Posting Freak
Posts: 6,070
Threads: 229
Joined: 2008-07
Number 3 looks like a pure theory problem. I suggest you look it up in your text book or something.
Number 4... I haven't dealt with those unfortunately. Srry =(
Senior Member
Posts: 316
Threads: 6
Joined: 2009-02
KajitiSouls Wrote:Number 3 looks like a pure theory problem. I suggest you look it up in your text book or something.
Number 4... I haven't dealt with those unfortunately. Srry =(
I just relooked at 4, and the 2nd part is easy, and the other two are doable.
I wish my Calc 2 professor wasn't one of the horsemen of the apocalypse at my school -.-;; I looked at my roommate's stuff for his calc 2 class and I was actually helping him with it, although I'm still going to make a far lower grade in the class.
Posting Freak
Posts: 4,302
Threads: 256
Joined: 2008-07
Gender: Male
Level: 251
Find
It's 1/2 but I wonder if there's an easier way than mine.
Member
Posts: 110
Threads: 3
Joined: 2008-07
Factor x^2 out of the expression under the square root and rework everything so that L'Hopital's Rule can be applied.
Posting Freak
Posts: 844
Threads: 139
Joined: 2008-08
came up with this question for my physics assignment:
What is the total mechanical energy associated with Earth's orbital motion?
again, no idea how to do it.... I decided to cheat and google the answer, and apparently 2.7*10^33 is not the right answer...
Posting Freak
Posts: 6,070
Threads: 229
Joined: 2008-07
2009-11-20, 07:14 PM
(This post was last modified: 2009-11-20, 07:16 PM by KajitiSouls.)
We needs constants to do that problem =O
Mearth = 5.97x10^24 kg (I've seen 5.98 in some books)
Msun = 1.99x10^30 kg
r = 6,371.0 km = 6.371x10^6 m
R = 1.50x10^6 km = 1.50x10^9 m
G = 6.674x10^(-11) N(m/kg)^2
Potential and Kinetic energy formulas:
KEtranslational = 1/2mv^2
KErotational = 1/2Iω^2, where I = mr^2 for masses distributed at distance r (read: point masses) on a plane, and I = 2/5mr^2 for spheres.
PEgravitational = -GM[SIZE="1"]1[/SIZE]M[SIZE="1"]2[/SIZE]/R
For the record, 2.7*10^33 is WAY too small!
I'm not sure how your teacher/book wants it done, but the rotational kinetic energy of earth orbiting the sun beats your answer by far. There's also the gravitational energy to consider and the rotation of the earth itself.
Posting Freak
Posts: 844
Threads: 139
Joined: 2008-08
^
I tried ur way and it doesn't work, after reading textbook, Etotal= -k, so I got it right with -2.7*10^33.
Another question:Two satellites are in geosynchronous orbit but in diametrically opposite positions. Into how much lower a circular orbit should one spacecraft descend if it is to catch up with the other after 10 complete orbits
I cant seem to get a number, I always get the answer with respect to radius.
Posting Freak
Posts: 6,070
Threads: 229
Joined: 2008-07
Horusmaster Wrote:^
I tried ur way and it doesn't work, after reading textbook, Etotal= -k, so I got it right with -2.7*10^33.
Another question:Two satellites are in geosynchronous orbit but in diametrically opposite positions. Into how much lower a circular orbit should one spacecraft descend if it is to catch up with the other after 10 complete orbits
I cant seem to get a number, I always get the answer with respect to radius.
How did you go about finding the answer? Reason why I didn't attempt to do it fully was because I knew there was tricky business with PEgravity. Never liked it.
As for your next question, the key word is geosynchronous. If you don't know what that means, it means that the satellite "hovers" at exactly the same spot above the planet. There's always a specific radius at which geosynchronous orbit is possible.
Posting Freak
Posts: 1,803
Threads: 85
Joined: 2008-07
Gender: Male
Sexual Orientation: Straight
Country Flag: south korea
IGN: LvlBoost
Server: Windia
Level: Pro
Job: Dawn Warrior
Guild: N/A
Guild Alliance: N/A
2009-12-02, 10:03 PM
(This post was last modified: 2009-12-02, 10:10 PM by xLeviathan.)
I need some help with Calculus problem again...it might be something related to Optimization, seeing as that's the section it's in.
It reads...
Math Textbook Wrote:Find, correct to two decimal places, the coordinates of the point on the curve y = tan x that is closest to the point (1,1).
So, you set solve y = tan x for x, which is....something, then use the distance formula for something. I have notes on it, but I'm kinda lost because they're on different levels.
Posting Freak
Posts: 4,302
Threads: 256
Joined: 2008-07
Gender: Male
Level: 251
Any point on the curve will be (x, tan x)
Distance formula: sqrt((x - 1)^2 + (tan x - 1)^2)
Differentiate and set to 0. Looks like lots of Chain Rule.
Posting Freak
Posts: 1,803
Threads: 85
Joined: 2008-07
Gender: Male
Sexual Orientation: Straight
Country Flag: south korea
IGN: LvlBoost
Server: Windia
Level: Pro
Job: Dawn Warrior
Guild: N/A
Guild Alliance: N/A
Differentiate the entire distance formula? FFS.
*Plugs in Wolfram|Alpha*. FFFFFFFF. :3
Posting Freak
Posts: 4,302
Threads: 256
Joined: 2008-07
Gender: Male
Level: 251
Well technically you can get rid of the square root, because the minimum point is the same either way. That makes it a lot easier to differentiate, and then all you have is
(x - 1)^2 + (tan x - 1)^2
2 (x - 1) + 2 (tan x - 1) (sec^2 x) = 0
x - 1 + (tan x - 1) (sec^2 x) = 0
which... you're better off solving calculator-ically anyway.
