Report, edit, etc...Posted by HolySin on 2006-10-10 at 23:47:56
Why not? Isn't trigonometry fun?
Report, edit, etc...Posted by Night on 2006-10-10 at 23:48:19
QUOTE(Syphon(MM) @ Oct 10 2006, 09:39 PM)
The vertex is the axis of symmetry. (Atleast in parabolas it is.)
[right][snapback]574782[/snapback][/right]
[right][snapback]574782[/snapback][/right]
I'm not saying that your wrong, Im just saying that you have just stated the Axis of symmetry (thx for correcting), we need the vertex, all you've said is the x value ( p ) we needed both x and y (p and q)
Here's an intersection question:
x>0 = y (y is binary)
-x²+9 = y
Intersection of these lines are... ?
ADDITION: Woops didnt see that another equation was made, well after his do mine. I'm not doing that question cuz it's triginometry and it sucks lol.
Report, edit, etc...Posted by HolySin on 2006-10-10 at 23:53:37
It's just related rates...
Report, edit, etc...Posted by CheeZe on 2006-10-10 at 23:58:23
Let me see if I have this set up correctly:
tan(y) = 10,000/x
thus,
1/cos(y)^2 dy = 10,000(-1/x^2) dx
Plug and solve for dx right?
*Runs off to find calculus book*
tan(y) = 10,000/x
thus,
1/cos(y)^2 dy = 10,000(-1/x^2) dx
Plug and solve for dx right?
*Runs off to find calculus book*Report, edit, etc...Posted by HolySin on 2006-10-11 at 00:12:01
I could be nit-picky and such, but yes, it's just using the derivative to find (dx/dt).
Report, edit, etc...Posted by Night on 2006-10-11 at 00:24:24
QUOTE(Night @ Oct 10 2006, 09:47 PM)
x>0 = y (y is binary)
-x²+9 = y
[right][snapback]574790[/snapback][/right]
-x²+9 = y
[right][snapback]574790[/snapback][/right]
Here you go, the two intersections: ( ~2.8 , 1 ) and ( -3 , 0 ).
Yes, x>0 = y can be graphed with a line. (Basically a straight line that is either y=1 (true) or y=0 (false) or a mix of the two)
Report, edit, etc...Posted by Zeratul_101 on 2006-10-11 at 00:26:56
did you just solve your own question...
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-11 at 01:03:43
QUOTE(HolySin @ Oct 10 2006, 08:41 PM)
Let me screw this up for you guys with the older sibling of algerbra, Very Basic Calculus:
An airplane at an altitude of 10,000 ft is flying at a constant speed on a line that will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60º and is increasing at a rate of 1º per second, find the speed of the airplane.
[right][snapback]574784[/snapback][/right]
An airplane at an altitude of 10,000 ft is flying at a constant speed on a line that will take it directly over an observer on the ground. If, at a given instant, the observer notes that the angle of elevation of the airplane is 60º and is increasing at a rate of 1º per second, find the speed of the airplane.
[right][snapback]574784[/snapback][/right]
dθ/dt = π/180 (radians/sec)
------
tan θ = 10000/x
x = 10000cotθ
dx/dθ = -10000csc[sup]2[/sup]θ
------
dx/dt = (dθ/dt)(dx/dθ) = π/180 * -10000csc[sup]2[/sup]θ [θ=60°=π/3 radians]
dx/dt = π/180 * -13333.333 ft/sec
dx/dt ~= -232.711 ft/sec
The airplane's speed is approximately 232.711 feet per second.
A pot of water is put on a stove which supplies heat to the water a constant rate of 15,000 Joules per second. Unfortunately, a rookie cook forgets to put a lid on the pot, so heat begins to escape from the water at a rate proportional to the amount of heat in the water. Assume that the amount of relative heat in the pot at time t = 0 is equal to 0 Joules, and that after 2 minutes the amount of heat in the pot is 28,000 Joules, find h(t), the amount of relative heat in Joules in the pot after t minutes. How much heat is in the pot after 10 minutes? After the pot is left sitting for an entire day (24 hours)?
Report, edit, etc...Posted by Syphon on 2006-10-11 at 08:11:04
QUOTE(HolySin @ Oct 10 2006, 10:47 PM)
Why not? Isn't trigonometry fun?
[right][snapback]574787[/snapback][/right]
[right][snapback]574787[/snapback][/right]
Yes sir.
ADDITION:
QUOTE(DT_Battlekruser @ Oct 11 2006, 12:03 AM)
dθ/dt = π/180 (radians/sec)
------
tan θ = 10000/x
x = 10000cotθ
dx/dθ = -10000csc[sup]2[/sup]θ
------
dx/dt = (dθ/dt)(dx/dθ) = π/180 * -10000csc[sup]2[/sup]θ [θ=60°=π/3 radians]
dx/dt = π/180 * -13333.333 ft/sec
dx/dt ~= -232.711 ft/sec
The airplane's speed is approximately 232.711 feet per second.
A pot of water is put on a stove which supplies heat to the water a constant rate of 15,000 Joules per second. Unfortunately, a rookie cook forgets to put a lid on the pot, so heat begins to escape from the water at a rate proportional to the amount of heat in the water. Assume that the amount of relative heat in the pot at time t = 0 is equal to 0 Joules, and that after 2 minutes the amount of heat in the pot is 28,000 Joules, find h(t), the amount of relative heat in Joules in the pot after t minutes. How much heat is in the pot after 10 minutes? After the pot is left sitting for an entire day (24 hours)?
