Question

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 √1.6t30.2t4+8100 dollars per year. The land is currently worth $580. Set up the integral needed to solve the problem, and then find the value of the land after 10 years to the nearest cent.

Answer 1

a) The expression of the value in time is

`V(t)=\sqrt{0.032t^(4)+1296}+544`

b) The value in ten years is $ 584.20.

Step-by-step explanation:

The rate of change in time of V is

`(dV)/(dT)=\frac{1.6*t^(3) }{\sqrt{0.2t^(4) +8100} }`

We have to solve this integral

`\int dV=\int (\frac{1.6*t^(3) }{\sqrt{0.2t^(4) +8100} })dt`

To solve this, we can define

`u=0.2t^(4) +8100\\\\du/dt=0.8t^(3) \\\\dt=(du)/(0.8t^(3))`

Replacing in the integral

`\int dV=\int (\frac{1.6*t^(3) }{\sqrt{0.2t^(4) +8100} })dt\\\\\int dV=\int ((1.6*t^(3) )/(√(u) ))*(du)/(8t^(3) ) \\\\\int dV=\int ((1.6 )/(√(u) ))*(du)/(8 )\\\\\int dV=0.2*\int ((1 )/(√(u) ))du\\\\V=0.2*(2*√(u) )+C=0.4*\sqrt{0.2t^(4)+8100 } +C\\\\V=\sqrt{0.4^(2) *(0.2t^(4)+8100)} +C=\sqrt{0.032t^(4)+1296}+C`

If V(0) = 580, we have

`V(0)=\sqrt{0.032*0^(4)+1296}+C=580\\\\√(1296)+C=580 \\\\C=580-36=544`

The expression of the value in time is

`V(t)=\sqrt{0.032t^(4)+1296}+544`

The value at ten years (t=10) is

`V(10)=\sqrt{0.032*10^(4)+1296}+544\\V(10)=√(320+1296)+544\\V(10)=\sqrt(1616)+544=40.20+544=584.20`