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## Calculus 8th Edition
Published by Cengage
# Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 22: 59
#### Answer
$A=\frac{\sqrt{3}}{4}x^2$, $x>0$
#### Work Step by Step
We have an equilateral triangle with sides $x$ and we need to find the area. We know that: $A=1/2*base*height=\frac{1}{2}xh$ We need to eliminate $h$. We construct a right triangle in the middle of the equilateral triangle with sides $h$, $x$, and $1/2x$. We use the Pythagorean Theorem with these three sides: $(\frac{1}{2}x)^2+h^2=x^2$ $h^2=x^2-(\frac{1}{2}x)^2$ $h=\pm\sqrt{x^2-(\frac{1}{2}x)^2}$ $h=+\sqrt{\frac{3}{4}x^2}=\frac{\sqrt{3}}{2}x$ (We eliminate the negative because lengths must be positive.) We plug in $h$ in the area formula: $A=\frac{1}{2}x\frac{\sqrt{3}}{2}x=\frac{\sqrt{3}}{4}x^2$ The domain is $x>0$ because lengths must be positive.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Financial Mathematics
1. Nov 9, 2005
### playboy
How do you find the annual effective rate of interest?
The question reads: You lend a freind $15 000 to be amortized by semiannual payments for 8 years, with interest at j2 = 9%. You deposit each payment in an account paying J12 = 7%. What annual effective rate of interest have you earned over the entire 8-year period? Ans = 8.17% Hmmm... i have absolutly no idea how to get the annuale effective rate of interest. My TA showed, (in another question) that its something like (1 + i)^n = 1 + r and solve for r? Please help somebody Thanks 2. Nov 10, 2005 ### hotvette Conceptually, it works like this. There is initial outlay of$15,000. The payments that come in annually are immediately invested. At the end of 8 years there is a total value of all investments. The 'effective' interest rate is the equivalent rate at which the initial outlay would compound at to achieve the same final result after 8 years. It might help to draw out a time line and treat each pmt and ensuing investment as a separate problem. Find out how much each is worth after the 8 years is up, sum the totals together, and then it's a straightforward back solution for a std compound interest problem.
By the way, you have 2 identical posts. If this was intentional, pls avoid that in the future.
P.S. One of the most useful classes (in terms of constantly using the material learned) I took in graduate school was called "Engineering Economy".
Last edited: Nov 10, 2005
3. Nov 10, 2005
### playboy
No, that was not intentional, i didn't know i did that :S... I will avoid that in the future!
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https://quizanswered.com/module-8-ac-amplifiers-based-on-jfets-multisim/
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Module 8: ac amplifiers based on jfets (multisim)
MULTISIM
Introduction
This experiment explores the uses of JFET as AC signal amplifiers. It also describes techniques for measuring the input and output impedance of circuits.
Remember that your lab report will need to include your measurements, calculations, screenshots, etc. as indicated at the end of this outline.
Procedure
1. Common Source Amplifier
1.1 Build the common source amplifier shown in Figure 8.1
Figure 8. 1: Common source amplifier
1.2 Using the oscilloscope, measure the voltage gain of the amplifier defined as Av = Vout/Vin.
2. Measurement of Input impedance
In this section we will learn a very useful technique to measure the input impedance of any circuit. This technique is based on placing a known resistor in series with the input of the circuit and measuring the voltage drop across the new resistor. This technique can also be used in live circuits and not just
simulations.
Figure 8. 2: Circuit to measure input impedance
2.1 Build the circuit shown in Figure 8.2 You will notice that this is the same circuit used in Figure 8.2 with the extra R4 resistor added to the input.
2.2 Measure with the oscilloscope the voltage at node V1 and the voltage at node V4.
2.3 Calculate input impedance as follows:
3. Measurement of output impedance
The following technique can be used to measure the output impedance of a circuit. In practical circuits, the best value of the load resistor must be selected by trial and error.
3.1 Measure Vout as shown in the circuit from Figure 8.1. We will name this voltage Vout
3.2 Connect a load resistor of 1 kΩ at the output of the same circuit. Measure the voltage across the load. We will call this voltage Vload
3.3 Calculate output impedance as:
(in the case of using a different value for the load resistor, change the 10 kΩ value in the equation to the appropriate value of resistor used).
