Rate of Flow - CLEP Calculus

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Question

A -gallon tank begins to drain at a rate of $0.3t^2\frac{gallon}{s^3}$ , where is a time in seconds.

How much water will remain in the tank after five seconds?

Answer

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:

$V(t)=\int Rdt$

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

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

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:

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

Therefore the integral is:

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

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

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Question

The volume in liters of water in a tank at time in seconds is $v = 600-4\sqrt{t}$ . What is the rate of flow from the tank at

Answer

To find the rate of flow, you need to differentiate the function. This is $v^{'}=\frac{-2}{\sqrt{t}}$ substituting in for gives the answser of .

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Question

A cylindrical tank with a radius of 20 centimeters and an arbitrary height is filling with water at a rate of 1.5 liters per second. What is the rate of change of the water level (its height)?

Answer

The first step we may take is to write out the formula for the volume of a cylinder:

$V = \pi r^{2}h$

Where r represents the radius and h the height.

With this known we can find the rate change of volume with respect to height, by deriving these functions with respect to height:

$dV = \pi r^{2}dh$

Since we're interested in the rate change of height, the dh term, let's isolate that on one side of the equation:

$dh = \frac{dV}{\pi r^{2}}$

dV, the rate change of volume, is given to us as 1.5 liters per second, or 1500 cm3/second. Plugging in our known values, we can thus solve for dh:

$dh = \left( \frac{1500\frac{cm^{3}}{s}}{\pi (20cm)^{2}}\right) = \frac{15}{4\pi }\frac{cm}{s}$

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Question

Make sure you identify what this question is asking.

A large vat contains of butter. The vat has a small leak, out of which, of butter escapes every hour. What is the rate of change in the volume of butter in the vat?

Answer

This question tells you that there is a leak of mL/hour, and then asks you to identify how quickly the leak is causing butter to be lost.

The key is to identify the units, mL/hour, see that there are mL/hour being moved, and recognize that the units are negative, as mL are being removed every hour.

Thus the rate of change in the volume of the butter is $-100\ mL/hr$ .

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Question

The area bouned by the curve and is being filled up with water at a rate of unit-squared per second. When the water-level is at how quickly is the water-level rising?

Answer

If , then $x=\pm\sqrt{2}/2$ . Therefore the width of the water at this level is $\sqrt{2}$ . Some small change in height is associated with a change of $\sqrt{2}dh$ in area. In other words:

$\frac{dA}{dh}=\sqrt{2}$

Taking the reciprocal yields:

$\frac{dh}{dA}=\frac{\sqrt{2}}{2}$

We also know that is increasing at 2 unit-squared per second, so:

$\frac{dA}{dt}=2$

Therefore:

$\frac{dh}{dt}=\frac{dh}{dA}\frac{dA}{dt}=\left(\frac{\sqrt{2}}{2}\right)\left(2\right)=\sqrt{2}$

Therefore, the height will be rising at $\sqrt{2}$ units per second.

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Question

If water is poured inside a cylindrical cup with radius of 3 inches at a rate of 2 cubic inches per second, how fast is the surface rising?

Answer

Write the volume of the cylinder.

$V=\pi r^2 h$

Substitute the radius.

$V=\pi (3)^2 h= 9\pi h$

Differentiate the volume with respect to time.

$\frac{dV}{dt}=9\pi \frac{dh}{dt}$

Substitute the given rate.

$2=9\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{2}{9\pi}$

The height is increasing at a rate of $\frac{2}{9\pi}$ inches per second.

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Question

Suppose a fish tank has a shape of a square prism with a length of 10 inches. If a hose filled the tank at 3 cubic inches per second, how fast is the water's surface rising?

Answer

Given the length of the cube is 10 inches, the length and the width is also 10 inches. However, the height of the water is unknown. Let's assume this height is .

Write the volume of the water in terms of .

Differentiate the volume equation with respect to time.

$\frac{dV}{dt}=100 \cdot \frac{dy}{dt}$

Substitute the rate of the water flow into $\frac{dV}{dt}$ .

$3=100 \cdot \frac{dy}{dt}$

$\frac{dy}{dt}=\frac{3}{100}$

The water is rising at a rate of $\frac{3}{100}$ inches per second.

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Question

The volume of water a pipe recieves is given as $V(t)=\sqrt{t}+60$ . What is the flow rate of the pipe at ? The volume is in liters.

Answer

Flow rate can be defined as the volume of a liquid passing through a surface per time . 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, $\frac{d}{dx}x^n=nx^{n-1}$ .

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

$\frac{dV}{dt}=\frac{1}{2\sqrt{t}}$ .

Plugging in for this equation, the flow rate of this pipe at is

$\frac{1}{8}\frac{liters}{second}$ .

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Question

The volume of water (in liters) in a pool at time (in minutes) is defined by the equation If Paul were to siphon water from the pool using an industrial-strength hose, what would be the rate of flow at in liters per minute?

Answer

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

Given

, we can apply the power rule,

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

Then .

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a river at time (in minutes) is defined by the equation . What is the river’s rate of flow at in liters per minute?

Answer

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

Given,

, we can apply the power rule,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ to find

.

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation . What is the tank’s rate of flow at in liters per minute?

Answer

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

Given,

, then using the power rule which states,

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

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a stream at time (in minutes) is defined by the equation . What is the stream’s rate of flow at in liters per minute?

Answer

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

Given,

and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

where , then

.

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a pool at time (in minutes) is defined by the equation . What is the pool’s rate of flow at in liters per minute?

Answer

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

Given

and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where , then

.

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation . What is the tank’s rate of flow at in liters per minute?

Answer

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

Given

and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where , then

.

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation $v(t)=12t^{2}+10t+9$ . If Daisy were to drain the tank, what would be the rate of flow at in liters per minute?

Answer

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

Given

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

$\frac{d}{dt}t^{n}=nt^{n-1}$ where , then .

Therefore, at ,

liters per minute.

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Question

The volume of water (in liters) in a river at time (in minutes) is defined by the equation $v(t)=4t^{3}+5t^{2}+7$ . What is the rate of flow at in liters per minute?

Answer

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

Given

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

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

Therefore, at ,

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

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Question

The volume of water (in liters) in a water slide at time (in minutes) is defined by the equation $v(t)=7t^{3}-4t^{2}+5t-10$ . What is the rate of flow at in liters per minute?

Answer

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

Given

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

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

Therefore, at ,

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

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Question

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

Answer

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.

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

Becomes:

Then,

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

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

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Question

The volume (in gallonws) of water in a sink after the drain is opened as a function of time can be written as: $V(t)=40e^{-0.5t}$ . What is the rate of flow out of the sink at

Answer

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

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

The rate of change at is then:

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

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 $\frac{20}{e}$

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Question

The volume of a sink with a newly open drain is a function of time, given as: $V(t)=40e^{-0.5t}$ .

Determine an equation that models the rate of change of flow into the sink.

Answer

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

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

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:

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

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

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