Question

The radius of a spherical watermelon is growing at a constant rate of 2 centimeters per week. The thickness of the rind is always one tenth of the radius. The volume of the rind is growing at the rate _________________ cubic centimeters per week at the end of the fifth week. Assume that the radius is initially zero.

Answer 1

first, what is r after 5 weeks ?

Answer 2

10

Answer 3

since the rate is 2. therefore you have to do 2*5

Answer 4

yes.
next use V = 4pi/3 r^3
and find dV/dt as a function of r and dr/dt

Answer 5

dV/dt = 4pir^2(dr/dt) i'm not sure if that's right

Answer 6

yes, that is correct.
dr/dt is given as a constant 2 cm/week
so
\[ \dot{V} = 4 \pi r^2 \cdot 2\]
now we need a value for r (of the "flesh" part, which excludes the rind)
the rind is 1/10 of r, which means the flesh has a radius of 0.9r
use that :
\[ \dot{V} =8 \pi (0.9 r)^2 8\]
where r=10 is the entire radius

Answer 7

i get the first formula but not the second. how did you get the second 8?

Answer 8

ok so when you differentiate V˙=4πr^2⋅2 you get 16pir right? so how did you get your formula?

Answer 9

sorry, the second 8 is a typo

Answer 10

\[ \dot{V} =8 \pi (0.9 r)^2 \]

Answer 11

oh ok so what do i have to do next again?

Answer 12

using your formula
dV/dt = 4pir^2(dr/dt)
with dr/dt =2
r = 0.9r (to adjust because we want to exclude the rind)
and r=10
we get
dV/dt = 4*pi * (0.9*10)^2 * 2
now multiply

Answer 13

i got 2035.75204

Answer 14

that should be the answer in cubic cm per week

Answer 15

oh ok. idk if my friend's answer's right, but she got 681.09 cm^3. i also googled the question and the person 680.75 cm^3/week but then again i don't know if they're right

Answer 16

do you have a link to the 680.75 answer ?

Answer 17

yeah i actually found three places with similar answers https://answers.yahoo.com/question/index?qid=20080922135603AAjyLZ0

http://openstudy.com/study#/updates/4eb437fce4b095a5c85ff7f0

https://answers.yahoo.com/question/index?qid=20090606185848AAJWzvc

Answer 18

ok, I was doing the volume of the flesh.
so we should redo the problem
V of the rind is the total volume - volume of the inside (flesh)
\[ V= \frac{4}{3} \pi r^3 - \frac{4}{3} \pi (0.9r)^3 \]
differentiate that

Answer 19

we can first simplify that a bit:
\[ V = \frac{4\pi}{3}(1-0.729) r^3 \\
V = \frac{4\pi}{3}\cdot 0.271\ r^3
\]

Answer 20

how did you get this part? (1−0.729)

Answer 21

rewrite (0.9r)^3 as 0.9^3 r^3 = 0.729 r^3
then factor out 4pi/3 r^3

Answer 22

you should get 681.097 which rounds to 681.10

Answer 23

ok i got that. thank you!

Answer 24

sorry for the long detour!

Answer 25

no it's okkk. don't worry :)

Answer 26

Refer to the attached Mathematica 9 calculation.

Answer 27

@robtobey Your derivative of V is with respect to r
you must adjust it by multiplying by dr/dt (apply the chain rule)
to find dV/dt (i.e. volume change per unit time (a week in this case) )