Laboratory Report
Create a laboratory report using Word or another word processing software that contains at least these elements: – Introduction: what is the purpose of this laboratory experiment? – Results for each section : Measured and calculated values, calculations, etc. following the outline. Include screenshots for the circuits and waveforms as necessary — You can press Alt + Print_Screen inside Multisim or if using Windows 7, you can use the “Snipping tool”. Either way, you can paste these figures into your Word processor.
– Conclusion : What area(s) you had difficulties with in the lab; what did you lean in this experiment; how it applies to your coursework and any other comments.
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https://workforce.libretexts.org/Bookshelves/Electronics_Technology/Book%3A_Trigonometry_and_Single_Phase_AC_Generation_for_Electricians_(Flinn)/03%3A_AC_Generation/03.5%3A_Frequency_and_Alternators
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# 3.5: Frequency and Alternators
In the last chapter, we learned the term cycle means from the point in a waveform to where the waveform starts to repeat itself. When we discuss the term frequency, we are referring to how many cycles can occur in one second. Frequency is measured in hertz (shout out to Heinrich Hertz) or CPS (cycles per second). Two factors affect the frequency in an alternator: rotation speed and the number of poles.
Figure 52. Sine wave cycle
## Rotation speed
As the armature rotates through the field, it starts to create a waveform (as we saw in the last chapter). One full mechanical rotation of the armature creates one full sine wave on a two-pole alternator. If the two-pole alternator spins three complete revolutions in one second, it will create three full sine waves in that one second. We would say that the frequency is at three cycles per second or three hertz (as the cool kids say).
A machine’s rotational speed is measured in rotations per minute or RPM. However, we are not concerned with minutes, but rather, with seconds when dealing with frequency. Therefore, RPM must be converted to rotations per second (RPS). As there are 60 seconds in a minute, all we have to do is to divide the RPM by 60 to convert it to RPS.
For example, if the armature is spinning at a rate of 1800 RPM on a two-pole alternator, we can say that it is spinning at 30 rotations per second. If this alternator has two poles, then in one second it will generate 30 cycles of voltage. It then could be said to have a frequency of 30 cycles per second or 30 Hertz. The frequency of an alternator is directly proportional to the rotational speed of the alternator.
## Number of Poles
If we add poles to the alternator, we can change the frequency. In a two-pole alternator, Side A of the armature (Figure 53) passes from north to south, and then south to north, to create one complete sine wave. I f we add two more poles, as in Figure 54, then Side A of the armature will move past two north poles and two south poles in one full mechanical revolution.
Figure 53. Two pole alternator
Two full sine waves are created in one complete mechanical revolution. If a two-pole alternator creates one cycle of voltage in one second (or one hertz of frequency), a four pole alternator will create two cycles of voltage in one second (or two hertz).
The frequency of an alternator is directly proportional to the number of poles in the alternator.
Figure 54. Four pole alternator
## Formula time!
Knowing that rotation speed is directly proportional to frequency and that the number of poles is directly proportional to frequency, we can use a formula. The formula looks like this:
$f= \dfrac{P}{2} \times \dfrac{N}{60} \tag{Frequency formula}$
where…
• $$f$$ = frequency in hertz
• $$P$$ = number of poles
• $$N$$ = rotational speed in RPM
We divide the number of poles by two because there will always be a set of two poles. You can’t have a north pole without a south. We divide the RPM by 60 because we are concerned with rotations per second, not rotations per minute. The formula in Figure 56 can be combined to look like this:
$f = \dfrac{PN}{120} \tag{Combined frequency formula}$
Video! This video will walk you through how frequency is related to the RPM and the number of poles of an alternator.
A YouTube element has been excluded from this version of the text. You can view it online here: https://pressbooks.bccampus.ca/trigf...ricians/?p=278
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https://www.physicsforums.com/threads/inequalities-math-homework-problem.190125/
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# Inequalities math homework problem
1. Oct 9, 2007
### xCanx
A rectangular solid is to be constructed with a special kind of wire along all
the edges. The length of the base is to be twice the width of the base. The
height of the rectangular solid is such that the total amount of wire used (for
the whole figure) is 40 cm. Find the range of possible values for the width of
the base so that the volume of the figure will lie between 2 cm3 and 4 cm3.
Can someone show me how to start off?
