Question

Got a calc 1 question,

Let x be the number of people, in thousands, inside an amusement park. The number of people

inside the park that have contracted a virus can be modeled by v(x) =

h(x)/3x

for 3 < x < 5.


The number of people in the park is increasing at a constant rate of 0.2 thousands of people per minute.

Using this model, what is the rate that people inside the park are contracting the virus with respect to time when there are four thousand people in the park?


(h(x) = 4e^(2x−6) − x^2 + 5) when x > 3

Answer 1

`v=\frac h{3x}\implies(\mathrm dv)/(\mathrm dt)=(3x(\mathrm dh)/(\mathrm dt)-3h(\mathrm dx)/(\mathrm dt))/(3x^2)`

By the chain rule,

`(\mathrm dh)/(\mathrm dt)=(\mathrm dh)/(\mathrm dx)(\mathrm dx)/(\mathrm dt)`

We have

`h(x)=4e^(2x-6)-x^2+5\implies(\mathrm dh)/(\mathrm dx)=8e^(2x-6)-2x`

and we're given that
`x` changes at a constant rate of
`(\mathrm dx)/(\mathrm dt)=0.2` thousand people per minute, which means

`(\mathrm dv)/(\mathrm dt)=\frac{3x\left(8e^(2x-6)-2x\right)\left(0.2\frac{\text{thousand people}}{\rm min}\right)-3h\left(0.2\frac{\text{thousand people}}{\rm min}\right)}{3x^2}`

At the moment
`x=4` thousand people are in the park, we have
`h(4)=4e^2-11`, so we find

`(\mathrm dv)/(\mathrm dt)=(12\left(8e^2-8\right)-3(4e^2-11))/(240)\frac{\text{thousand people}}{\rm min}=(7(4e^2-3))/(80)\frac{\text{thousand people}}{\rm min}`

or approximately 2.324 thousand people per minute.