Respuesta :
Answer:
a) The expression of the value in time is
[tex]V(t)=\sqrt{0.032t^{4}+1296}+544[/tex]
b) The value in ten years is $ 584.20.
Explanation:
The rate of change in time of V is
[tex]\frac{dV}{dT}=\frac{1.6*t^{3} }{\sqrt{0.2t^{4} +8100} }[/tex]
We have to solve this integral
[tex]\int dV=\int (\frac{1.6*t^{3} }{\sqrt{0.2t^{4} +8100} })dt[/tex]
To solve this, we can define
[tex]u=0.2t^{4} +8100\\\\du/dt=0.8t^{3} \\\\dt=\frac{du}{0.8t^{3}}[/tex]
Replacing in the integral
[tex]\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[/tex]
If V(0) = 580, we have
[tex]V(0)=\sqrt{0.032*0^{4}+1296}+C=580\\\\\sqrt{1296}+C=580 \\\\C=580-36=544[/tex]
The expression of the value in time is
[tex]V(t)=\sqrt{0.032t^{4}+1296}+544[/tex]
The value at ten years (t=10) is
[tex]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[/tex]
