Question

to melt a snowball, with initial radius of 12 cm, its radius decreases at a constant rate of 0.5 cm / hr. if it starts to melt when t = 0, the rate of decline in the volume of the snowball after 12 hours is:

the answer is : A. 50pi cm3/hr B. 60 pi .. C. 72 pi .. D. 120pi .. E. 144pi ..

Answer 1

\[v=\frac{ 4 }{ 3 }*R ^{3}\]

Answer 2

\[V=\frac{4}{3}\pi r^3\] if i remember correctly
taking derivatives you get
\[V'=4\pi r^2r'\]

Answer 3

you are told \(r'=-0.5\) and you need \(V'\)

Answer 4

oh and i guess after 12 hours \(r=6\) so make the replacements and you are done

Answer 5

why 6? can you explain me

Answer 6

is 4pi(36)*0.5 = 72

Answer 7

why not 4pi(0.5*12)^2=144

Answer 8

\(r=6\) cm because it starts at 12 cm and decreases at a rate of 0.5 cm per hour

Answer 9

do you get this, OP?

Answer 10

is 4pi(36)*0.5 = 72

why not 4pi(0.5*12)^2=144