2. Oct 9, 2007
### symbolipoint
w=width, L=length, h=height;
w+L+h=40, L=2w
wLh>2 and wLh<4
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https://17calculus.com/integrals/volume/washer-disc/
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## 17Calculus Integrals - Volume of Revolution Using The Washer-Disc Method
##### 17Calculus
This page covers single volume integrals when an area is rotated about a vertical or horizontal line. The area is defined by equations in the form $$y=f(x)$$ or $$x=f(y)$$ and we use the washer (disc) method. For other ways to calculate volume, see the links in the related topics panel.
If you want a full video lecture on this topic, we recommend this video and this instructor.
### Prof Leonard - Volume of Solids By Disks and Washers Method [2hr-47mins-48secs]
video by Prof Leonard
Alternate Names Disc Method Disk Method Ring Method
We choose to use the term washer-disc method to refer to this technique. We think it covers the two most commonly used and most descriptive names. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.
What Is A Washer?
If you are not familiar with a washer (other than to wash clothes), this wiki page has pictures and explains what a washer is. In short, it is a disc with a circular hole in it whose center is the same as the full disc.
Overview
When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)
On this page, we discuss the washer-disc method where the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane. This is a critical step to setting up your integral correctly. If you didn't completely understand this from the main volume integrals page, you can go over it again here.
### Describing A Region In The xy-Plane
To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by $$f(x)$$ (red line), $$g(x)$$ (blue line) and $$x=a$$ (black line).
[Remember that an equation like $$x=a$$ can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically
Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing $$g(x)$$, no matter where we draw it. Similarly, the arrow always exits the area by crossing $$f(x)$$, no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions $$f(x)$$ and $$g(x)$$ intersect. You should be able to do that. We will call that point $$(b,f(b))$$. Also, we will call the left boundary $$x=a$$. So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow
$$g(x) \leq y \leq f(x)$$
arrow leaves through $$f(x)$$ and enters through $$g(x)$$
$$a \leq x \leq b$$
arrow sweeps from left ($$x=a$$) to right ($$x=b$$)
Horizontally
We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of $$f(x)$$ and $$g(x)$$ in terms of $$y$$. ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations $$f(x) \to F(y)$$ and $$g(x) \to G(y)$$.
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line $$x=a$$. However, there is something strange going on at the point $$(b,f(b))$$. Notice that when the arrow is below $$f(b)$$, the arrow exits through $$g(x)$$ but when the arrow is above $$f(b)$$, the arrow exits through $$f(x)$$. This is a problem. To overcome this, we need to break the area into two parts at $$f(b)$$.
Lower Section - - This section is described by the arrow leaving through $$g(x)$$. So the arrow sweeps from $$g(a)$$ to $$g(b)$$.
Upper Section - - This section is described by the arrow leaving through $$f(x)$$. The arrow sweeps from $$f(b)$$ to $$f(a)$$.
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow
lower section
$$a \leq x \leq G(y)$$
arrow leaves through $$G(y)$$ and enters through $$x=a$$
$$g(a) \leq y \leq g(b)$$
arrow sweeps from bottom ($$y=g(a)$$) to top ($$y=g(b)$$)
upper section
$$a \leq x \leq F(y)$$
arrow leaves through $$F(y)$$ and enters through $$x=a$$
$$f(b) \leq y \leq f(a)$$
arrow sweeps from bottom ($$y=f(b)$$) to top ($$y=f(a)$$)
Type 1 and Type 2 Regions
Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
### Krista King Math - type I and type 2 regions [1min-39secs]
video by Krista King Math
washer-disc method
x-axis rotation
y-axis rotation
Now we will discuss each of these plots separately and explain each part of the plots.
Getting Started
Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r.
Once those steps are done, you are ready to set up your integral.
The volume integral using the washer-disc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washer-disc. If we start with a full disc (no hole in the middle), the volume is the surface area times the thickness. Since the disc is a circle, the area of a circle is $$\pi R^2$$ where $$R$$ is the radius of the circle. The volume is $$\pi R^2 t$$ where $$t$$ is the thickness. We choose to use a capital R here as the radius of the disc.
Now, with a washer, we take the disc we just discussed and put a circular hole in it with it's center the same as the full disc. (Think of a CD or DVD disc.) Now the volume is reduced by what we have taken out of the center. This empty space has volume $$\pi r^2 t$$, where $$r$$ is the radius of the small hole. The thickness, $$t$$, is the same as the full disc.
So now we have what we need to put together an equation for the washer-disc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get $$V = \pi R^2 t - \pi r^2 t = \pi t(R^2-r^2)$$. [Notice the special case when there is no hole in the middle, can be thought of as $$r=0$$ giving the volume of the disk as just $$V=\pi R^2 t$$.]
summary of the washer-disc method
the representative rectangle is perpendicular to axis of revolution
$$R$$ is the distance from the axis of rotation to the far end of the representative rectangle
$$r$$ is the distance from the axis of rotation to the closest end of the representative rectangle
x-axis rotation equation
$$\displaystyle{ V = \pi \int_{a}^{b}{R^2-r^2~dx} }$$
y-axis rotation equation
$$\displaystyle{ V = \pi \int_{c}^{d}{R^2-r^2~dy} }$$
Note - Notice that $$R$$ and $$r$$ are distances, so they are always positive (although since we square them, the sign doesn't make any difference in the equations).
washer-disc method with x-axis rotation
If you feel like you need further explanation of this, here is a video that tries to explain this method by drawing in three dimensions. In this video, notice that the axis of rotation runs along one side of the figure and, consequently, $$r=0$$.
### Khan Academy - Disc Method [10min-4secs]
Okay, now let's work some problems using the washer-disc method revolving an area about the x-axis.
x-axis rotation practice
area: $$y=1/x, x=1, x=3$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3487 video solution
$$2\pi/3$$
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area: $$y=x^2, y=4x-x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 32\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3494 video solution
$$V = 32\pi/3$$
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area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Krista King Math - 324 video solution
video by Krista King Math
$$V=8\pi$$
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area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{3\pi}{10}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
### Krista King Math - 877 video solution
video by Krista King Math
$$\displaystyle{V=\frac{3\pi}{10}}$$
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area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
$$V=4\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### PatrickJMT - 1174 video solution
video by PatrickJMT
$$V=4\pi/21$$
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area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
$$8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
In the second video, he doesn't finish the integration, so here are the details.
$$\displaystyle{ \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi [(4^2)/2 - (0^2)/2] = 8\pi}$$
### PatrickJMT - 1359 video solution
video by PatrickJMT
$$8\pi$$
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area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 9\pi^2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Dr Chris Tisdell - 2003 video solution
video by Dr Chris Tisdell
$$V = 9\pi^2$$
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area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$4.5\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2272 video solution
video by Michel vanBiezen
$$4.5\pi$$
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area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2273 video solution
video by Michel vanBiezen
$$2\pi/15$$
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area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$V=18\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### MIP4U - 2276 video solution
video by MIP4U
$$V=18\pi$$
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area: $$y=\sqrt{x}, y=x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1184 video solution
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area: $$y=\sqrt{x}, y\geq0, x=1$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y\geq0, x=1$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1182 video solution
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Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Problem Statement
Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Solution
### Khan Academy - 1183 video solution
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Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the x-axis or the y-axis. A twist you will see is when the axis of revolution is another line. On this site, we will discuss only axes that are parallel to one of the coordinate axes.
In this case, the equations that will change are the ones that describe the distance from the axes of rotation. We suggest that you set up a sum from the parallel coordinate axis to the axis of rotation and then solve for whatever variable you need. This concept, especially, requires you to think over in your mind several times and look at examples.
parallel to x-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: $$y=4$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
$$V = 256\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3490 video solution
$$V = 256\pi/15$$
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area: $$y=2-x^2, y=1$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3491 video solution
$$V = 16\pi/15$$
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area: $$y=x, y=x^2$$ axis of rotation: $$y=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
$$V = 8\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3497 video solution
$$V = 8\pi/15$$
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area: $$y=1/x, y=0, x=1, x=3$$ axis of rotation: $$y=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3498 video solution
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
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area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Solution
1. Draw the plot and the example rectangle and label r and R. See the hint or the animation. We drew the example rectangle perpendicular to the axis of revolution since we were told to use the washer-disc method. The distance r is the distance from the axis of revolution to closest end of the rectangle. The distance R is from the axis of revolution to the farthest end of the example rectangle.
2. Choose The Integral - The volume integral we need is $$V = \pi\int_a^b{R^2-r^2~dx}$$. We chose this integral because we are told to use the washer-disc method, so we need an integral with r and R. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine r and R - From the plot, let's start on the axis of revolution and move down to the x-axis of revolution. This distance is 5 units. Moving back up to the end of the rectangle that lands on $$y=x$$, we move y units. What we are left with is r, so $$r=5-y$$. However, we need to replace y with expression for y in terms of x. Since $$y=x$$ is the line that we are working with, we have $$r=5-x$$.
