IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: mousa on September 24, 2009, 07:15:40 pm
-
hey everyone , well i got stuck on those questions , astar, nid.. anybody..
help me out!..plz i need workin out with reasoning..
1, let p,with coorinates (p,q), be a fixed point on the 'curve ' with equation y=mx+c and let Q with coordinates (r,s), be any other point on y=mx+c . use the fact thar the coordinates of p and Q satisfy the equation y=mx+c to show that the gradient of PQ is m for all positions of Q.
2 there are some values of a, b and c for which the equation ax+by+c=0 does not represent a straight line. give an example of such values
3 two lines have equations y=m1x+c1 and y=m2x+c2 and m1m2=-1. prove that the lines are perpendicular.
Thanks in advance..
-
Will answer his tom definitely....have an exam tom..so sorry...tom for sure
-
I answered these questions somewhere else. m1m2=-1 is the def of perpendicular.
-
the answer for the third one is dere in the question itself..
if m1*m2 = -1
the lines are perpendicular...
-
the answer for the third one is dere in the question itself..
if m1*m2 = -1
the lines are perpendicular...
nope the answer aint in the question...........we did this q in schl a few days ago.....the teacher gave us a weird sort of proof....ill post it when i find it :)
-
this is what i remember from my add. math book
"two non-vertical lines with gradients m1 and m2 are perpendicular if m1*m2 = -1"
-
oh no adi adi it's not as simple as that.....
I'll do it in sometime...I'm just back from an exam which i am sure to flunk......I mixed my reagents and my practicals was such a mess >:( >:( >:( :( :( :'( :'( :'(
-
oh no adi adi it's not as simple as that.....
I'll do it in sometime...I'm just back from an exam which i am sure to flunk......I mixed my reagents and my practicals was such a mess >:( >:( >:( :( :( :'( :'( :'(
ooohh1!! nid, chem practical?
-
ohh..okk...sure..
-
but i am sure that m1*m2 means perpendicular
-
the third one is 0...
a=0 b=0 c=0
-
ooohh1!! nid, chem practical?
yeah....why can't they name the damn things with the name of the substance rather than complicate it....FA1, FA2....F**k u...
and here's the m1m2=-1 proof
Slope of a given line= tan theta (delta y/ delta X)
So if there are two slopes m1 and m2
1 having inclination of theta, then m1=tan theta....Other line has inclination pi/2 + theta......i hope you can visualize this
then m2= tan (pi/2 +theta)=-cot theta(-1/tan theta) => m1m2=-1
-
let p,with coorinates (p,q), be a fixed point on the 'curve ' with equation y=mx+c and let Q with coordinates (r,s), be any other point on y=mx+c . use the fact thar the coordinates of p and Q satisfy the equation y=mx+c to show that the gradient of PQ is m for all positions of Q.
ok..
Same logic...
m=tan theta
P=(p,q)
Q=(r,s)
tan theta= s-q / r-p
For any value of (r,s) on y=mx+c, (tan theta) will be constant, hence m will be constant
-
Maybe they meant to start from pre first principles ie the principle before the principle
-
Maybe they meant to start from pre first principles ie the principle before the principle
didn't get that ???
-
ok mousa here
check the diagram and tell me if you get it or not
-
I think this should be able to clear your doubt
-
I think this should be able to clear your doubt
i cant view. i guess have a problem.. but still i can reead it, Thanks alot alotttt nid ... ;D
-
i cant view. i guess have a problem.. but still i can reead it, Thanks alot alotttt nid ... ;D
Can't view what??
-
Can't view what??
the attatchement.. i mean, i cant open it.. i am getting an
error..
-
ok wait....I'll attach a new format
-
i hope you r able to see this
-
dont click on the image..click on the attachment symbol...and when asked..click save
-
i hope you r able to see this
aha.. now i can open it properly .....your awesome nid.. good explanation...Thanks
-
aha.. now i can open it properly .....your awesome nid.. good explanation...Thanks
anytime my frnd :D
-
chk this for the markscheme image
http://img200.yfrog.com/i/3oa.jpg/
heres the q again
To a cyclist travelling due south on a straight horizontal road at 7 m/s, the wind appears to be
blowing from the north-east. Given that the wind has a constant speed of 12 m/s, find the direction
from which the wind is blowing.
I took ages to think about this.
Relative to the cyclist the vertical and horizontal components of wind are the same since the angle is 45
so
vertically relative to the cyclis, if x is the angle the wind makes with the vertical, Vv=12cosx-7
and horizonatally, Vh=12sinx
These are the same so solve
12cosx-7=12sinx
12cosx-12sinx=7
For how to solve see
http://www.astarmathsandphysics.com/a_level_maths_notes/C4/a_level_maths_notes_c4_trigonometry.html
-
Q --> A curve has the equation y # (ax ! 3) ln x, where x p 0 and a is a positive constant. The
normal to the curve at the point where the curve crosses the x-axis is parallel to the line 5y + x = 2.
Find the value of a.
My ans --> y = (ax+3)lnx
On x-axis, y = 0
ax + 3 = 0 ? x is -ve ?no soln
But lnx = 0 ? x =1
dy/dx = alnx + (ax+3).(1/x)
dy/dx (m of tangent) = a + 3
what to do next???
5y+x=2 so y=-(1/5)x+2/5 so gradient =-(1/5)
dy/dx=alnx+a+3/x
y=0 when x=1 since y(1)=(A1+3)LN(1)=0
AND DY/DX AT THIS POINT=a*ln1+1+3/1=a+3=-1/5 so a =-1/5-3=-16/5
-
My ans --> y = (ax+3)lnx
On x-axis, y = 0
ax + 3 = 0 ? x is -ve ?no soln
But lnx = 0 ? x =1
dy/dx = alnx + (ax+3).(1/x)
dy/dx (m of tangent) = a + 3
----
gradient of normal = -1/5 so gradient of tangent =5
so gradient of tangent = 5
a+3 = 5 so a=2
a = 2
correct answer?