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maths help!!

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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, 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

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?

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