The math help thread - Printable Version

The math help thread - Printable Version

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The math help thread - Salguod - 2009-12-10

Unless I'm reading it wrong, I believe the key is that she gets $7500 every 6 months, and therefore twice a year. The first increase is enough to tie with Jack's income and the second increase of the year is the extra. Year 1: Jack= 15000, Jill = 7500+ 7750 = 15250. She gets $250 more every year, so there's the $2500.

The math help thread - Corn - 2009-12-10

Salguod Wrote:Unless I'm reading it wrong, I believe the key is that she gets $7500 every 6 months, and therefore twice a year, so there isn't that big gap in initial income.

Oh....

I can't believe I missed that.

Wow, thanks.

The math help thread - Hazzy - 2009-12-14

http://www.youtube.com/watch?v=LkCNJRfSZBU#t=01m11s

"Number crunch"
That suggest math is involved, not just a random guess.

Where'd he get that number?

The math help thread - Russt - 2009-12-14

Statistics.

But 32.33 repeating sounds pretty fake. Biggrin

The math help thread - Horusmaster - 2009-12-14

studying for physics exam, I stumple upon this equation on the textbook that doesn't make sense:

first it says:

m1(v1i-v1f)=m2(v2f-v2i)

then it says:

m1(v1i^2-v1f^2)=m2(v2f^2-v2i^2)

I don't get how they come up with the 2nd equation...

The math help thread - Nikkey - 2009-12-14

Horusmaster Wrote:studying for physics exam, I stumple upon this equation on the textbook that doesn't make sense:

first it says:

m1(v1i-v1f)=m2(v2f-v2i)

then it says:

m1(v1i^2-v1f^2)=m2(v2f^2-v2i^2)

I don't get how they come up with the 2nd equation...

Only applies if m1 = m2, if I understand the formula correct. Err, sec. Nah. Mind dragging out what it says?

The math help thread - Michaels2 - 2009-12-14

Horusmaster Wrote:studying for physics exam, I stumple upon this equation on the textbook that doesn't make sense:

first it says:

m1(v1i-v1f)=m2(v2f-v2i)

then it says:

m1(v1i^2-v1f^2)=m2(v2f^2-v2i^2)

I don't get how they come up with the 2nd equation...

Formula #1 looks like conservation of momentum to me, while #2 looks like conservation of energy.
#1 = loss of momentum from 1st body is gained by 2nd body.
#2 = loss of energy (kinetic) from 1st body is gained by 2nd body.

The math help thread - FallenMemory - 2009-12-14

Find the Deritive:

1. g(t)= e^(1+3t)^2
2. y= te^-t^2

for these problems (e) is that little number 2.71 or something. The exponents inside are confusing me. I know I have to use chain rule though.

The math help thread - Horusmaster - 2009-12-14

FallenMemory Wrote:Find the Deritive:

1. g(t)= e^(1+3t)^2
2. y= te^-t^2

for these problems (e) is that little number 2.71 or something. The exponents inside are confusing me. I know I have to use chain rule though.

1. g'(t)=e^(1+3t)^2*2(1+3t)*3
2. y'=e^-t^2+te^-t^2*2(-t)*-1

Also I got a probability question:
There's a 100% chance i get <60% for psych.
There's a 36% chance i get <50% for psych.
There's a 35% chance i get <60% for calc.
There's a 3% chance i get <50% for calc.
There's a 60% chance i get <60% for physics.
There's a 20% chance i get <50% for physics.

I will get kicked out of uni if I fail 2 course (get <50%). or if I fail 1 course and get <60% on 2 other courses.
What's my chance of getting kicked out.

The math help thread - FallenMemory - 2009-12-14

In relation to the problem I posted above there's these 3 ones I think I have the answer to, but I want a check just to be sure.

Find the Deritive:
1. z= sqrt(w)/w^2+1
2. g(x)= 25x^2/e^x
3. f(z)= z^2+1/sqrt(z)

I've got them down to the last step, but I think I'm messing up the simplification.

