Question

Got a calc 1 question,

Answer 1

$v=\dfrac h{3x}\implies\dfrac{\mathrm dv}{\mathrm dt}=\dfrac{3x\frac{\mathrm dh}{\mathrm dt}-3h\frac{\mathrm dx}{\mathrm dt}}{3x^2}$

By the chain rule,

$\dfrac{\mathrm dh}{\mathrm dt}=\dfrac{\mathrm dh}{\mathrm dx}\dfrac{\mathrm dx}{\mathrm dt}$

We have

$h(x)=4e^{2x-6}-x^2+5\implies\dfrac{\mathrm dh}{\mathrm dx}=8e^{2x-6}-2x$

and we're given that $x$ changes at a constant rate of $\frac{\mathrm dx}{\mathrm dt}=0.2$ thousand people per minute, which means

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

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

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

or approximately 2.324 thousand people per minute.