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the x-axis which is 5 units. However, when $$x < 3$$ we need to go a little further in the same direction to get the full distance R. Let's put that aside for a minute and think about the part of the graph for $$x > 3$$. In this case, we need to go back up y units, so $$R=5-y$$. This looks the same as r but in this case, we are landing on the plot $$y=x^2-3x$$. So $$R=5-(x^2-3x) = -x^2+3x+5$$. Let's plug in a few values and compare the values to graph to see if they match.
$$x=3$$ $$R=-3^2+3(3)+5=5$$ 𞀄 $$x=4$$ $$R=-4^2+3(4)+5=1$$ 𞀄
So far, so good. Let's plug in a few values $$x < 3$$ and see what we get.
$$x=0$$ $$R=-0^2+3(0)+5=5$$ 𞀄 $$x=2$$ $$R=-2^2+3(2)+5=7$$ 𞀄
So it looks like we have the correct equation for R. We can do the same with r to check our equation. This does not guarantee that we have the right equations but it may give an indication if they are incorrect. I usually check both endpoints and at least one other point, two other points is even better.
4. Set up and evaluate the integral - If we look at the animation above, we can tell that the rectangle sweeps across the area from $$x=0$$ to $$x=4$$. So our integral is $$\displaystyle{ V = \pi \int_0^4{ (-x^2+3x+5)^2 - } }$$ $$\displaystyle{ (5-x)^2 ~ dx }$$. Let's evaluate it.
$$\displaystyle{ V = \pi\int_0^4{ (-x^2+3x+5)^2 - (5-x)^2 ~ dx } }$$ $$\displaystyle{ V = \pi\int_0^4{ (x^4-3x^3-5x^2-3x^3+ } }$$ $$9x^2+15x-5x^2+15x+25) -$$ $$(25-10x+x^2) ~dx$$ $$\displaystyle{ V = \pi\int_0^4{ x^4-6x^3-2x^2+40x ~dx } }$$ $$\displaystyle{ V = \pi\left[ \frac{x^5}{5} - \frac{6x^4}{4} -\frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }$$ $$\displaystyle{ V = \pi\left[ \frac{4^5}{5} - \frac{6(4)^4}{4} -\frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right] - 0 }$$ $$\displaystyle{ V = \pi\left[ \frac{1024}{5} - \frac{1536}{4} -\frac{128}{3}+\frac{640}{2} \right] }$$ $$\displaystyle{ V = \pi \frac{1472}{15} }$$
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
$$\displaystyle{V=11\pi/30}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
Solution
### Krista King Math - 1172 video solution
video by Krista King Math
$$\displaystyle{V=11\pi/30}$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
$$V=25\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
Solution
In the video, he set up the integral but did not evaluate it. His integral was $$\displaystyle{ \int_{0}^{1}{\pi[(2+x)^2 - \pi(2+x^3)^2]dx} }$$. This evaluates to $$\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3} - x^4 - \frac{x^7}{7} \right]_{0}^{1} }$$
### PatrickJMT - 1175 video solution
video by PatrickJMT
$$V=25\pi/21$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
$$V=97\pi/42$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### PatrickJMT - 1176 video solution
video by PatrickJMT
$$V=97\pi/42$$
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area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$23\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2275 video solution
video by Michel vanBiezen
$$23\pi/15$$
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Now, let's work some problems with the y-axis as the axis of rotation. Here is the plot that contains all the information you need to work these problems.
washer-disc method with y-axis rotation
y-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3488 video solution
$$V = 8\pi$$
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area: $$y=x^{2/3}, x=0, y=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
$$V = \pi/4$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3489 video solution
$$V = \pi/4$$
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area: $$y^2 = x, x=2y$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 64\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3496 video solution
$$V = 64\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 36\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3495 video solution
$$V = 36\pi\sqrt{3}/5$$
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area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
$$V=162\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 343 video solution
video by Krista King Math
$$V=162\pi$$
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area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{\pi}{2}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 878 video solution
video by Krista King Math
$$\displaystyle{V=\frac{\pi}{2}}$$
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area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
$$25\pi/2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2271 video solution
video by Michel vanBiezen
$$25\pi/2$$
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area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
$$\pi/6$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2274 video solution
video by Michel vanBiezen
$$\pi/6$$
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area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
$$V=32\pi/5$$
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### MIP4U - 2277 video solution
video by MIP4U
$$V=32\pi/5$$
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area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
$$V=5\pi/24$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2280 video solution
video by Michel vanBiezen
$$V=5\pi/24$$
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area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2281 video solution
video by Michel vanBiezen
$$V=8\pi$$
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area: $$y=x^2, y=0, x=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=0, x=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1185 video solution
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parallel to y-axis rotation practice
area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=3$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
$$V = 24\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3492 video solution
$$V = 24\pi\sqrt{3}/5$$
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area: $$x=y^2, x=1$$ axis of rotation: $$x=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3493 video solution
$$V = 16\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=6$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
$$V = 84\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3499 video solution
$$V = 84\pi\sqrt{3}/5$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$x=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
$$V = 29\pi\/30$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3500 video solution
$$V = 29\pi\/30$$
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area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
$$V = 3\pi/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Krista King Math - 1173 video solution
video by Krista King Math
$$V = 3\pi/5$$
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area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
$$V=832\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
Solution
### MIP4U - 1916 video solution
video by MIP4U
$$V=832\pi/15$$
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area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
$$V=(\pi/12)[3\pi-4]$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2279 video solution
video by Michel vanBiezen
$$V=(\pi/12)[3\pi-4]$$
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https://www.clutchprep.com/physics/practice-problems/139925/explain-the-distinction-between-an-ohmic-and-non-ohmic-material-in-terms-of-how-
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crawl-data/CC-MAIN-2021-39/segments/1631780057036.89/warc/CC-MAIN-20210920101029-20210920131029-00525.warc.gz
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|
Resistors and Ohm's Law Video Lessons
Concept
# Problem: a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
###### FREE Expert Solution
a)
Ohmic materials are materials that obey Ohm's law: show linear relationship between the current and the voltage, and their resistances do not change with the variation in voltage.
83% (26 ratings)
###### Problem Details
a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.
b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
| 338
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http://www.fmxfeeds.com/2018/09/quick-algorithm-get-ideal-size-square-like-for-a-board-game-having-an-arbitrary-but-even-number-of-fields/
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# Quick Algorithm: Get Ideal Size (Square like) For a Board Game Having an Arbitrary (but Even) Number of Fields
Say you are developing a game like Chess, Go, Checkers, Tic-Tac-Toe or Memory. In each of those games the game board is a rectangle looking playfield of different size (rows x columns). Tic-Tac-Toe is 3×3, Checkers is 8×8, while Go can be 19×19 or 13×13 and similar.
In a game with an arbitrary number of game fields you might want to have the board look as closely to square as possible (rectangle where height and width are the same). Think of Memory. Let’s say we have 24 cards, that is 12 pairs. If you want to place them in a rectangle looking grid, most similar to square, you would go for 4 x 6 (or 6 x 4) board size (as it would look more square like than 3 x 8 and 2 x 12 or 1 x 24 would be too wide).
Therefore, the question: having an arbitrary number of game field pairs, what is the ideal, most square looking, grid size?
And the answer is an algorithm (some math knowledge required) like this one:
TGridSize = record
Rows, Columns : integer;
end;
function CalcGridSize(const numberOfPairs : Cardinal) : TGridSize;
//look for ideal rectangle dimensions (square is ideal)
//to present fields, number of fields = 2 * numberOfPairs
var
fieldCount : integer;
i : integer;
begin
fieldCount := 2 * numberOfPairs;
result.Rows := 1;
result.Columns := fieldCount;
if Sqrt(fieldCount) = Trunc(Sqrt(fieldCount)) then
begin
result.Rows := Trunc(Sqrt(fieldCount));
result.Columns := Trunc(Sqrt(fieldCount));
Exit;
end;
for i := Trunc(Sqrt(fieldCount)) downto 2 do
if (fieldCount mod i) = 0 then
begin
result.Rows := i;
result.Columns := fieldCount div i;
Exit;
end;
end; (*CalcGridSize*)
And here are some results:
var
gridSize : TGridSize;
i : integer;
begin
for i := 1 to 20 do
begin
gridSize := CalcGridSize(i);
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https://metanumbers.com/6011
| 1,708,554,300,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947473558.16/warc/CC-MAIN-20240221202132-20240221232132-00537.warc.gz
| 409,762,550
| 7,370
|
# 6011 (number)
6011 is an odd four-digits prime number following 6010 and preceding 6012. In scientific notation, it is written as 6.011 × 103. The sum of its digits is 8. It has a total of one prime factor and 2 positive divisors. There are 6,010 positive integers (up to 6011) that are relatively prime to 6011.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 4
• Sum of Digits 8
• Digital Root 8
## Name
Name six thousand eleven
## Notation
Scientific notation 6.011 × 103 6.011 × 103
## Prime Factorization of 6011
Prime Factorization 6011
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 6011 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ 8.70135 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 6011 is 6011. Since it has only one prime factor, 6011 is a prime number.
