Here's a challenge for you...

Before I start, this is not me trying to get you to do my homework, I've done this already.
Try this on for size:
A parcel of land is bounded by a stream y = x + 3Pi/4 + (Pi/Sqrrt(2))*cos(x + 3Pi/4), and 3 wire fences x = 0, y = 0, and x = 2Pi. An 1 unit represents 100 metres. The y-axis points to the north.
There are a few questions first that are pretty easy. Skip to question d) if you want a challenge.
a) Find the exact coordinates of the 4 corners of the parcel of land.
b) Sketch the parcel of land. Label the boundaries with equations, and the corners with their coordinates.
ci) Write down a definite integral for the area of the parcel of land.
cii) Find the exact area (in m^2) of the parcel of land.

And the hard one:
di) A square building is to be erected on the parcel of land as far to the west as possible. Building regulations require it to be at least 15 metres from all boundaries.
Determine the floor area (correct to the nearest m^2) of the largest building possible.
To see the answer, scroll down and highlight the box:















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| 1529 |

No takers?

...so the cross section of the building must be a square, not a rectangle, right?

Does the attached drawing reflect your last question? The clue to the problem is then to determine the coordinates of point P such that the distance CP (red segment) is no less than 0.15 -which corresponds to 15 m if 1 unit is 100 m.

Unless I have overlooked something, the brute force approach has led me to 2 coupled non-linear equations for 2 unknowns and I'm at a loss about how to solve them. If there's a more direct and easier path I have yet to find it.. :confused:

krtxmrtz, your sketch is way off. First of all, there is a boundary at x=2*pi, not at y=2*pi. Second, your stream looks nothing like mine. See below.

Black lines are x- and y-axes, blue is the stream, red is boundary at x=2*pi, and the gray square is where the building sits.

Sorry krtxmrtz, logophobic is on the ball today :p
And yeah, it has to be a square.
Now all you have to do is find the shaded region of that square. I guess this question isn't that hard, but it looks it at first. Once you wrap your head around it it's OK.

I think your forumla is wrong because I got a straight line for my stream too...

Edit: Whoops! Blond moment.... I didn't meant to forget it was in radians....:lol:

This is odd. Now I'm getting a straight line... That's strange that 3 of us can't get it right when someone else was able to. Unless logophobic was an examiner for the VCE 2005 Maths Methods Exam so he knows what it's meant to look like!:eek2:
Hang on, I'll try to figure out what the problem is.

OK so it's a really simple solution. Chances are you're doing what I did and you just have your window settings wrong. Have X go from 0 to 2Pi and Y go from 0 to 10 and it should work. It's the same graph, but when you zoom out lots it looks straight.
If changing the axes doesn't work then you're just putting the function in wrong. What I have above is fine.

Quote:

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Originally Posted by Logophobic

krtxmrtz, your sketch is way off. First of all, there is a boundary at x=2*pi, not at y=2*pi.

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Gorblimey! I must remember to cleanse my glasses from time to time...

Quote:

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Originally Posted by Logophobic

Second, your stream looks nothing like mine. See below...

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Actually both streams are correct, only I zommed in the central part because of the mistake I made. I attach the original sketch.

Now it's much easier, all that needs to be done is find the height of the function's minimum by equating its derivative to zero:

f(x) = x + 3*Pi/4 + (Pi/Sqr(2))*cos(x + 3*Pi/4)

f'(x) = 1 - (Pi/Sqr(2))*sin(x + 3*Pi/4) = 0

whereby:

x + 3*Pi/4 = sin^-1[Sqr(2)/Pi]

And because the sine is a periodic function, the multiple solutions are

x + 3*Pi/4 = 2*k*Pi + 0.4669...
and
x + 3*Pi/4 = (2*n + 1)*Pi - 0.4669...

with k and n taking any integer value (..., -2, -1, 0, 1, 2, ...)

and the one corresponding to the minimum we are interested in is obviously for n=0:

x + 3*Pi/4 = Pi - 0.4669... -> x = 0.3185...

The actual length of the square is then f(0.3185...) - 2*0.15 = 0.6910... - 0.3 =0.3910...

The area of the building's cross section is:

Area = (0.3910...)² = 0.1529...

and because 1 unit is 100 m, we finally arrive at:

Area (m) = 100²*0.1529... = 1528.94...

Congratulations! You win.....
The privelege of doing my maths exam on Monday.


Please?

Quote:

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Originally Posted by metalmidget

Congratulations! You win.....
The privelege of doing my maths exam on Monday.


Please?

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Great! So you'll pay for my flight tickets from Europe? :D

Ummm.... Sure! The payment's in the mail.

Just wondering.... What programs do you all use to create these images? I use Microsoft Math. Does what I need for school. Never tried it for anything more advanced than trig graphs.

Quote:

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Originally Posted by 03myersd

Just wondering.... What programs do you all use to create these images? I use Microsoft Math. Does what I need for school. Never tried it for anything more advanced than trig graphs.

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One that I've heard people speak highly of is 'Origin', but I've never used it myself. For graphs such as the above I use Excel, and sometimes PowerPoint. They are quite handy.

I drew the curve on a form in VB, then pasted it into Paint and drew the rest of the image by hand.

Origin is pretty good. Not cheap, but pretty good. For straightforward stuff like this though, Excel is fine.

I use a TI-84+ :)
Serves my needs.

Hmmm.... A small selection to choose from.... Think I will stick with my Microsoft Math. No point in getting a calculator as I have one already. Just a basic casio one but it does all the graphing functions I need. Except choosing a base for log.....

Mine can't do that either but you can force it to.
logb(n) = x
b^x = n
loge(b^x) = loge(n)
xloge(b) = loge(n)
x = loge(n)/loge(b), therefore
logb(n) = loge(n)/loge(b)
yay!

Right.... im confused....

Quote:

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Originally Posted by 03myersd

Hmmm.... A small selection to choose from.... Think I will stick with my Microsoft Math. No point in getting a calculator as I have one already. Just a basic casio one but it does all the graphing functions I need. Except choosing a base for log.....

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If you want to find some number, n, in base b, this would be logb(n) which reads as "log base b of n", or "log of n to the base b", however you want to say it. Using the transposition above, logb(n) is the same as loge(n)/loge(b). Or if you want to, log10(n)/ log10(b). So if you want to know what 7 is in base 2, it's log2(7) = loge(7)/loge(2), which gives ~2.807. Which means that 2^2.907 ~ 7.
Easy!

Quote:

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Originally Posted by metalmidget

Mine can't do that either but you can force it to.
logb(n) = x
b^x = n
loge(b^x) = loge(n)
xloge(b) = loge(n)
x = loge(n)/loge(b), therefore
logb(n) = loge(n)/loge(b)
yay!

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It's easier to read if you use tags to crete subindices. If you don't know how to use them, just click the quote button and see how I've implemented subindices in this post.

log_bx = log_zx / log_zb

where z is any number, but of course you want z to be either e or 10, which are the available logs in calculators:

log_bx = log₁₀x / log₁₀b
or
log_bx = log_ex / log_eb = ln x / ln b

WOW!!! Thanks!!!! This will make my exams so much easier now that I can check my log graphs with my calculator! Thanks!!!