Question

A restaurant purchased kitchen equipment on January​ 1, 2017. On January​ 1, 2019, the value of the equipment was ​$14 comma 550. The value after that date was modeled as follows. ​V(t)equals 14 comma 550 e Superscript negative 0.158 t ​a) What is the rate of change in the value of the equipment on January​ 1, 2019

Answer 1

Answer:

$\frac{dV(t)}{dt} =$ - 1675.38

Step-by-step explanation:

In 2017, the vakue of the kitchen equipment was $14550

V(0)=$14550

Its value after then was modelled by $V(t)=14550e^{-0.158t$

We are required to find the rate of change in value on January 1, 2019

$V(t)=14550e^{-0.158t$

$\frac{dV(t)}{dt} =\frac{d}{dt}14550e^{-0.158t$

$\frac{dV(t)}{dt} =14550 \frac{d}{dt}e^{-0.158t$

$\\Let u= -0.158t,\frac{du}{dt}=-0.158$

$\frac{dV(t)}{dt} =14550 \frac{d}{du}e^u\frac{du}{dt}$

$\frac{dV(t)}{dt} =14550 X -0.158 e^{-0.158t}=-2298.9e^{-0.158t}$

In 2019, i.e. 2 years after, t=2

The rate of change of the value

$\frac{dV(t)}{dt} =-2298.9e^{-0.158X2}$

=$\frac{dV(t)}{dt} =-2298.9e^{-0.316}$= - 1675.38