the thickness x of ice on the surface of a pond increases with time t according to the equation dx/dt = k/x.?

Question

where k is a positive constant.
(a) find x as a function of t, given thant x=0 when t=0.
(b)if the ice is 1cm thick after 2 days,how long will it be before it is 2cm thick?

Answer 1

a Separate the variables:

x.dx = k.dt

Integrating both sides gives

0.5x^2 = kt + C

now find C

0 = 0 + C therefore C must be 0

x = (2kt)^0.5

b when t = 2, x = 1, find t when x = 2
0.5*1^2 = 2k
k = 0.25

0.5*2^2 = 0.25t

2 = 0.25t

t = 8 days
easy when u think about really

Answer 2

a) dx/dt = k/x which leads to ∫xdx = ∫kdt
x²/2 = kt + c
When t = 0, x = 0 so c = 0
Thus x²/2 = kt
x² = 2kt
x = √(2kt)

b) 1 = √(2 x k x 2) so k = 1/4
2 = √((1/2)(t))
4 = (1/2 ) t
t = 8
Will take 8 days before ice is 2cm thick.

Answer 3

a) Separate the variables:

x.dx = k.dt

Integrating both sides gives

0.5x^2 = kt + C

now find C

0 = 0 + C therefore C must be 0

x = (2kt)^0.5

b) when t = 2, x = 1, find t when x = 2

0.5*1^2 = 2k

k = 0.25

----

0.5*2^2 = 0.25t

2 = 0.25t

t = 8 days

Answer 4

a) x = (2kt)^1/2

b) t = 8 days

Answer 5

I get a D in math. =((((( Maybe B is four days? I dunno. =((((((