Calculus

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Into Infinity United States. Dec 13 2009 20:38. Posts 1884

i need help for my upcoming calculus final. a couple problems i dont get... i'd appreciate it if anyone can help...

the teacher gave the answers, but obviously i don't know how to show the work for it.

edit: for #4, the top x is supposed to be a dx.

answers here:

+ Show Spoiler +

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Last edit: 13/12/2009 20:43

SugoGosu Korea (South). Dec 13 2009 20:41. Posts 1793

rofl i forgot all of calc. last semester after it finished. shit....

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genjix China. Dec 13 2009 20:46. Posts 2677

1. find a formula for the area of that box + circles.

then differentiate and find the maxima- thats where dy/dx = 0

2. (5+1)*(1+1)*(5-1)/2 (area of a trapezoid)

3. look it up the formulas and whatnot

4. split that fraction into 2 parts... integ(1/x) = lnx

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Into Infinity United States. Dec 13 2009 20:49. Posts 1884

for number 2, you're saying 6*2*4/2? the answer is 16 though.

nixxxbg Bulgaria. Dec 13 2009 21:12. Posts 436

Here's prob 2

domyouji Zimbabwe. Dec 13 2009 21:17. Posts 435

for some reason that field looks like a weiner

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Jonoman92 United States. Dec 13 2009 21:17. Posts 280

For problem one I think I can tell you how to start:

It's an optimization problem:
-first off 2x + 2(pi)(y/2)=440 (that is the total perimeter and is your constraint) (This is the straight sides (2x) plus the semicircle curves, (y/2) is the radius, 2(pi)(r) is circumference)
-secondly the equation you want to optimize is just xy=Area

From here I think you need to use substitution to plug one equation into the other only using one variable (at least this is how I would do it, it's not the only way.) ie: -(2(pi)(y/2) - 440)/2=x
Dunno if what I just said is actually right.

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nixxxbg Bulgaria. Dec 13 2009 21:23. Posts 436

Prob 3:
Let u = sin(x); du = cos(x)dx

Then: INT u^2 du = u^3/3 + const = (sin(x)^3)/3 + const

nixxxbg Bulgaria. Dec 13 2009 21:27. Posts 436

Prob 4:

d/dx (log(1-x)) = -1 / (1-x)

Then:

INT 1/(1-x) dx = -log(1-x) between 0 and 3/4 = -(log(1-3/4) - log(1)) = -log(1/4) = -(log(1) - log(4)) = log(4)

**Note1: if you dont see this, let u = 1-x, du = -dx

**Note2: log(1) = 0

Patrocle France. Dec 16 2009 13:31. Posts 623

PM me if you are still lost.

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Hawkansson Sweden. Dec 18 2009 21:35. Posts 36

Your mistake in the answer for problem 3 is that u forget the inner derivative of sinx that is exactly cosx.

DER( (sinx)^3/3) = (sinx)^2 * cosx

Hawkansson Sweden. Dec 18 2009 22:11. Posts 36

On December 13 2009 20:17 Jonoman92 wrote:
For problem one I think I can tell you how to start:

It's an optimization problem:
-first off 2x + 2(pi)(y/2)=440 (that is the total perimeter and is your constraint) (This is the straight sides (2x) plus the semicircle curves, (y/2) is the radius, 2(pi)(r) is circumference)
-secondly the equation you want to optimize is just xy=Area

From here I think you need to use substitution to plug one equation into the other only using one variable (at least this is how I would do it, it's not the only way.) ie: -(2(pi)(y/2) - 440)/2=x
Dunno if what I just said is actually right.

Think that's right. After that

y*x = -y*(2(pi)(y/2) - 440)/2 = (440y - pi*y^2)/2

which is the function to maximize, and how do we maximize a negative second degree, yeah we derive it and set it to zero and there's our answer..