Posting Freak
Posts: 1,803
Threads: 85
Joined: 2008-07
Gender: Male
Sexual Orientation: Straight
Country Flag: south korea
IGN: LvlBoost
Server: Windia
Level: Pro
Job: Dawn Warrior
Guild: N/A
Guild Alliance: N/A
Since there's obviously going to be multiple roots given the nature of the equation, I'm assuming I use the one most relevant to the problem? E.g. the one closest to (1,1), which seems to be like .824...blahblah.
Posting Freak
Posts: 4,302
Threads: 256
Joined: 2008-07
Gender: Male
Level: 251
Posting Freak
Posts: 1,803
Threads: 85
Joined: 2008-07
Gender: Male
Sexual Orientation: Straight
Country Flag: south korea
IGN: LvlBoost
Server: Windia
Level: Pro
Job: Dawn Warrior
Guild: N/A
Guild Alliance: N/A
Thanks. I can't believe you know this off the top of your head like this. Amazing.
Posting Freak
Posts: 2,969
Threads: 56
Joined: 2009-07
[COLOR="Green"]I got this type of problem in calc today. I know the formula to use, but can't figure out what to do.
y=X^3, X[SIZE="1"]0[/SIZE]=3
Find the instantaneous velocity of the object at X[SIZE="1"]0[/SIZE].
The formula is:
lim...y={f(X) - f(X[SIZE="1"]0[/SIZE])} / (X - X[SIZE="1"]0[/SIZE])
X->X[SIZE="1"]0[/SIZE]
The numbers and equation for y are just placeholders, because they were on a test and I don't have it back yet. What exactly am I supposed to be doing here? :f6:[/COLOR]
Posting Freak
Posts: 2,621
Threads: 86
Joined: 2009-10
I'm a stupid algebra II student. 8D
I need a step by step showing of how to do;
x^2 - kx + 100 = 0
"Find the vale of k that would make the left side of the equation a perfect trinomial"
We are working these by completeing the square,
I tried that and got something like x-k/2 = -k/2 + 10i
....
Posting Freak
Posts: 6,070
Threads: 229
Joined: 2008-07
2009-12-03, 12:44 AM
(This post was last modified: 2009-12-03, 01:04 AM by KajitiSouls.)
DualReaver Wrote:[COLOR="Green"]I got this type of problem in calc today. I know the formula to use, but can't figure out what to do.
y=X^3, X[SIZE="1"]0[/SIZE]=3
Find the instantaneous velocity of the object at X[SIZE="1"]0[/SIZE].
The formula is:
lim...y={f(X) - f(X[SIZE="1"]0[/SIZE])} / (X - X[SIZE="1"]0[/SIZE])
X->X[SIZE="1"]0[/SIZE]
The numbers and equation for y are just placeholders, because they were on a test and I don't have it back yet. What exactly am I supposed to be doing here? :f6:[/COLOR]
That looks like the precursor to derivatives!
Using magical derivative magic, we find out that y' = 3x^2. y' also describes the slope of y at any point x, or in the case of movement, y' describes instantaneous velocity. Plugging in for x[SIZE="1"]0[/SIZE], we find that y' = 27.
Okay, doing it the proper way, using the formula for derivatives (which is a total PITA and impractical later on), we say that f(x) = y and plug in the numbers!
Code: y' = lim(x->x[SIZE="1"]0[/SIZE]) [f(x) - f(x[SIZE="1"]0[/SIZE])] / (x - x[SIZE="1"]0[/SIZE])
y' = lim(x->x[SIZE="1"]0[/SIZE]) [x^3 - 27] / (x - 3)
y' = lim(x->x[SIZE="1"]0[/SIZE]) x^2 + 3x + 9
y' = 3^2 + 3*3 + 9
y' = 27
Tay Wrote:I'm a stupid algebra II student. 8D
I need a step by step showing of how to do;
x^2 - kx + 100 = 0
"Find the vale of k that would make the left side of the equation a perfect trinomial"
We are working these by completeing the square,
I tried that and got something like x-k/2 = -k/2 + 10i
....
Uh, I would assume k = 20 in that case.
(x - 10)^2 = x^2 - 20x + 100
As for a step-by-step treatment...
Code: Perfect Trinomial = a^2 + 2ab + b^2
a = x
b^2 = 100
b = 10, since 10*10 = 100
2ab = 2 * x * 10
keeping in mind that the term 2ab can be negative. It doesn't matter much whether a or b is negative, since squaring any number (except i) is always positive.
If you can't recognize square values, then this can get pretty hard o.O
Posting Freak
Posts: 4,302
Threads: 256
Joined: 2008-07
Gender: Male
Level: 251
KajitiSouls Wrote:since squaring any number (except an imaginary or complex number) is always positive. Nitpick.
DualReaver Wrote:I got this type of problem in calc today. I know the formula to use, but can't figure out what to do.
y=X^3, X0=3
Find the instantaneous velocity of the object at X0.
The formula is:
lim...y={f(X) - f(X0)} / (X - X0)
X->X0
The numbers and equation for y are just placeholders, because they were on a test and I don't have it back yet. What exactly am I supposed to be doing here? You plug in the numbers and find the limit.
Posting Freak
Posts: 2,969
Threads: 56
Joined: 2009-07
Russt Wrote:Nitpick.
You plug in the numbers and find the limit.
![[Image: ygpeyfk.png]](http://mathurl.com/ygpeyfk.png)
Thanks, it looks like I had the right sort of idea when I did it. Thanks for showing me that I didn't fail.
|