The rate at which ice in a skating pond is melting is given by dV/ft = -4t - 2
Where V is volume of the ice in cubic inches and t is time in days after March 1st. Skaters can use the pond as long as the volume is greater than 3 cubic ft. If the ice is 5cubic ft in march 1st, when is the last day it will be safe on the pond?
if dV/ft = -4t - 2
V = -2t^2 - 2t + c
if March 1 ---> t = 0 , and when t=0 , V = 5
5 = 0-0 + c
V = -2t^2 - 2t + 5
so we want to know when V = 3
3 = -2t^2 - 2t + 5
2t^2 + 2t - 2 = 0
t^2 + t - 1 = 0
t = (-1 ± √5)/2 = .618 or a negative
Bad news, the pond is only safe on March 1
check:
when t = 1 ---> March 2nd
V = -2 - 2 + 5 = 1 cubic foot, not safe
To find the last day the ice will be safe on the pond, we need to determine when the volume of the ice becomes less than or equal to 3 cubic ft.
The given rate of melting is dV/dt = -4t - 2.
We can solve this differential equation by integrating both sides.
∫ dV = ∫ (-4t - 2) dt
Integrating the right side:
V = -2t^2 - 2t + C
To find the constant of integration (C), we use the initial condition given on March 1st: V = 5 cubic feet when t = 0.
Substituting into the equation, we get:
5 = -2(0)^2 - 2(0) + C
C = 5
So, the equation becomes:
V = -2t^2 - 2t + 5
Now, we can set V to 3 (the safe volume) to find when the ice is no longer safe on the pond.
3 = -2t^2 - 2t + 5
Rearranging the equation:
2t^2 + 2t - 2 = 0
Simplifying further:
t^2 + t - 1 = 0
Now we can solve for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Substituting a = 1, b = 1, and c = -1:
t = (-1 ± √(1^2 - 4(1)(-1))) / (2(1))
Simplifying further:
t = (-1 ± √(1 + 4)) / 2
t = (-1 ± √5) / 2
Since we are looking for a positive value of t, we take the positive square root:
t = (-1 + √5) / 2
Therefore, the last day it will be safe on the pond is approximately t = 0.618 (rounded to three decimal places) days after March 1st.