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# C1 question Watch

1. question:
can someone tell me why in the mark scheme it does the negative recipcocal thing because it says 'parallel' and not 'perpendicular'. am i missing something??
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2. (Original post by cooldudeman)
question:
can someone tell me why in the mark scheme it does the negative recipcocal thing because it says 'parallel' and not 'perpendicular'. am i missing something??
Its asking for the normal, this is perpendicualr to the tangent so you do the reciprocal
3. (Original post by hazbaz)
Its asking for the normal, this is perpendicualr to the tangent so you do the reciprocal
what the fish, i dont remember any of this... maths is a pisstake
4. ah, i remember this question on a past paper i have done

they tell you y=k(root x) so y = kx^1/2

they tell you that its parallel to the line 2x + 3y = 0 so y=-2/3x

if lines are parallel gradients are equal, to get the gradient of a curve differentiate.

so dy/dx = 1/2kx^-1/2

make it equal to -2/3 (thats the gradient ( y=mx + c) )

after that it should be easy
5. The grad of the normal to any curve is given by: , where m is the gradient of the tangent to the curve at any point.
6. The normal is parallel to 2x+3y=0. This equation can be rearranged to give y=-2/3x, hence giving the gradient they specify in the MS.
7. (Original post by sohail.s)
ah, i remember this question on a past paper i have done

they tell you y=k(root x) so y = kx^1/2

they tell you that its parallel to the line 2x + 3y = 0 so y=-2/3x

if lines are parallel gradients are equal, to get the gradient of a curve differentiate.

so dy/dx = 1/2kx^-1/2

make it equal to -2/3 (thats the gradient ( y=mx + c) )

after that it should be easy
um i just tried that now and i got k=8/3.....
the answer is 6. i did on the last part, -2/3=k/4
8. (Original post by cooldudeman)
um i just tried that now and i got k=8/3.....
the answer is 6. i did on the last part, -2/3=k/4
hmm,

dy/dx = 1/2kx^-1/2

so 0.5kx^-1/2 = -2/3 (x=4 sub that in)

k/4 = -2/3 thats what we get ( but this is the gradient of the tangent, for the normal its -1/m )

so k/4 = 3/2 therefore k = 6
9. (Original post by sohail.s)
hmm,

dy/dx = 1/2kx^-1/2

so 0.5kx^-1/2 = -2/3 (x=4 sub that in)

k/4 = -2/3 thats what we get ( but this is the gradient of the tangent, for the normal its -1/m )

so k/4 = 3/2 therefore k = 6
its just the reciprocal part that i dont understand because the gradient '-2/3' came from the equation, 2x + 3y = 0 and isnt this the equation of a 'normal' line?

dy/dx gave (or always gives... not sure on that) the gradient of the tangent and -2/3 is the normal gradient so is it like this?:
you have to change the normal one into atangent one so -2/3 is 3/2.
is that why you have to change it?????
10. (Original post by cooldudeman)
its just the reciprocal part that i dont understand because the gradient '-2/3' came from the equation, 2x + 3y = 0 and isnt this the equation of a 'normal' line?

dy/dx gave (or always gives... not sure on that) the gradient of the tangent and -2/3 is the normal gradient so is it like this?:
you have to change the normal one into atangent one so -2/3 is 3/2.
is that why you have to change it?????
dy/dx gives the gradient function of a tangent to a point (x=whatever)

your misunderstanding it mate, the questions says the NORMAL at point p is parallel to the line 2x + 3y = 0 as far as we're concerned this line is just an ordinary line that has nothing to do with the curve (it could be a normal or a tangent but we're not interested in this- we are just interested in the gradient of the line )

TO find the normal to a point on a curve you find dy/dx which is the gradient of the tangent and then take the -ve reciprocal to find the gradient of the normal.

SO if dy/dx = 1/2kx^-1/2 and at that the point the gradient of tangent to that point is 0.5kx^-1/2 = -2/3
the gradient of the normal would equal -1/ -2/3 which gives 3/2

Just remember when differentiating you are finding the gradient function of a TANGENT to that point, once you have the function substitute values of x into the function to find the value of the tangent and then -ve reciprocal to find the gradient of the normal

Please forgive me for any typos
11. (Original post by sohail.s)
dy/dx gives the gradient function of a tangent to a point (x=whatever)

your misunderstanding it mate, the questions says the NORMAL at point p is parallel to the line 2x + 3y = 0 as far as we're concerned this line is just an ordinary line that has nothing to do with the curve (it could be a normal or a tangent but we're not interested in this- we are just interested in the gradient of the line )

TO find the normal to a point on a curve you find dy/dx which is the gradient of the tangent and then take the -ve reciprocal to find the gradient of the normal.

SO if dy/dx = 1/2kx^-1/2 and at that the point the gradient of tangent to that point is 0.5kx^-1/2 = -2/3
the gradient of the normal would equal -1/ -2/3 which gives 3/2

Just remember when differentiating you are finding the gradient function of a TANGENT to that point, once you have the function substitute values of x into the function to find the value of the tangent and then -ve reciprocal to find the gradient of the normal

Please forgive me for any typos
ohh i think i get it nowwww. thanks a lot.
lol if you dont mind, can you or anyone else help me with the second part?
i dont know how to find point Q...
12. (Original post by cooldudeman)
ohh i think i get it nowwww. thanks a lot.
lol if you dont mind, can you or anyone else help me with the second part?
i dont know how to find point Q...
Q is the point where the normal crosses the x-axis, so where y=0. Just sub that into the normal equation and solve for x.

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