## Divisors of 6011
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 6012 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 3006 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 77.5306 Returns the nth root of the product of n divisors H 1.99967 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6011 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 6011) is 6012, the average is 3006.
## Other Arithmetic Functions (n = 6011)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 6010 Total number of positive integers not greater than n that are coprime to n λ 6010 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 789 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 6,010 positive integers (less than 6011) that are coprime with 6011. And there are approximately 789 prime numbers less than or equal to 6011.
## Divisibility of 6011
m n mod m
2 1
3 2
4 3
5 1
6 5
7 5
8 3
9 8
6011 is not divisible by any number less than or equal to 9.
## Classification of 6011
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non hypotenuse
• Prime Power
• Square Free
## Base conversion 6011
Base System Value
2 Binary 1011101111011
3 Ternary 22020122
4 Quaternary 1131323
5 Quinary 143021
6 Senary 43455
8 Octal 13573
10 Decimal 6011
12 Duodecimal 358b
16 Hexadecimal 177b
20 Vigesimal f0b
36 Base36 4mz
## Basic calculations (n = 6011)
### Multiplication
n×y
n×2 12022 18033 24044 30055
### Division
n÷y
n÷2 3005.5 2003.67 1502.75 1202.2
### Exponentiation
ny
n2 36132121 217190179331 1305530167958641 7847541839599391051
### Nth Root
y√n
2√n 77.5306 18.1823 8.80515 5.69888
## 6011 as geometric shapes
### Circle
Radius = n
Diameter 12022 37768.2 1.13512e+08
### Sphere
Radius = n
Volume 9.09764e+11 4.5405e+08 37768.2
### Square
Length = n
Perimeter 24044 3.61321e+07 8500.84
### Cube
Length = n
Surface area 2.16793e+08 2.1719e+11 10411.4
### Equilateral Triangle
Length = n
Perimeter 18033 1.56457e+07 5205.68
### Triangular Pyramid
Length = n
Surface area 6.25827e+07 2.55961e+10 4907.96
## Cryptographic Hash Functions
md5 e3b80d30a727c738f3cff0941f6bc55a 83a08a7fdc4059c7ac2e647e546add90679b9ac4 c37fd5582393747ef03b83ad095a5650d2f5335acc65eaa7db54c4b2a21d1092 3dc2b7303e1342405a887f123ed4e5615ff7f0a3a805f9e157d23fb8fedee7bd4eb3302d39c67b8b66304f3309bb8e0720dcac9649bfa295eb8cc8e7a4451918 ac22fadd6e29870db1b5de10b3d7f60f5317d167
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| 3.78125
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CC-MAIN-2024-10
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latest
|
en
| 0.814264
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https://origin.geeksforgeeks.org/plot-multiple-data-sets-on-the-same-chart-in-excel/
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Skip to content
# Plot Multiple Data Sets on the Same Chart in Excel
• Last Updated : 29 Jun, 2021
Sometimes while dealing with hierarchical data we need to combine two or more various chart types into a single chart for better visualization and analysis. This type of chart having multiple data sets is known as “Combination charts”.
In this article, we are going to see how to make combination charts from a set of two different charts in Excel using the example shown below.
Example: Consider a famous coaching institute that deals with both free content in their YouTube channel and also have their own paid online courses. There are broadly two categories of students in this institute :
1. The students who enrolled in the coaching but are learning from YouTube free video content.
2. The students who enrolled as well as bought paid online courses.
So, the institute asked their Sales Department to make a statistical chart about how many paid courses from a pool of courses which the institute deals with were sold from the year 2014 to the last year 2020 and also show the percentage of students who have enrolled in these paid courses.
