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

Answer:

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.

Explanation:

The rate of change in time of V is

$\frac{dV}{dT}=\frac{1.6*t^{3} }{\sqrt{0.2t^{4} +8100} }$

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

$u=0.2t^{4} +8100\\\\du/dt=0.8t^{3} \\\\dt=\frac{du}{0.8t^{3}}$

Replacing in the integral

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

If V(0) = 580, we have

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

The expression of the value in time is

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

The value at ten years (t=10) is

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