Since the change in volume is due solely to the downpour, the rate change of the volume is equal to the rate of downpour.

For this probelm we will use the power rule to solve which states,

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

Applying this rule we are able to find the rate of downpour.

\(\displaystyle V'(t)=\frac{d}{dt}(15+\sqrt[3]{t^2})=\frac{2}{3t^{\frac{1}{3}}}\)

The volume of a cylinder is \(\displaystyle V=\pi r^{2}h\). By implicit differentiation while holding the radius constant,

\(\displaystyle \frac{dV}{dt}=\pi r^2\frac{dh}{dt}\)

The rate of volume change is 4 and the radius is 2 so

\(\displaystyle 4=\pi*4\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{1}{\pi}\)

The equation for volume for the container is \(\displaystyle V=\frac{s^{2}h}{3}\). Implicit differentiation while holding side length constant gives

\(\displaystyle \frac{dV}{dt}=\frac{s^2}{3}\frac{dh}{dt}\)

With a side length of 3 and a flow rate of 10

\(\displaystyle 10=\frac{9}{3}\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=30\) 

Begin by finding the capacity of the tank:

\(\displaystyle V_{max}=(10ft)^3=1000ft^2\)

Now, determine the amount of water initially in the tank:

\(\displaystyle V_0=1ft(10ft)^2=100ft^3\)

From here, integrate the flow equation to determine the amount of water at any point in time:

\(\displaystyle V(t)=\int Flow=\int (10\frac{ft^3}{s}+3t^2\frac{ft^3}{s^3})dt\)

\(\displaystyle V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+C\)

This constant of integration can be found by using the initial condition:

\(\displaystyle V(0)=V_0=10(0)\frac{ft^3}{s}+0^3\frac{ft^3}{s^3}+C\)

\(\displaystyle C=100ft^3\)

\(\displaystyle V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+100ft^3\)

Now, to find when the tank is full, set the equation equal to the max capacity of the tank:

\(\displaystyle V(t_f)=10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}+100ft^3=1000ft^3\)

\(\displaystyle 10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}=900ft^3\)

\(\displaystyle t_f=9.31s\)

The flow of money out of the bank account is the negative of the rate of change (increase) in the amount of money.

Thus, the value we are looking for is 

\(\displaystyle -m'(15)=-(-0.0001e^{15})=0.0001\cdot3268984.4=326.90\).

So the net rate of flow out of the bank account is $326.90/hr.

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

\(\displaystyle V(t)=\int Rdt\)

\(\displaystyle V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt\)

\(\displaystyle V(t)=-0.1t^3\frac{gallon}{s^3}+C\)

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

\(\displaystyle V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C\)

\(\displaystyle C=40 gallons\)

Therefore the integral is:

\(\displaystyle V(t)=40gallons -0.1t^3\frac{gallon}{s^3}\)

\(\displaystyle V(5s)=40gallons -0.1(5s)^3\frac{gallon}{s^3}\)

\(\displaystyle V(5s)=27.5gallons\)

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

\(\displaystyle V(t)=\int Rdt\)

\(\displaystyle V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt\)

\(\displaystyle V(t)=-0.1t^3\frac{gallon}{s^3}+C\)

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

\(\displaystyle V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C\)

\(\displaystyle C=40 gallons\)

Therefore the integral is:

\(\displaystyle V(t)=40gallons -0.1t^3\frac{gallon}{s^3}\)

Now to find when the tank is drained, solve for when the equation is zero:

\(\displaystyle V(t_f)=0=40gallons-0.1t_f^3\frac{gallon}{s^3}\)

\(\displaystyle t_f^3=400s^3\)

\(\displaystyle t_f=7.37s\)

Begin by finding the volume of the tank:

\(\displaystyle V_{tank}=\pi r^2h\)

\(\displaystyle V_{tank}=2000\pi ft^3\)

Now, to find the amount the tank has filled over a given time, integrating the flow function with respect to time:

\(\displaystyle V(t)=\int (3t^2\frac{ft^3}{s^3}+6t\frac{ft^3}{s^2})dt\)

\(\displaystyle V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}+C\)

To find the constant of integration, use the initial condition of an empty tank:

 \(\displaystyle V(0)=0=0^3\frac{ft^3}{s^3}+3(0)^2\frac{ft^3}{s^2}+C\)

\(\displaystyle C=0\)

\(\displaystyle V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}\)

Now, to find the time when the tank is full, set the equation equal to the max volume of the tank:

\(\displaystyle V(t_f)=t_f^3\frac{ft^3}{s^3}+3t_f^2\frac{ft^3}{s^2}=V_{tank}=2000\pi ft^3\)

\(\displaystyle t_f=17.51s\)

Since we're given volume and need to calculate flow, we take the derivative of volume \(\displaystyle Q\) with respect to time \(\displaystyle t\).

\(\displaystyle Q=\frac{1}{3}t^3+2.5t^2+4t+7\)

To find the derivative of \(\displaystyle Q{}\) with respect to \(\displaystyle t{}\), we need to use the power rule. 

Recall that the power rule for differentiation is given as

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} (At^n)=nAt^{n-1}{}\), where \(\displaystyle A,n{}\) are constants and \(\displaystyle t{}\) is a variable. 

Using this rule for 

\(\displaystyle Q=\frac{1}{3}t^3+2.5t^2+4t+7{}\)

\(\displaystyle \frac{dQ}{dt}=t^2+5t+4\)

At \(\displaystyle t=4,{}\)

\(\displaystyle \frac{dQ(4)}{dt}=4^2+5*4+4= 40 \frac{m^3}{s}{}\). 

We know the units are in \(\displaystyle \frac{m^3}{s}{}\) because we determined the rate of flow, which is volume of fluid moving per unit time. 

We will use the equation for the volume of a cone and its derivative to solve this problem.  

The equation for the volume of a cone is

\(\displaystyle V=\frac{\pi }{3}r^{2}h\)

where \(\displaystyle r\) is the radius of the cone at the top and \(\displaystyle h\) is the height of the cone.

Since we do not know \(\displaystyle \frac{dr}{dt}\), we must eliminate \(\displaystyle r\) from the volume equation before we differentiate.

We use the ratio of height to radius to eliminate \(\displaystyle r\).

\(\displaystyle \frac{r}{h}=\frac{4}{8}=\frac{1}{2}\)

\(\displaystyle r=\frac{1}{2}h\)

The volume equation is now

\(\displaystyle V=\frac{\pi }{3}\left (\frac{1}{2}h \right )^{2}h=\frac{\pi }{12}h^{3}\)

We will now take the derivative of the equation with respect to time or \(\displaystyle t\).  We will use the power and chain rules to find the derivative.

\(\displaystyle \frac{dV}{dt}=\frac{d}{dt}\left [\frac{\pi }{12}h^{3}\right ]=\frac{\pi }{12}*3h^{2}\frac{dh}{dt}\)

\(\displaystyle \frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}\)

We are given the following parameters

\(\displaystyle \frac{dV}{dt}=1cm^{3}/min\) and we need to find \(\displaystyle \frac{dh}{dt}\) at \(\displaystyle h=3cm\).

Subsituting this information into the first derivative

\(\displaystyle \frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}\)

\(\displaystyle 1=\frac{\pi }{4}(3)^{2}\frac{dh}{dt}\)

\(\displaystyle 1=\frac{9\pi }{4}\frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{4 }{9\pi}\)