Math Is Fun Forum

wow!!  12 pages???  that's crazy!!!  i'd be like, UP YOURS TEACHER!!!

ohhh... so that's what "total present value" is.  my mistake.

btw... why hasen't n e one said something nice bout my avatar?? u guys don't like it??

hi yaz nick!!

hmmm... i'm not too sure how to compute compund interest but this is how i would do it:

use the compund interest formula:
A = P(1 + r/n)^(nt)
where t is in years, P is principal invested, and r is annual interest compunded n times per year.

our initial investment (P) is $5000, the interest is r=.06 compunded annually n=1, and t=3 (because we will be receiving more money after three years).  plug these guys into the formula:

5000(1 + .06/1)^(1*3) = 5955.08

after three years, we will receive $5000 more so add that to 5955.08 which gives us 10955.08.  we use the formula again but this time taking P=10955.08 and t=1.

10955.08(1+.06)^(1) = 11612.38

in the fourth year, we will receive an aditional $5000 so the total present value is:
11612.38 + 5000 = 16612.38

applying logarithms into a complicated function can make the function much easier.  for example:

y= [x^(3/4)√ (x² + 1)]/(3x + 2)^5

apply log to both sides

lny = 3/4lnx + 1/2lx(x² + 1) - 5ln(3x + 2)

all we gotta do now is add and subtract which is much easier than the original one.

hi yaz lloyd!

i can't remember how to do these exactly, but i think it should go something like this:

2tan²θ + 7tanθ = -3

2tan²θ + 7tanθ + 3 = 0

2(tan²θ + 7/2tanθ + 3/2) = 0

2(tanθ + 1/2)(tanθ + 3) = 0

tanθ = -1/2           tanθ = -3
θ = tan-¹(-1/2)      θ = tan-¹(-3)

hmm... i guess u need a calculator to find θ

hmm... well the book did give one lousy example:

find a vector function that represents the curve of intersection of the cylinder x² + y² = 1 and the plane y + z = 2

answer: the projection of C (curve of intersection) onto the xy plane is the circle x² + y² = 1, z=0.  so we know that we can write
                                     x=cost  y=sint       0≤t≤2pi
from the equation of the plane we have z = 2 - y = 2 - sint

so we can write parametric equations for C as
       x=cost  y=sint  z= 2 - sint     0≤t≤2pi
the corresponding vector equation is
       r(t) = costi + sintj + (2-sint)k     0≤t≤2pi

i hope someone can make sense out of it.

hi yaz matilde!!

oh geeez.... this is a tuff one!!  i will still give it a try!!!

∫e^(2x)sin²xdx                            sin²x = 1/2(1 - cos2x)

∫e^(2x)[1/2(1 - cos2x)]dx            let u=2x so du=2dx

1/2∫e^(u)[1/2(1 - cosu)]du

1/4∫e^(u)(1 - cosu)du

1/4∫e^(u)du - 1/4∫e^(u)cosudu

-1/4(∫e^(u)cosudu - ∫e^(u)du)                use integration by parts on ∫e^(u)cosudu
                                                             f(u) = e^(u)          g'(u) = cosu
                                                             f'(u) = e^(u)         g(u) = sinu
                                                            ∫f(u)g'(u)du = f(u)g(u) - ∫g(u)f'(u)du

-1/4[(e^(u)sinu - ∫e^(u)sinudu) - ∫e^(u)du]     use integration by parts on ∫e^(u)sinudu
                                                                     f(u) = e^(u)             g'(u) = sinu
                                                                     f'(u) = e^(u)            g(u) = -cosu

-1/4[{e^(u)sinu - (-e^(u)cosu + ∫e^(u)cosudu)} - ∫e^(u)du]

-1/4[(e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu) - ∫e^(u)du]

observe that ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu - ∫e^(u)cosudu so

∫e^(u)cosudu + ∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
2∫e^(u)cosudu = e^(u)sinu + e^(u)cosu
∫e^(u)cosudu = 1/2(e^(u)sinu + e^(u)cosu)

finally...

-1/4{[1/2(e^(u)sinu + e^(u)cosu)] - e^(u)}

you can plug u=2x back in there if ya want.  i may have done it all wrong or there may be an easier way of doing it   who knowns???

ahhh... thank you mathsisfun the sailor, math, and word person!!!

is this what your looking for??
http://www.mathsisfun.com/equation_of_line.html

ya me too!

i feel comforted now.

my name comes from my ex, carlos.  he broke up w/ me cuz i wasn't spending nuff time w/ him.  now i'm desperately trying to get back with him.  so when i registered my name i was thinking of him.

hmm... it does sound like the book flowers for algernon.  it's a good book imo!!!