Calculus 1 : Rate of Flow

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #1 : How To Find Rate Of Flow

The volume \displaystyle (v) in liters of water in a tank at time  \displaystyle t in seconds is \displaystyle v = 600-4\sqrt{t}.  What is the rate of flow from the tank at \displaystyle t=16

Possible Answers:

\displaystyle -2 liters per second

\displaystyle \frac{1}{2} liters per second

\displaystyle 4 liters per second

\displaystyle \frac{-1}{2} liters per second

Correct answer:

\displaystyle \frac{-1}{2} liters per second

Explanation:

To find the rate of flow, you need to differentiate the function.  This is \displaystyle v^{'}=\frac{-2}{\sqrt{t}} substituting \displaystyle 16 in for \displaystyle t gives the answser of \displaystyle -0.5.

Example Question #2 : How To Find Rate Of Flow

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?

Possible Answers:

\displaystyle \frac{3}{25}

\displaystyle \frac{3}{100}

\displaystyle \frac{3}{1000}

\displaystyle \frac{3}{75}

\displaystyle \frac{3}{50}

Correct answer:

\displaystyle \frac{3}{100}

Explanation:

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 \displaystyle y.

Write the volume of the water in terms of \displaystyle y.

\displaystyle V=10(10)y= 100y

Differentiate the volume equation with respect to time.

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

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

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

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

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

Example Question #3 : How To Find Rate Of Flow

Make sure you identify what this question is asking.

A large vat contains \displaystyle 1000 L of butter. The vat has a small leak, out of which, \displaystyle 100 mL \displaystyle (0.1 L) of butter escapes every hour. What is the rate of change in the volume of butter in the vat?

Possible Answers:

\displaystyle 100\ mL/hr 

\displaystyle -100\ mL/hr

\displaystyle 0\ mL/hr 

\displaystyle -0.01\ mL/hr 

\displaystyle 100\ L/hr 

Correct answer:

\displaystyle -100\ mL/hr

Explanation:

This question tells you that there is a leak of \displaystyle 100 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 \displaystyle 100 mL/hour being moved, and recognize that the units are negative, as \displaystyle 100 mL are being removed every hour.

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

Example Question #4 : How To Find Rate Of Flow

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)?

Possible Answers:

\displaystyle \frac{15\pi }{8}\frac{cm}{s}

\displaystyle \frac{15}{4\pi }\frac{cm}{s}

\displaystyle \frac{15}{8\pi }\frac{cm}{s}

\displaystyle \frac{15\pi}{4 }\frac{cm}{s}

Correct answer:

\displaystyle \frac{15}{4\pi }\frac{cm}{s}

Explanation:

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

\displaystyle 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:

\displaystyle 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:

\displaystyle 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:

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

Example Question #4 : How To Find Rate Of Flow

The volume \displaystyle v of water (in liters) in a pool at time \displaystyle t (in minutes) is defined by the equation \displaystyle v(t)=7t^2+5t-12. If Paul were to siphon water from the pool using an industrial-strength hose, what would be the rate of flow at \displaystyle t=2 in liters per minute?

Possible Answers:

\displaystyle 25

\displaystyle 23

\displaystyle 20

\displaystyle 33

\displaystyle 32

Correct answer:

\displaystyle 33

Explanation:

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^2+5t-12, we can apply the power rule,

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

Then \displaystyle v'(t)=14t+5

Therefore, at \displaystyle t=2

\displaystyle v'(2)=14(2)+5=28+5=33 liters per minute. 

Example Question #6 : How To Find Rate Of Flow

The volume \displaystyle v of water (in liters) in a river at time \displaystyle t (in minutes) is defined by the equation \displaystyle v(t)=8t^3+9t^2-20t+6. What is the river’s rate of flow at \displaystyle t=3 in liters per minute?

Possible Answers:

\displaystyle 300

\displaystyle 250

\displaystyle 150

\displaystyle 225

\displaystyle 90

Correct answer:

\displaystyle 250

Explanation:

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

Given, 

\displaystyle v(t)=8t^3+9t^2-20t+6, we can apply the power rule,

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

\displaystyle v'(t)=24t^2+18t-20.  

Therefore, at \displaystyle t=3

\displaystyle v'(3)=24(3)^2+18(3)-20=216+54-20=250 liters per minute.

Example Question #2 : How To Find Rate Of Flow

The volume \displaystyle v of water (in liters) in a tank at time \displaystyle t (in minutes) is defined by the equation \displaystyle v(t)=4t^3+7t^2+10t. What is the tank’s rate of flow at \displaystyle t=1 in liters per minute?

Possible Answers:

\displaystyle 36

\displaystyle 32

\displaystyle 34

\displaystyle 30

\displaystyle 37

Correct answer:

\displaystyle 36

Explanation:

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+7t^2+10t, then using the power rule which states,

\displaystyle f(x)=x^n\rightarrow f'(x)=nx^{n-1} thus \displaystyle v'(t)=12t^2+14t+10.

Therefore, at \displaystyle t=1,

\displaystyle v'(1)=12(1)^2+14(1)+10=12+14+10=36 liters per minute.

Example Question #1 : How To Find Rate Of Flow

The area bouned by the curve \displaystyle y=x^2 and \displaystyle y=1 is being filled up with water at a rate of \displaystyle 2 unit-squared per second. When the water-level is at \displaystyle y=0.5 how quickly is the water-level rising?

Canvas

Possible Answers:

\displaystyle \sqrt{\sqrt{2}} units per second

\displaystyle \sqrt{2} units per second

\displaystyle 2\sqrt{2} units per second

\displaystyle \frac{\sqrt{2}}{2} units per second

\displaystyle 2 units per second

Correct answer:

\displaystyle \sqrt{2} units per second

Explanation:

If \displaystyle y=0.5, then \displaystyle x=\pm\sqrt{2}/2. Therefore the width of the water at this level is \displaystyle \sqrt{2}. Some small change \displaystyle dh in height is associated with a change of \displaystyle \sqrt{2}dh in area. In other words:

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

Taking the reciprocal yields:

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

 

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

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

 

Therefore:

\displaystyle \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 \displaystyle \sqrt{2} units per second.

Example Question #3 : How To Find Rate Of Flow

The volume \displaystyle v of water (in liters) in a stream at time \displaystyle t (in minutes) is defined by the equation \displaystyle v(t)=7t^3+5t^2-10t+6. What is the stream’s rate of flow at \displaystyle t=3 in liters per minute?

Possible Answers:

\displaystyle 109

\displaystyle 209

\displaystyle 219

\displaystyle 210

\displaystyle 190

Correct answer:

\displaystyle 209

Explanation:

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+5t^2-10t+6 and the power rule 

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

where \displaystyle n\neq0,  then

\displaystyle v'(t)=21t^2+10t-10

Therefore, at \displaystyle t=3,

\displaystyle v'(3)=21(3)^2+10(3)-10=189+30-10=209  liters per minute. 

Example Question #4 : How To Find Rate Of Flow

The volume \displaystyle v of water (in liters) in a pool at time \displaystyle t (in minutes) is defined by the equation \displaystyle v(t)=6t^3-4t^2-8t+9. What is the pool’s rate of flow at \displaystyle t=4  in liters per minute?

Possible Answers:

\displaystyle 248

\displaystyle 258

\displaystyle 238

\displaystyle 218

\displaystyle 288

Correct answer:

\displaystyle 248

Explanation:

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

Given 

\displaystyle v(t)=6t^3-4t^2-8t+9 and the power rule

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

\displaystyle v'(t)=18t^2-8t-8.

Therefore, at \displaystyle t=4

\displaystyle v'(t)=18(4)^2-8(4)-8=288-32-8=248  liters per minute. 

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