Rotating on a slant

[vpetersen]

Hi, I'm taking a summer Calculus course and my professor assigned the following take home assignment from Stewart.
http://www.stewartcalculus.com/data/CAL ... 02_stu.pdf
I'm having trouble finding the relationship between delta x and delta u. I've redrawn out a larger picture but I feel that my geometry knowledge isn't helping me out much. I think that it must have something to do with arctan since the final area formula begins with 1/(1+m^2). Beyond this I'm pretty lost. Any help would be appreciated.

 

[tkhunny]

Two Obvious Solutions:

1) Rotate the axis so it is horizontal, like you are used to.

2) Theorem of Pappus. You'll need a Centroid.

 

[JeffM]

Hi, I'm taking a summer Calculus course and my professor assigned the following take home assignment from Stewart.
http://www.stewartcalculus.com/data/CAL ... 02_stu.pdf
I'm having trouble finding the relationship between delta x and delta u. I've redrawn out a larger picture but I feel that my geometry knowledge isn't helping me out much. I think that it must have something to do with arctan since the final area formula begins with 1/(1+m^2). Beyond this I'm pretty lost. Any help would be appreciated.

You need to say explicitly when a problem needs to be solved by a particular method rather than expect volunteers to click on a link to get an exact problem statement. Unfortunately, I cannot help you because I have forgotten 90% of the calculus I once knew. If no one answers soon, repost in a way that makes clear exactly how you are to calculate your Riemann sum. Many will think your question has been answered when they see responses so reposting, though normally inappropriate, may be necessary. By the way, now that you know that I remember little calculus, it seems to me that your question about delta u simply involves applying the Pythogorean Theorem to the equation of the slant line, but don't give much weight to my opinion.

PS Unless someone clicks on your link, they do not know what delta u even represents.

 

[vpetersen]

Two Obvious Solutions:

1) Rotate the axis so it is horizontal, like you are used to.

2) Theorem of Pappus. You'll need a Centroid.

Thank you for your suggestion. I'm familiar with the using the Theorem of Pappus, but I'm not sure what you mean by the term exso...

 

[vpetersen]

Jeff, I included the link to the problem because it involves a picture that I'm not sure how to properly express my question without using that as a reference. Thank you for your suggestion though, I'll keep it in mind for next time. Thank you

 

[galactus]

I do not believe a centroid was meant to be used here since we're only in section 7.2. But, use whatever works.

Note, the first part of the problem asks us to derive the formula for AREA.

So, here is the first part. This is probably the most involved part of the project.

Once this is shown, the volume formula can be derived easier.

The given hint suggests finding the area of the slanted approximating rectangle(s).

The height of the rectangle is the distance between a point and a line. The distance marked with a ? on the upper side of the rectangle. Call this length D

The formula for the distance between a point and a line is $\displaystyle D=\frac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}$

In this case, our general line has standard equation $\displaystyle -mx+y-b=0$

Thus, the distance from point $\displaystyle (x_{i}, y_{i})$ to the line y=mx+b is:

$\displaystyle \boxed{D=\frac{f(x_{i})-mx-b}{\sqrt{m^{2}+1}}}$

Now, the width of the rectangle at the other little question mark:

At the question mark we have the arc length of the tangent line to the curve C.

Call this ds. ds is often used for arc length. Make a right triangle with ds as the hypotenuse and dx as the base.

Using the arc length formula, $\displaystyle \frac{ds}{dx}=\sqrt{1+[f'(x)]^{2}}$

Now, make a triangle with ds as the hypoteneuse and du as the base this time.

Notice the angles $\displaystyle \alpha, \;\ \beta$ down at the bottom of the diagram?.

They come in to play here. I hadn't noticed them at first, so I scratched my head a little longer than I should have.

