Value Depreciation, solving: $dV/dt=-1080\:e^{-0.06t}$

Question

The change in value of a car is modeled with

$$dV/dt=-1080e^{-0.06t}$$

t is years since the car was purchased and V is in dollars. If the original purchase price of the car was $18,000 find the car's value after 5 years.

Answer 1

From $$ dV/dt=-1080e^{-0.06t} $$ you may integrate each variable separately giving $$ \int_{V(0)}^{V(t)} dV=-1080\int_0^t e^{-0.06s}ds $$ or$$ V(t)-V(0)=18\:000\:e^{-0.06t}-18\:000 $$ but we know that $V(0)=18\:000$ then

$$ V(t)=18\:000\:e^{-0.06t},\quad t\geq0. $$

You are asked to find $V(5)$, which is now easy to obtain.

Answer 2

$$\frac{\text{d}V}{\text{d}t} = -1080\ e^{-0.06 t'}$$

We integrate both term with respect to time

$$\int_{V_0}^{V(t)} \frac{\text{d}V}{\text{d}t}\ \text{d}t= -1080\int_0^t e^{-0.06 t'}\ \text{d}t'$$

And we get

$$V(t) - V(0) = -1080\left(-\frac{1}{0.06}e^{-0.06t} - \left(-\frac{1}{0.06}e^{-0.06\cdot 0}\right)\right)$$

id est

$$V(t) - V(0) = 18000 e^{-0.06t} - 18000$$

Knowing that $V(0) = 18000$ we can compute the fraction in the right side and we get:

$$v(t) = 18000 - 18000 + 18000e^{-0.06t}$$

$$V(t) = 18000e^{-0.06t}$$

Now it's easy to finish because you know $t = 5$ thence

$$\boxed{V(5) = 18000e^{-0.06\cdot 5} = 18000e^{-0.3} = 13334.7\ \$ }$$