[right][snapback]574816[/snapback][/right]
------
tan θ = 10000/x
x = 10000cotθ
dx/dθ = -10000csc[sup]2[/sup]θ
------
dx/dt = (dθ/dt)(dx/dθ) = π/180 * -10000csc[sup]2[/sup]θ [θ=60°=π/3 radians]
dx/dt = π/180 * -13333.333 ft/sec
dx/dt ~= -232.711 ft/sec
The airplane's speed is approximately 232.711 feet per second.
A pot of water is put on a stove which supplies heat to the water a constant rate of 15,000 Joules per second. Unfortunately, a rookie cook forgets to put a lid on the pot, so heat begins to escape from the water at a rate proportional to the amount of heat in the water. Assume that the amount of relative heat in the pot at time t = 0 is equal to 0 Joules, and that after 2 minutes the amount of heat in the pot is 28,000 Joules, find h(t), the amount of relative heat in Joules in the pot after t minutes. How much heat is in the pot after 10 minutes? After the pot is left sitting for an entire day (24 hours)?
[right][snapback]574816[/snapback][/right]
If 2t = 28,000h, then (assuming it's a linear increase, which I do.) h = 14,000t.
10 minutes would be 140,000 joules. An entire day is 20,160,000?
I'm pretty sure I shouldn't assume linearity, should I?
Report, edit, etc...Posted by Zeratul_101 on 2006-10-11 at 10:09:40
probably not, he did mention porportionality, so it'd probly be more j-curve like. at least thats what i think.
Report, edit, etc...Posted by Lord_Agamemnon(MM) on 2006-10-11 at 12:39:41
Ouch. It makes my head hurt. I hate recursive series. 
Report, edit, etc...Posted by Deathawk on 2006-10-11 at 14:21:30
5 sojs + 4 sojs = 10 high runes. Solve for sojs.
Report, edit, etc...Posted by Lord_Agamemnon(MM) on 2006-10-11 at 15:24:37
QUOTE(Deathawk @ Oct 11 2006, 01:21 PM)
5 sojs + 4 sojs = 10 high runes. Solve for sojs.
[right][snapback]574984[/snapback][/right]
[right][snapback]574984[/snapback][/right]
Um. 1 soj = 10/9 high rune...
Report, edit, etc...Posted by HolySin on 2006-10-11 at 18:45:32
QUOTE(Syphon(MM) @ Oct 11 2006, 06:10 AM)
I'm pretty sure I shouldn't assume linearity, should I?
[right][snapback]574885[/snapback][/right]
[right][snapback]574885[/snapback][/right]
Maybe this picture will help!
[attachmentid=21442]
Report, edit, etc...Posted by Syphon on 2006-10-11 at 21:00:53
...Double helix?
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-11 at 21:11:25
QUOTE(Syphon(MM) @ Oct 11 2006, 05:10 AM)
Yes sir.
ADDITION:
If 2t = 28,000h, then (assuming it's a linear increase, which I do.) h = 14,000t.
10 minutes would be 140,000 joules. An entire day is 20,160,000?
I'm pretty sure I shouldn't assume linearity, should I?
[right][snapback]574885[/snapback][/right]
ADDITION:
If 2t = 28,000h, then (assuming it's a linear increase, which I do.) h = 14,000t.
10 minutes would be 140,000 joules. An entire day is 20,160,000?
I'm pretty sure I shouldn't assume linearity, should I?
[right][snapback]574885[/snapback][/right]
The addition of heat is linear, but as soon as there is heat in the water, it starts escaping at a rate proportional to the amount of heat in the water.
Report, edit, etc...Posted by Syphon on 2006-10-11 at 21:39:22
QUOTE(DT_Battlekruser @ Oct 11 2006, 08:11 PM)
The addition of heat is linear, but as soon as there is heat in the water, it starts escaping at a rate proportional to the amount of heat in the water.
[right][snapback]575298[/snapback][/right]
[right][snapback]575298[/snapback][/right]
What exactly is that rate?
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-11 at 22:14:13
Here's a hint: The derivative of heat with respect to time, dh/dt, is equal to the derivative of the input minus the derivative of the output.
Therefore we have dh/dt = 15000 - kh
We also know that h(0) = 0 and that h(2) = 28000.
Therefore we have dh/dt = 15000 - kh
We also know that h(0) = 0 and that h(2) = 28000.
Report, edit, etc...Posted by Chef on 2006-10-11 at 22:57:15
Not even cosplay is more nerdy than this thread. I challange any of you to find something on the internet more nerdy than this. Easy, time comsuming, and boring math... and you're not even doing it to get answers for your homework.
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-11 at 23:06:01
Time consuming? I did the airplane problem in like 3 mintes flat.
Report, edit, etc...Posted by HolySin on 2006-10-11 at 23:09:55
I find math fun, may or may not be time consuming, and easy or challenging...
Report, edit, etc...Posted by Chef on 2006-10-11 at 23:11:02
Three minutes is a long time for a single peice of information =P It's like taking three minutes to detemine how many fingers you have at the moment.
Report, edit, etc...Posted by Syphon on 2006-10-12 at 08:18:27
QUOTE(SexyPinkPrincess @ Oct 11 2006, 09:56 PM)
Not even cosplay is more nerdy than this thread. I challange any of you to find something on the internet more nerdy than this. Easy, time comsuming, and boring math... and you're not even doing it to get answers for your homework.
[right][snapback]575410[/snapback][/right]
[right][snapback]575410[/snapback][/right]
Any website related to physical activity. Especially ones with boards.
Report, edit, etc...Posted by DT_Battlekruser on 2006-10-12 at 21:36:50
Asnwer the problem, anyone?
Report, edit, etc...Posted by CheeZe on 2006-10-12 at 23:32:27
DTBK. You are a showoff. Stop it.