maths help!!

[nid404]

ok mousa here

check the diagram and tell me if you get it or not

[nid404]

I think this should be able to clear your doubt

[mousa]

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 ...

[nid404]

i cant view. i guess  have a problem.. but still i can reead it, Thanks alot alotttt nid ...

Can't view what??

[mousa]

Can't view what??

 the attatchement.. i mean, i cant open it.. i am getting  an
 error..

[nid404]

ok wait....I'll attach a new format

[nid404]

i hope you r able to see this

[Ghost Of Highbury]

dont click on the image..click on the attachment symbol...and when asked..click save

[mousa]

i hope you r able to see this

aha.. now i can open it properly .....your awesome nid.. good explanation...Thanks

[nid404]

aha.. now i can open it properly .....your awesome nid.. good explanation...Thanks

anytime my frnd

[astarmathsandphysics]

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, V_v=12cosx-7

and horizonatally, V_h=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

[astarmathsandphysics]

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

[Ghost Of Highbury]

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?