C1 question

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

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

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?
if it is then, the gradient should already be 'normal'.

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?:
(a normal gradient) -2/3= k/4 (a tangent gradient)
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

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...