@ horusmaster, Ima try to figure out that probability one. I'll edit them in if I get an answer.

The math help thread - Russt - 2009-12-15

Horusmaster Wrote:1. g'(t)=e^(1+3t)^2*2(1+3t)*3
2. y'=e^-t^2+te^-t^2*2(-t)*-1

Also I got a probability question:
There's a 100% chance i get <60% for psych.
There's a 36% chance i get <50% for psych.
There's a 35% chance i get <60% for calc.
There's a 3% chance i get <50% for calc.
There's a 60% chance i get <60% for physics.
There's a 20% chance i get <50% for physics.

I will get kicked out of uni if I fail 2 course (get <50%). or if I fail 1 course and get <60% on 2 other courses.
What's my chance of getting kicked out.

For each of the three classes you have three disjoint possibilities: more than 60% (call this 'A'), between 50 and 60% (B), and below 50% ©. Also, for the final result you have two possibilities: pass (P) and fail (F).
F = 2 C and 1 anything, or 1 C and 2 B
P = 3 B or higher, or 1 C, 1 A, 1 B or higher

So, making a tree:
0% psych-A
64% psych-B
-> 65% calc-A (P)
-> 32% calc-B
--> 40% physics-A (P)
--> 40% physics-B (P)
--> 20% physics-C (F)
-> 3% calc-C
--> 40% physics-A (P)
--> 40% physics-B (F)
--> 20% physics-C (F)
36% psych-C
-> 65% calc-A
--> 40% physics-A (P)
--> 40% physics-B (P)
--> 20% physics-C (F)
-> 32% calc-B
--> 40% physics-A (P)
--> 40% physics-B (F)
--> 20% physics-C (F)
-> 3% calc-C (F)

Summing up all the F's...
0.64 * 0.32 * 0.2
0.64 * 0.03 * 0.6
0.36 * 0.65 * 0.2
0.36 * 0.32 * 0.6
0.36 * 0.3
= 0.2764 = 27.64% chance.

The math help thread - Horusmaster - 2009-12-18

thx ruust, that's kinda higher than I expected... but oh well, what's done is done

I just wrote the algebra exam, while i'm 100% sure I'll get >60%
there's this question that I have no idea how to do:

prove: gcd (5^98+3, 5^99+1) =14

The math help thread - Nikkey - 2009-12-18

You could use Fermat's little Theorem and prove that
[Image: ycu3vhk.png]

(And that both equal to 0)

Though that doesn't prove that there aren't a higher number which divides both numbers.

The math help thread - Russt - 2009-12-18

a = 5^98+3
b = 5^99+1

b isn't divisible by 5, so gcd(a, b) = gcd(5a, b)
5a = 5^99+15

5a-b is 14, so the gcd must be a factor of 14 (1, 2, 7, or 14). Both numbers are even, so they're divisible by 2 -> gcd is divisible by 2. Now to prove that it equals 14, you have to show that they're divisible by 7.

... -shrugs-

The math help thread - Hazzy - 2009-12-18

Code:
if d>=P value: reduction = 0.7*(L/1300 + 8/9*n)*(d-P) + (n*d)

else:

-if l < 15: reduction = 1.3*(L/550 + n)*(d-P) + (n*d)

-else: (13/(L-2))*(L/550 + n)*(d-P) + (n*d)

I want to rewrite this to express d in terms of n, P, and L. Absolutely no idea how.

