- #1
me_maths
- 4
- 0
tricky problem please HELP :( -SOLVED by "courtrigrad"-
[SOLVED, thanks to "courtrigrad"!]
I've tried to solve this problem for over 3 or 4 hours but I can't figure out how, please help me.
Ok so it's a curve ( parable? ) y = (ax - b) / ( 1 - x^2 )
You are supposed to solve what 'a' and 'b' should be if the curve goes through the point (-2,1) AND that the tangents* direction of that same point (-2,1) is 45 degrees ( so the tangents 'k' value should be 1, which is 45 degrees ).
*I'm not sure but I might have the wrong name in English for 'tangent', but it is a line with the same 'k' value as the single point in the curve... if you know what I mean?
THE BELOW IS WHAT I'VE TRIED TO DO:
Ok, so we all know that to get the 'k' value of a point in a curve you must begin with taking the derivate of the curve, which in this case would be:
MAOL s.43 : D f/g = ( gDf - fDg ) / ( g^2 )
#1. y' = ( ( 1-x^2 )*a - ( ax-b )*( 2x ) ) / ( 1 - x^4 )
and then you should put the derivate in a function of the x value, something like: y'(x) = ... , which would in this case be: y'(-2) = ... .
so we take that y'(-2) = ... and equal it to 1, because that's the 'k' value we had to have for that point.
and thus we should get an equation with both 2 unknown variables, 'a' and 'b' , in which we should somehow figure out what they should be to get this to work... which is where I fail.
The answer for this problem is that 'a' and 'b' both = 1.
I've tried to do the countings myself etc but I can't come up with it...
Could someone please help me!?
[SOLVED, thanks to "courtrigrad"!]
I've tried to solve this problem for over 3 or 4 hours but I can't figure out how, please help me.
Ok so it's a curve ( parable? ) y = (ax - b) / ( 1 - x^2 )
You are supposed to solve what 'a' and 'b' should be if the curve goes through the point (-2,1) AND that the tangents* direction of that same point (-2,1) is 45 degrees ( so the tangents 'k' value should be 1, which is 45 degrees ).
*I'm not sure but I might have the wrong name in English for 'tangent', but it is a line with the same 'k' value as the single point in the curve... if you know what I mean?
THE BELOW IS WHAT I'VE TRIED TO DO:
Ok, so we all know that to get the 'k' value of a point in a curve you must begin with taking the derivate of the curve, which in this case would be:
MAOL s.43 : D f/g = ( gDf - fDg ) / ( g^2 )
#1. y' = ( ( 1-x^2 )*a - ( ax-b )*( 2x ) ) / ( 1 - x^4 )
and then you should put the derivate in a function of the x value, something like: y'(x) = ... , which would in this case be: y'(-2) = ... .
so we take that y'(-2) = ... and equal it to 1, because that's the 'k' value we had to have for that point.
and thus we should get an equation with both 2 unknown variables, 'a' and 'b' , in which we should somehow figure out what they should be to get this to work... which is where I fail.
The answer for this problem is that 'a' and 'b' both = 1.
I've tried to do the countings myself etc but I can't come up with it...
Could someone please help me!?
Last edited: