We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given 

\(\displaystyle v(t)=5t^3-4t^2+9t+10\) and the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) where \(\displaystyle n\neq0\), then 

\(\displaystyle v'(t)=15t^2-8t+9\). 

Therefore, at \(\displaystyle t=3\), 

\(\displaystyle v'(3)=15(3)^2-8(3)+9=135-24+9=120\)  liters per minute. 

Flow rate can be defined as the volume of a liquid passing through a surface per time \(\displaystyle t\).  This means that in order to solve this equation we must differentiate the volume equaiton we are given with respect to time.  By doing so we will obtain the change in volume per unit time, or the flow rate.  To take the derivative of this equation, we must use the power rule,  \(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\).  

We also must remember that the derivative of an constant is 0.  Differentiating the volume equaiton given, we obtain 

\(\displaystyle \frac{dV}{dt}=\frac{1}{2\sqrt{t}}\).

Plugging in \(\displaystyle t=16 seconds\) for this equation, the flow rate of this pipe at \(\displaystyle t=16 seconds\) is

 \(\displaystyle \frac{1}{8}\frac{liters}{second}\).

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given

\(\displaystyle v(t)=12t^{2}+10t+9\) and the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) where \(\displaystyle n\neq0\), then \(\displaystyle v'(t)=24t+10\) .

Therefore, at \(\displaystyle t=2\),

\(\displaystyle v'(2)=24(2)+10=48+10=58\) liters per minute. 

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given 

\(\displaystyle v(t)=4t^{3}+5t^{2}+7\) and the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) where \(\displaystyle n\neq0\), then \(\displaystyle v'(t)=12t^{2}+10t\) .

Therefore, at \(\displaystyle t=1\), 

\(\displaystyle v'(1)=12(1)^{2}+10(1)=12+10=22\) liters per minute. 

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given 

\(\displaystyle v(t)=7t^{3}-4t^{2}+5t-10\) and the power rule 

\(\displaystyle \frac{d}{dt}t^{n}=nt^{n-1}\) where \(\displaystyle n\neq0\), then \(\displaystyle v'(t)=21t^{2}-8t+5\) .

Therefore, at \(\displaystyle t=3\), 

\(\displaystyle v'(3)=21(3)^{2}-8(3)+5=189-24+5=170\) liters per minute. 

A spring-fed lake has a volume modeled by V(t). Find the rate of flow after 50 seconds and tell whether water is flowing into or out of the lake.

\(\displaystyle V(t)=16000t-5t^2\)

To find the rate of flow from a volume function, differentiate the volume function and evaluate at the given value of t.

In other words, we need to take V(t) and find V'(5).

So, this

\(\displaystyle V(t)=16000t-5t^2\)

Becomes:

\(\displaystyle V'(t)=16000-10t\)

Then,

\(\displaystyle V'(50)=16000-10(50)=15,500\frac{liters}{second}\)

Since we have a positive flow rate, we can say that water is flowing into the lake, thereby increasing the volume of the lake.

The rate of change in volume of the sink with respect to time is given as the derivative of the volume function:

\(\displaystyle \frac{dV}{dt}=\frac{d}{dt}(40e^{-0.5t})=-20e^{-0.5t}\)

The rate of change at \(\displaystyle t=2\) is then:

\(\displaystyle \frac{dV}{dt}(2)=-20e^{-0.5(2)}=\frac{-20}{e}\)

However, keep in mind that this problem asks for the flow out of the sink, so a negative change in the volume means a positive outflow. Therefore, the flow out of the sink is \(\displaystyle \frac{20}{e}\)

The flow of volume into the sink can be found as the derivative of the volume function:

\(\displaystyle \frac{dV}{dt}=\frac{d}{dt}(40e^{-0.5t})=-20e^{-0.5t}\)

Note that as a negative value, water is flowing out of the tank. However, the problem asked for not the flow, but the rate of change of the flow, which can be found by deriving the flow function:

\(\displaystyle \frac{d}{dt}(-20e^{-0.5t})=10e^{-0.5t}\)

As the sign is opposite to that of the flow, it means that the flow slows over time.

The volume of a sphere, in terms of its diameter, is given as:

\(\displaystyle V=\frac{\pi}{6}D^3\)

Since the change in volume is equivalent to the rate of flow, we can find the latter by deriving the above equation:

\(\displaystyle \frac{dV}{dt}=\frac{\pi}{2}D^2\frac{dD}{dt}\)

The diameter is \(\displaystyle 10cm\) and its rate of change is  \(\displaystyle 0.1\frac{cm}{s}\).

Therefore, the rate of flow is:

\(\displaystyle \frac{\pi}{2}(10cm)^2\left(.1\frac{cm}{s}\right)=5\frac{cm^3}{s}\)

Rate of flow can be found by taking the time derivative of the volume function:

\(\displaystyle \frac{dV}{dt}=\frac{d}{dt}(20-20e^{-2t})=40e^{-2t}\)

To determine whether the flow increases or decreases over time, take the derivative of it and check the sign:

\(\displaystyle \frac{d}{dt}(40e^{-2t})=-80e^{-2t}\)

For all \(\displaystyle t>0\), the function is negative, so flow decreases over time.