OpticNerve_
Veteran X
Just one thing, instead of using radius of 3 like the problem states, i need to use 5.
I'm kind of stuck on this one, i hate IMP
. Any of you math gurus able to help me out?
Need help with a math problem
- Thread starter OpticNerve_
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Just one thing, instead of using radius of 3 like the problem states, i need to use 5.
I'm kind of stuck on this one, i hate IMP
. Any of you math gurus able to help me out?OpticNerve_
Veteran X
member...
fartiusstinkius
Veteran XV
Wow the wording on that problem is complete shit. Each dot is a tree right? I hope so.
If I'm understanding the question I would do it like this.
Okay the tree trunk area increases by 1.5 square inches a year.
pi*r^2=1.5 ----> r=sqrt(1.5/pi)
That means that r increases by sqrt(1.5/pi) inches a year.
That is the equivalent of sqrt(1.5/pi)/12 feet per year.
The initial circumference of the tree is 2.5 inches which means the intial radius of all the trees is 2.5/pi inches or 2.5/12pi feet.
So the equation for the radius of a tree is r=2.5/12pi + (sqrt(1.5/pi)/12)t; where t is the time in years and r is the radius in feet.
The line of sight will be cut off once the tree at (1,0) grows and its radius reaches that green line of sight. The other tree at (2,1) will cut off the other half of the vision at that time because it has the same distance to the line and grows at the same rate.
The radius of the tree, when it cuts off your vision will be 3.2 feet (.32 units * 10 feet/unit). So now you have the information necessary to solve the problem for t.
3.2 = 2.5/12pi + (sqrt(1.5/pi)/12)t
-----> t = (3.2 - 2.5/12pi)/((sqrt(1.5/pi)/12)
t = 54.4 years
Well I just realized that you wanted it with a garden radius of 5.
For that, all you do is that same stuff as above but you have to find the new small distance that the tree needs to grow (in the last problem is was 3.2 feet). You can tell, if you draw out a picture that it is the first and last tree that will cut off the sight first. So you have to find the small length using that information.
Okay so you know that the slope of the line will be 1/5. Last time the slope was 1/3. The small length we need is the distance from that line to the point (1,0). The shortest distance is the line from that point that crosses perpendicular to that line.
The slope of that line would be -5 (the negative reciprocal of 1/5)
It would go through the point (1,0) so we can find the equation for the line.
y = mx + b
We need to find b.
0 = -5(1) + b (just plugged in value for slope and the points for (1,0)
b = 5
So the equation is y = -5x + 5
No you have to find the intersection of this line and the first line which is
y = (1/5)x
Plug that into the first equation and you get.
(1/5)x = -5x + 5
solving for x = 5/5.2 which is approximately .9615
Pluggin in the second equation you get that y = (1/5)(5/5.2)
or y = 1/5.2
Okay now you know that the tree must grow from it's position (1,0) to (5/5.2 , 1/5.2).
Now you must find the distance in between those points. Using the pythagorean theorem:
d = sqrt((5/5.2 - 1)^2 + (1/5.2)^2)
d = .1961 units
Which means that this time the tree must grow 1.961 feet for your vision to be obscured.
The next steps are the same as above so I'm skipping straight to the equation.
r = 2.5/12pi + (sqrt(1.5/pi)/12)t
Plugging in the values:
1.961 = 2.5/12pi + (sqrt(1.5/pi)/12)t
-----> t = (1.961 - 2.5/12pi)/((sqrt(1.5/pi)/12)
t = 32.9 years
So it should take 32.9 years of growth of the tree to obscure your vision.
If I fucked something up than that's because of the retarded wording of the problems or because I fucked up my numbers. Either way I don't give a shit I put enough thought into this problem already.
edit: I have to say. That was the slowest fucking quick reply I've ever posted.
If I'm understanding the question I would do it like this.
Okay the tree trunk area increases by 1.5 square inches a year.
pi*r^2=1.5 ----> r=sqrt(1.5/pi)
That means that r increases by sqrt(1.5/pi) inches a year.
That is the equivalent of sqrt(1.5/pi)/12 feet per year.
The initial circumference of the tree is 2.5 inches which means the intial radius of all the trees is 2.5/pi inches or 2.5/12pi feet.
So the equation for the radius of a tree is r=2.5/12pi + (sqrt(1.5/pi)/12)t; where t is the time in years and r is the radius in feet.
The line of sight will be cut off once the tree at (1,0) grows and its radius reaches that green line of sight. The other tree at (2,1) will cut off the other half of the vision at that time because it has the same distance to the line and grows at the same rate.
The radius of the tree, when it cuts off your vision will be 3.2 feet (.32 units * 10 feet/unit). So now you have the information necessary to solve the problem for t.
3.2 = 2.5/12pi + (sqrt(1.5/pi)/12)t
-----> t = (3.2 - 2.5/12pi)/((sqrt(1.5/pi)/12)
t = 54.4 years
Well I just realized that you wanted it with a garden radius of 5.
For that, all you do is that same stuff as above but you have to find the new small distance that the tree needs to grow (in the last problem is was 3.2 feet). You can tell, if you draw out a picture that it is the first and last tree that will cut off the sight first. So you have to find the small length using that information.
Okay so you know that the slope of the line will be 1/5. Last time the slope was 1/3. The small length we need is the distance from that line to the point (1,0). The shortest distance is the line from that point that crosses perpendicular to that line.
The slope of that line would be -5 (the negative reciprocal of 1/5)
It would go through the point (1,0) so we can find the equation for the line.
y = mx + b
We need to find b.
0 = -5(1) + b (just plugged in value for slope and the points for (1,0)
b = 5
So the equation is y = -5x + 5
No you have to find the intersection of this line and the first line which is
y = (1/5)x
Plug that into the first equation and you get.
(1/5)x = -5x + 5
solving for x = 5/5.2 which is approximately .9615
Pluggin in the second equation you get that y = (1/5)(5/5.2)
or y = 1/5.2
Okay now you know that the tree must grow from it's position (1,0) to (5/5.2 , 1/5.2).
Now you must find the distance in between those points. Using the pythagorean theorem:
d = sqrt((5/5.2 - 1)^2 + (1/5.2)^2)
d = .1961 units
Which means that this time the tree must grow 1.961 feet for your vision to be obscured.
The next steps are the same as above so I'm skipping straight to the equation.
r = 2.5/12pi + (sqrt(1.5/pi)/12)t
Plugging in the values:
1.961 = 2.5/12pi + (sqrt(1.5/pi)/12)t
-----> t = (1.961 - 2.5/12pi)/((sqrt(1.5/pi)/12)
t = 32.9 years
So it should take 32.9 years of growth of the tree to obscure your vision.
If I fucked something up than that's because of the retarded wording of the problems or because I fucked up my numbers. Either way I don't give a shit I put enough thought into this problem already.
edit: I have to say. That was the slowest fucking quick reply I've ever posted.
fartiusstinkius
Veteran XV
Well I figured all I would do is whore tribalwar anyway. Might as well do something productive.
fartiusstinkius
Veteran XV
You get that optic? Don't want my post to go to waste.
fartiusstinkius
Veteran XV
okay i'm out i hope you understand that optic
OpticNerve_
Veteran X
Just about. Quick question, wth does sqrt stand for? Square root? Other than that big thanks go out to you
fartiusstinkius
Veteran XV
no problem. it's good exercise for my brain anyway
fartiusstinkius
Veteran XV
I wish I was good at writing...we all have our strengths.naptown said:I wish I was good at math.
edit: hey pressing alt + s is the same as post on quick post...saves me even more hz when i'm quick posting