$\displaystyle \frac{du}{ds}=cos(\beta-\alpha)$

Now, using the subtraction formula for cos, we get:

$\displaystyle \frac{du}{ds}=cos(\beta-\alpha)=cos\beta cos\alpha+sin\beta sin\alpha$

But $\displaystyle tan\alpha=f'(x) \;\ (\text{slope of f(x)}), \;\ m=tan\beta \;\ (\text{slope of line y=mx+b})$

Thus, $\displaystyle \frac{ds}{dx}=sec\alpha=\sqrt{1+[f'(x)]^{2}}$

We want $\displaystyle \frac{du}{dx}=\frac{du}{ds}\cdot\frac{ds}{dx}$

Divide du/ds by $\displaystyle cos\alpha$ and we get:

$\displaystyle \frac{du}{dx}=\frac{cos\beta cos\alpha+sin\beta sin\alpha}{cos\alpha}=\frac{1+tan\beta tan\alpha}{\sqrt{1+tan^{2}\beta}}=\frac{1+mf'(x)}{\sqrt{1+m^{2}}}$

Now, the area of each rectangle is $\displaystyle D\cdot\frac{du}{dx}$

$\displaystyle \left[\frac{f(x)-mx-b}{\sqrt{1+m^{2}}}\right]\left[\frac{1+mf'(x)}{\sqrt{1+m^{2}}}\right]$

$\displaystyle =\left[f(x)-mx-b\right]\left[\frac{1+mf'(x)}{m^{2}+1}\right]$

Now, finally, add up all the rectangle area:

$\displaystyle \frac{1}{m^{2}+1}\int_{p}^{q}\left[f(x)-mx-b\right]\left[1+mf'(x)\right]dx$

The volume is found by rotating each rectangle and adding them up.

$\displaystyle \pi\int D^{2}du$

 

[mmm4444bot]

I'm having trouble finding the relationship between delta x and delta u.

I can relate. I remember struggling through some of Stewart's "projects", especially when they were handed out as overnight assignments.

delta x and delta u are the bases of two different right triangles which share a common hypotenuse (called ds by galactus).

The angles (one in each triangle) where base meets hypotenuse are alpha and beta-alpha.

So, the relationship between delta x and delta u can be expressed using the angles and parts of these two triangles.

Here's a different read on the same part of Stewart's "project":

http://answers.yahoo.com/question/index ... 302AAxTSOg

 

[vpetersen]

I just want to double check I'm understanding. Is this how the triangles should look? I just don't see the relation between the du/ds and the angle beta. Thank you so much for all of you're help. I understand all the other steps you took for the most part except for how we can just start incorporating beta. Sorry part of the picture got cut off, but I think it's still clear the the triangles are still sharing the same hypotenuse.

 

[mmm4444bot]

Is this how the triangles should look?

No.

Your two triangles do not share a common hypotenuse.

Also, delta x is not the base on either of your triangles, and neither is delta u.

There is a line segment on Stewart's figure (labeled with a question mark) that lies on the tangent line.

You labeled this segment ds, on your diagram. THAT segment is the hypotenuse of both triangles.

You can use this hypotenuse to form one triangle that has delta x as its base; form the second triangle with delta u as its base.

If you're still stuck, let me know, and I will upload a diagram.

 

[galactus]

At the bottom left diagram on the problem sheet, finish drawing the rectangle in at the little question mark at what I called ds.

That little portion of the rectangle that is cut off because of overlap. if you draw that in it may help to see it.

Draw the triangles in there.

 

[mmm4444bot]

I want to start relearning how to Paint in Windows7. (Why can't Microsoft leave well-enough alone? :x )

So, I played with Stewart's diagram for you.

The smallest angle inside the purple triangle is ? - ?, calculated by some basic geometry.

[attachment=0:1nk64zaq]triangles.JPG[/attachment:1nk64zaq]

 

[galactus]

Very nice, mmm.

 

[vpetersen]

Wow, thank you so much for the picture, that helps a ton. I was having such a hard time visualizing it. Thanks again to everyone helping me out with this problem, I really appreciate the help.
THANK YOU!!

