Qualification > Math

maths C2 URGENT

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Phosu:
oh yah i saw that question in the text book...didn't know how to solve so i skipped
but a point can have 3 coordinate, if you're considering the z axis, but then again we haven't been taught into such details
im sure its something simple, that we're just not getting it.

anyways i have one more question
could someone plz solve C2 question 8 Jun05,

nid404:
assuming they are vectors A & P and using their first eqn...which i also think is wrong
U can solve the quadratic...
x=(-b +/- root of (b2-4ac))/ 2a
u get t as 2

AP2= 6(4)-24(2)+125=101
AP=root of 101= 10..

strange...

astarmathsandphysics:
the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)

astarmathsandphysics:
Which board is this? Not cie?

halosh92:

--- Quote from: astarmathsandphysics on February 21, 2010, 04:54:57 pm ---the fixed point A has coordinates (8,-6,5) and the variable point P has coordiantes ( t, t , 2t)
a) show that AP^2= 6t^2 - 24t + 125
b) hence find the value of "t" for which the distance AP is least
c) determine this least distance
a)AP=sqrt((t-8)^2+(t--6)^2+(5-2t)^)
=sqrt(t^2-16t+64+t^2+12t+36+25-20t+4t^2)=sqrt(6t^2-24t+125)
b)AP^2=6t^2-24t+125
diffferentiate and put=0 and solve
12t-24=0 so t=2
c)AP^2 =6*2^2-24*2+125=24-48+125=101
least distance=sqrt(101)

--- End quote ---

thanks alooot nid and astar  ;D ;D ;D
yea its is cie board!

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