... ><

 

[Estel]

dV/dr = (4/3)(pi)r^2.3
= 4(pi)r^2
= S
dV/dt = dV/dr.dr/dt
= S.dr/dt

dV/dt = 3dA/dt
S.dr/dt = 3dA/dt
dr/dt = 3[dA/dr]/S
= [3.4(pi)r^2]/[4(pi)r^2]
= 3
Now r=24
time for r to reach r/8, i.e. 3, t=7 sec.

For v/8:
r^3 = [(1/2)24]^3
i.e. time for r to reach 12
From above, it will take 4 seconds.

 

[mojako]

Note:
"and its volume decreases at a rate equal to three times its surface area at thet time"
means
dV/dt = 3S, not
dV/dt = 3dA/dt

Anyway from my quick calculation it gives the same answer.