 

[galactus]

Re:

I want to start relearning how to Paint in Windows7. (Why can't Microsoft leave well-enough alone? :x )

So, I played with Stewart's diagram for you.

The smallest angle inside the purple triangle is ? - ?, calculated by some basic geometry.

[attachment=0:2r3j98of]triangles.JPG[/attachment:2r3j98of]

mmm, can you give me some tips on how you done that so nice?. How did you make the Delta symbol and get the lines looking so nice?. Not to mention those arrows.
I still have the old Paint in 2003, but I doubt if that matters.

 

[vpetersen]

Galactus,
Thank you for all the help, but I'm a little confused as to how the volume is just the pi*Integral of D^2 du?

 

[mmm4444bot]

mmm, can you give me some tips on how you done that so nice?

I will be glad to. I'll try to post it on the Odds & Ends board, within a couple days, but I can say quickly now that I started with an enlarged screenshot of Stewart's diagram pasted into Paint. Then I slide stuff around, cropping the rest. In other words, I did not create the lines, delta notations, or arrows in Paint; those came directly from the original image. I added only the colored triangles.

 

[galactus]

Oh, cool. Very clever.

I downloaded the pdf of Stewart's problem from the link that vpeterson provided. I tired copy and pasting the image into paint, but it said it was 'secured' and I could not copy or do much of anything.

 

[galactus]

Galactus,
Thank you for all the help, but I'm a little confused as to how the volume is just the pi*Integral of D^2 du?

You've used washers. That's basically all it is.

Take a slice with thickness du. Those rectangles we have been talking about. Think of them as infinitely thin slices revolving about that line of rotation, y=mx+b.

$\displaystyle \pi\int_{p}^{q}\left(\underbrace{\frac{f(x)-mx-b}{\sqrt{m^{2}+1}}}_{\text{D}}\right)^{2}\overbrace{\frac{1+mf'(x)}{\sqrt{m^{2}+1}}}^{\text{du}}dx$

Now, simplify a little.

On another note, I was thinkin', this could be done by rotating coordinate axes as well. But, that is probably future material.

This is a fun problem. It also enables the calc student to get a good idea of what calc is all about in practice. What the rectangles, du, dx, slices and all that are about. You can do a lot of things with it if you have a good understanding.

I have rotated on a slant before, but this probllem learned me something as well. I normally used the centroid method that tkh mentioned earlier.

If you nose around the site, I think there are some older posts regarding rotating about a line.
viewtopic.php?f=3&t=21443&hilit=centroid

Check and see if we get the same result using this method. Let's do that. The result from the link is $\displaystyle \frac{2\sqrt{2}\pi}{15}$

Let's see if we get that using this method. In the link, we are rotating $\displaystyle \frac{x^{2}}{2}$ about $\displaystyle y=x$.

So, entering into the derived formula above, we have $\displaystyle f(x)=\frac{x^{2}}{2}, \;\ m=1, \;\ b=0, \;\ f'(x)=x, \;\ p=0, \;\ q=2$

$\displaystyle \pi\int_{0}^{2}\frac{(\frac{x^{2}}{2}-x)^{2}(1+x)}{2^{\frac{3}{2}}}dx=\frac{2\sqrt{2}\pi}{15}$

I'll be dang....it checks out.

As I said, I learned something here as well. An easier way than using calc III topics.

 

[kjbohn]

i am stuck on where you knew how to use the sum and difference formula for du/ds? did u use parallel lines or congruent triangles?

 

[mmm4444bot]

i am stuck on where you knew how to use the sum and difference formula for du/ds?

My guess is that he used right-triangle trigonometry, followed by that identity.

The green triangle is a right triangle.

With respect to its angle (? - ?), we have:

Adjacent = du

Hypotenuse = ds

You already know what the ratio Adjacent/Hypotenuse represents, yes? :wink:

And the sum-and-difference formula used is the identity for the cosine of a difference of angles.