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core 1 maths help?

The point P on the curve y = k √ x has x-coordinate 4. The normal to the curve at P is parallel to the line 2x + 3y = 0.
(i) Find the value of k.
Reply 1
Avatar for ANJXLA
ANJXLA
12
how many marks is this worth?
Original post by ANJXLA
how many marks is this worth?


6 marks:smile:
Original post by NoahMal
3y=-2x
y=(-2/3)x
you know what dy/dx at x=4 must be (-2/3)

so you differentiate y

re write the equation as y=k(x)^1/2

dy/dx = (1/2)(k)(x)^-1/2

dy/dx=(1/2)k(4)^-1/2

dy/dx=k/4

therefore (-2/3) = k/4 and k = (-8/3)


Thankyou!:smile:
Reply 4
Avatar for Zacken
Zacken
22
Well this was rather pointless... the OP just got the answer without learning anything.
Reply 5
Avatar for Naruke
Naruke
11
Original post by Zacken
Well this was rather pointless... the OP just got the answer without learning anything.


Don't think he cares, since he won't be getting a lunch-time detention tomorrow lol :smug:

Seems he will be getting a detention haha
(edited 7 years ago)
Original post by NoahMal
3y=-2x
y=(-2/3)x
you know what dy/dx at x=4 must be (-2/3)


Why isn't this part -3/2 instead of -2/3? They gave the normal so surely you want the gradient of the tangent? I'm sure you're correct but i don't understand how
Original post by surina16
Why isn't this part -3/2 instead of -2/3? They gave the normal so surely you want the gradient of the tangent? I'm sure you're correct but i don't understand how


He's wrong, and besides the tangent to the curve would have gradient +3/2 and not -3/2 like you said.

Original post by NoahMal
...


So you post a full solution yet you get it wrong too...
(edited 7 years ago)
Original post by RDKGames
He's wrong, and besides the tangent to the curve would have gradient +3/2 and not -3/2 like you said.



So you post a full solution yet you get it wrong too...


Oh yeah sorry that was a stupid error :colondollar::facepalm:

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