It is estimated that t years from now, the value of a small piece of land, V(t), will be increasing at a rate of 1.6t^3/(0.2t^4+8100)^(1/2)dollars per year. The land is currently worth $420. Find V(t) and also find the value of the land after 10 years to the nearest cent.
dv/dt = 1.6t^3/√(0.2t^4+8100)
Let u = 0.2t^4+8100
du = 0.8t^3 dt
Now you have
dv/dt = 2du/√u
v(t) = 4√u + c
= 4√(0.2t^4+8100) + c
Now plug in your numbers to find c, and then find v(10).
To find the value of the land after t years, we need to find the antiderivative of the rate function with respect to t, and then evaluate it.
Given that the rate of increase is 1.6t^3/(0.2t^4+8100)^(1/2) dollars per year, we can express the value function V(t) as the definite integral of this rate function with respect to t, starting from 0 (the current time) up to t years:
V(t) = ∫[0 to t] 1.6t^3/(0.2t^4+8100)^(1/2) dt
Let's find this antiderivative and solve for V(t):
First, make a substitution: u = 0.2t^4 + 8100
du = 0.8t^3 dt
We can rewrite V(t) as:
V(t) = ∫[0 to t] (1.6t^3)/(0.2t^4+8100)^(1/2) dt
= 8∫[0 to t] (t^3/u)^(1/2) du
= 8∫[0 to t] (t^3u^(-1/2)) du
= 8∫[0 to t] (t^3 u^(-1/2)) du
= 8t^3 ∫[(0.2t^4 + 8100)^(-1/2)] du
= 8t^3 (0.2t^4 + 8100)^1/2 + C
Since we know the initial value of the land is $420 (V(0) = 420), we can solve for C:
420 = 8(0^3)(0.2(0^4) + 8100)^(1/2) + C
420 = 8(0)+C
C = 420
Therefore, the value function is:
V(t) = 8t^3 (0.2t^4 + 8100)^(1/2) + 420
To find the value of the land after 10 years, substitute t = 10 into the value function:
V(10) = 8(10)^3 (0.2(10)^4 + 8100)^(1/2) + 420
Evaluating this expression will give you the value of the land after 10 years, to the nearest cent.