### Table :
Here, the first data is “Number of Paid courses sold” and the second one is “Percentage of Students enrolled”. Now our aim is to plot these two data in the same chart with different y-axis.
### Implementation :
Follow the below steps to implement the same:
Step 1: Insert the data in the cells. After insertion, select the rows and columns by dragging the cursor.
Step 2: Now click on Insert Tab from the top of the Excel window and then select Insert Line or Area Chart. From the pop-down menu select the first “2-D Line”.
From the above chart we can observe that the second data line is almost invisible because of scaling. The present y-axis line is having much higher values and the percentage line will be having values lesser than 1 i.e. in decimal values. Hence, we need a secondary axis in order to plot the two lines in the same chart. In Excel, it is also known as clustering of two charts.
The steps to add a secondary axis are as follows :
1. Open the Chart Type dialog box
`Select the Chart -> Design -> Change Chart Type`
Another way is :
`Select the Chart -> Right Click on it -> Change Chart Type`
2. The Chart Type dialog box opens. Now go to the “Combo” option and check the “Secondary Axis” box for the “Percentage of Students Enrolled” column. This will add the secondary axis in the original chart and will separate the two charts. This will result in better visualization for analysis purposes.
The combination chart with two data sets is now ready. The secondary axis is for the “Percentage of Students Enrolled” column in the data set as discussed above.
Now various formatting can be carried out in this secondary axis using the Format Axis window on the right corner of Excel.
`Select the secondary Axis -> Right Click -> Format Axis -> Format Axis Dialog Box`
Changing the Bounds of Secondary Axis
You can further format the above chart by making it more interactive by changing the “Chart Styles”, adding suitable “Axis Titles”, “Chart Title”, “Data Labels”, changing the “Chart Type” etc. It can be done using the “+” button in the top right corner of the Excel chart.
Finally, after all the modification, the chart with multiple data sets looks like :
We can infer from the above chart that in the year 2019, the percentage of students who enrolled in the online paid courses are relatively less but in 2020 more students have enrolled in paid courses than free content on YouTube.
My Personal Notes arrow_drop_up
Recommended Articles
Page :
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https://homework.cpm.org/category/CC/textbook/cca2/chapter/4/lesson/4.2.3/problem/4-95
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crawl-data/CC-MAIN-2024-22/segments/1715971058611.55/warc/CC-MAIN-20240523050122-20240523080122-00183.warc.gz
| 255,003,461
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### Home > CCA2 > Chapter 4 > Lesson 4.2.3 > Problem4-95
4-95.
Solve the equations below.
1. $\sqrt { x + 15 } = 5 + \sqrt { x }$
Square both sides.
$\left(\sqrt{x+15}\right)^2=\left(5+\sqrt{x}\right)^2$
$x+15=\left(5+\sqrt{x}\right)\left(5+\sqrt{x}\right)$
$x+15=25+10\sqrt{x}+x$
Isolate the square root of $x$ on one side of the equation.
$-10=10\sqrt{x}$
Divide both sides by $10$.
$-1=\sqrt{x}$
What can you take the square root of and get $−1$?
No real solutions.
1. $(y−6)^2+10=3y$
Expand the $(y−6)^2$.
Rearrange the equation so that it equals zero.
Solve by factoring and using the Zero Product Property, or use the Quadratic Formula.
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https://brainmass.com/math/graphs-and-functions/functions-kuhn-tucker-condition-7945
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|
text/html
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crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00530.warc.gz
| 117,297,621
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|
Purchase Solution
# Functions: K-T Condition
Not what you're looking for?
Consider the following program:
Maximize f(x,y)=x^2+4xy+y^2
subject to g(x,y)=x^2+y^2-1=0
##### Solution Summary
A function is maximized using Kuhn-Tucker condition. The maximized function results are determined.
##### Solution Preview
Solution. Let us denote the gradient vector of the function f(x,y) by Df(x,y). We rewrite the original program as follows.
Minimize F(x,y)=-f(x,y)=-x^2-4xy-y^2
. g(x,y)=x^2+y^2-1=0.
Since DF(x,y)=(-2x-4y,-4x-2y)', Dg(x,y)=(2x,2y)', by K-T ...
Solution provided by:
###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
| 575
| 2,406
|
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| 3.875
| 4
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CC-MAIN-2024-26
|
latest
|
en
| 0.878535
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