Question

At a certain point in time, each dimension of a cube of ice is 30 cm and is decreasing at the rate of 2 cm/hr, so how fast is the ice melting (losing volume)?

Answer 1

Volume of cube = (side)^3
so V = s^3
then take the derivative of that :)

Answer 2

3s^2

Answer 3

this is related rates

\[\frac{dv}{dt} = \frac{dv}{dl} \times \frac{dl}{dt}\]

Answer 4

don't forget chain rule! :)

Answer 5

so, 6s?

Answer 6

not quite :)
I meant that the derivative of V = s^3
should have been V' = 3s^2 (s')

Answer 7

l is the side length

and you know dl/dt = 2

you know V = l^3

find dV/dl

by differentiating volume...

then subtitute your given value of l

Answer 8

so, after v'=3s^2, i do?

Answer 9

V' = 3s^2 (s')

you were given s' = -2
and that s = 30
so plug that in and you'll get

V' = 3 (30)^2 (-2)
=

Answer 10

-5400

Answer 11

yup :)

Answer 12

<3