Question

A spherical snoowball is melting at a rate proportional to its surface area. That iis, the rate at which its volume is decreasing at any instant is proportional to its surface area at that rate.
Prove that the radius of the snowball is decreasing at a constant rate.

Answer 1

what formulas do you spose we should use for this?

Answer 2

What formulas can you use?

Answer 3

differentiation?

Answer 4

my first thought is to try to recall the formulas for volume and surface area of a sphere

Answer 5

differentiation is a method, which we will most likely use as well

Answer 6

oksy, so consider volume of a sphere= 4/3 pi r^3 ?

Answer 7

oh no

Answer 8

surface area

Answer 9

good,

\[V = \frac 43 \pi~r^3\]

now we take the derivatve of that:

\[V' = \frac {4(3)}{3} \pi~r^2~r'\]

Answer 10

so we're not using pi r^2 ?

Answer 11

why 43?

Answer 12

oh sorry it's a fraction

Answer 13

im thinking we will; now that we have the V' we are told that V' is equal to the formula for surface area (if im reading this right)

4(3) is just the result of taking an implicit derivative. the 3s cancels out and we are left with the formula for surface area, times r'

Answer 14

\[V' = 4\pi r^2~r'\]
we are told:
\[V' = 4\pi~r^2\]
so sub it in
\[4\pi~r^2 = 4\pi r^2~r'\]
now solve for r'

Answer 15

is the latex coding showing up correctly on your side?

Answer 16

no it's not but I think I'll figure it out.Thank you. :)

Answer 17

V' = 4 pi r^2 r'

we are told that V' = Surface area = 4 pi r^2, so sub it in

4 pi r^2 = 4 pi r^2 r'

then solve for r'

Answer 18

what is r'?

Answer 19

the acceeration rate of the radius?

Answer 20

r' is dr/dt --> the rate the radius is changing over time

acceleration is denoted as 2nd derivative