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I drew a rectangle with a diagonal in it and tried to keep the
diagonal a constant # like sqrt(200), but any # is fine.
Then I realized if height or width approaches 0, but is just
tiny bit over zero like 0.01, then you will get the
minimum area or product, but still have a diagonal length
of sqrt(200). So the minimum product approaches zero,
I think.
Slope = -2
X intercept = 2
Y intercept = 4
The x intercept must be greater than one.
Call x intercept J.
Here is the equation for the area.
Just find the J above one that makes the expression the smallest:
I get (-2,-1) and (2,1) by graphing a sketch.
And I guess majicWaffle is right,
also (i, -2i) and (-i, 2i), whatever these complex thingys mean, I don't remember.
Maybe I should have said that
The larger b is, then the parabola is bent more upward as you go right?
Just a guess.
If b is negative then the parabola is bent more downward as you go right?
What about 2 and -5? Negatives are probably disallowed, huh?
mathsyperson, how do you know the unaccounted for part is the broken off
piece? ganesh said one-eight is above the water, but he didn't say one-eighth was
currently above the water with respect to the original size?
I think it's a trick question.
I am guessing at the method, but here's how I got 3.4489379%.
.0575 ÷ 12
add one
raise to the 12th power (current result is effective rate about 1.059 or 5.9%)
Then work backwards with these ideas.
Subtract .024
raise to 1/12th power to get monthly rate
subtract one
multiply by 12
Bingo! You've got it. 0.034...
3.4489379%
Width is numerator; height of zeros is denominator,
so 1 is
o
which means 1/1.
Just over a half because 75K is half of 150K and if you make the 150K smaller in
the denominator and the 75K bigger in the numerator, then the fraction gets
bigger. That's why I say "over" a half. 76/147 is about 75/150.
Hi mathsyperson,
Yeah. I think in retrospect, I just got extremely lucky in the curvy calculus stuff.
And also it occurs to me that maybe I was taught it back in 1985 or 6, but my
memory is so bad, I may have been reinventing something I'd seen before.
My major was Electrical Engineering, but I took a lot of Calculus courses, but
just barely passed somehow. Now I am starting all over from nearly scratch again.
Thanks, I got it.
But if you don't know the class width, you probably can't do it ahead of
time. Like divide by
I don't understand.
What about y = x from 10 to 20, so the average should be 15?
Oh you mean divide by 10 and factor out the
If you want the average y value (height) of a
function, then you can get the area under the
curve and divide by the interval width.
Well I was hoping to find a way to do this a
little differently, but I may need some help.
Instead of integrating first and then dividing,
I would like to do something in preparation
before the integration, so after the definite
integral, there will be no division.
If anyone can help. Thank you.
If this can be done, I think it would help me
to grasp underlying principles better for
maybe other applications.
I can't believe the progress you made!!!!!!
That's amazing.
(Google Latex math for graphical math symbols, notice _ for subscript and ^ for superscript, \frac{numerator}{denominator}, etc. I just learned some today.
You've helped me a lot, and I also have been trying to do what you have done,
except I only got as far as the following, except I had the x and y axis
backwards from you:
Curved surface area of steradian = 1 =
∫2πf(y)√1+(dx/dy)² dy.
Tell me if this is the same thing? This is way over my head unless I learn a lot.
The equation for that line will be y = H - (H/R)x, this is true for all cones. This line intercepts the x axis at x = R.
Now multiply your line function by (2 pi x) and then integrate from zero to R. You will get:
2pix(H - (H/R)x) = 2piH ∫x - x^2 / R dx (remember to factor out constants?)
I finally figured out that you are making a series of hollow pipes that get
shorter and shorter as you approach R.
Thanks for teaching me so well!!!!!!
I am very grateful to everyone for spending so much time helping me.
Thank you very much for the compliment.
You made my day!
I'll check out that link in a minute.
And yes, I did find it in a table, and I got 26.40253635
for the length from 0 to 5 for
What is the length of the curvy line of a sine wave graphed from 0 to 2π radians?
I'll work on this and wish me luck...:)
I cheated and used a standard mathematical tables integrals answer.
Plugging in 5 and 0, and subtracting revealed a line length of 26.40253635
This might actually be right, since it is a little longer than going
straight from (0,0) to (5,25), and that's because it is a parabola.
Maybe it's right! Hurray! Maybe it's not. I'm not sure yet.
This is probably all wrong, but I am learning to use LaTeX!!!
Even if this is wrong, I don't know how to integrate it?
Anyone know where I can learn this one?
...
I found "Power Substitution" on internet.
I might learn this!
...
Well If I let