The math help thread - Russt - 2009-12-18

Several roots.

reduction = 0.7*(L/1300 + 8/9*n)*(d-P) + (n*d)
= 0.7*(L/1300 + 8/9*n)*d + n*d - 0.7*(L/1300 + 8/9*n)*P
reduction + 0.7*(L/1300 + 8/9*n)*P = 0.7(L/1300 + 146/63*n)*d
1. d = (reduction + 0.7P*(L/1300 + 8/9*n)) / 0.7(L/1300 + 146/63*n)
(only if resulting expression is greater than or equal to P)

reduction = 1.3*(L/550 + n)*(d-P) + n*d
= 1.3*(L/550+n)*d - 1.3*(L/550+n)*P + n*d
reduction + 1.3P*(L/550+n) = (1.3*L/550 + 2.3*n)*d
2. d = (reduction + 1.3P*L/550 + 1.3*n)) / (1.3*L/550 + 2.3*n)
(only if L < 15 and the resulting expression is less than P)

Same thing with the third one I'm too lazy
-goes to play S4-

The math help thread - Matt - 2009-12-20

Got two more variable problems. I always screw up on these for some reason - I've solved similar ones with numbers, but I just get lost when we change to variables.

Find the maximum volume of a right-circular cone placed upside-down in a right circular cone of radius R and height H. http://img451.imageshack.us/img451/6751/optimizationfi3.jpg

Find the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W.
http://img685.imageshack.us/img685/530/rectyrec.png

Thanks

The math help thread - KajitiSouls - 2009-12-20

Matt Wrote:Find the maximum volume of a right-circular cone placed upside-down in a right circular cone of radius R and height H. http://img451.imageshack.us/img451/6751/optimizationfi3.jpg

Fun stuff this be.

Volume of a right-circular cone = 1/3*Π*r^2*h

The dimensions of the upside-down cone can be written in terms of the standing cone's dimensions.

Code:
r = r; 0 ≤ r ≤ R

h = H - H/R*r

  = H*(1 - r/R)

V = 1/3*Π*R^2*H; volume of the large cone

v = 1/3*Π*r^2*H*(1 - r/R); volume of the small cone

Then we find out at which value r within 0 ≤ r ≤ R can yield the largest coefficient value for the other two variables, H and R. To do this, we take the derivative of the volume function and find out where it equals 0. Points where the derivative equals 0 indicates minimums and maximums.

Code:
v' = 2/3*Π*r*H - Π*r^2*H/R

v' = Π*r*H*(2/3 - r/R)

If r = 2/3*R, [color=Red]v = 4/81*Π*H[/color]

If r = R, v = 0

If r = 0, v = 0

Matt Wrote:Find the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W.
http://img685.imageshack.us/img685/530/rectyrec.png

Now this one is considerably trickier.

To determine the new dimensions of the circumscribed rectangle based on the original rectangle's dimensions L and W, and the variable θ, we'll apply geometry and trigonometry.

Code:
sin(θ) = o/h

cos(θ) = a/h

For both cases, h will either be L or W.

X = sin(θ)*W + cos(θ)*L; new rectangle's length

Y = sin(θ)*L + cos(θ)*W; new rectangle's width

V = X*Y

  = sin(θ)^2*L*W + cos(θ)^2*L*W + sin(θ)*cos(θ)*(W^2 + L^2)

  = L*W + sin(θ)*cos(θ)*(W^2 + L^2)

We'll do the same trick as with the cone problem: find the derivative of the volume function based off of θ.

Code:
V' = (-sin(θ)^2 + cos(θ)^2)*(W^2 + L^2)

If θ = Π/4 = 45°, V' = 0.

[color=Red]V = L*W + 1/2(W^2 + L^2)[/color]

The math help thread - HooKarez - 2009-12-20

Find the volume of the largest right circular cylinder that can be inscribed in a sphere of radius r.

I need better calc skills. This is pissing me off that I can't solve this. :[

The math help thread - Horusmaster - 2009-12-20

just wondering what's right circular cylinder?

edit: nvm just a fancy name for cylinder >.>
working on it now
$[Image: f4158dab7968d0898a95f83495604c73.png]$
[Image: Untitled-10.png]

let R be radius of sphere
and r be radius of cylinder
so basically h=2Rsin@ and r=Rcos@

so volume= piRcos^2@2Rsin@

and differentiate and solve