The Student Room Group

Just a quick question (differentiation)

Scroll to see replies

Original post by Zacken
Sounds like whatever answers you're looking at are ****ed to hell. :lol:


So I'm not going crazy :lol: Well I'll get cracking with part (b) I'll update you in a second...(or a more like 5 minutes)
Reply 21
Original post by homeland.lsw
K well this is another one...yuck

The sketch shows the curve y=x25 y = x^2 - 5
and the normal at (2,6) (2 , -6)

a) Find the equation of the normal at the point
(2,6) (2 , -6)

So I did
y=x^2 -5
dy/dx = 2x

2x=0
2(2)= 4 (The two coming from the given point)

so 4 is the gradient of the tangent (?)
Then -1/4 is the gradient of the normal (negative reciprocal)

So y= -1/4x + c
-6= -1/4 (2) + c
-11/2= c

Therefore
y= -1/2x + 11/2 for part a?

@Zacken


Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?
Original post by QT123
Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?


Yup this is the question direct from the book...
image.jpg

@Zacken (in case you were curious)

Edit sorry for the upside downedness...
Reply 23
Original post by QT123
Are you sure you have written the equation for the curve/the coordinates of the point correctly because (2, -6) does not lie on that curve?


Good point... not sure how I didn't spot that. :colondollar:

Can you post a picture of the question, homeland?
Reply 24
Original post by homeland.lsw
Yup this is the question direct from the book...
image.jpg

@Zacken (in case you were curious)

Edit sorry for the upside downedness...


This book seems very incompotent, the sketch given doesn't even look like x^2 - 5 at all. :laugh:

The actual question should be y=x25xy = x^2 - 5x. :tongue:
for the benefit of everyone's neck...

Original post by Student403
for the benefit of everyone's neck...



Thank you :h:.
Original post by Zacken
This book seems very incompotent, the sketch given doesn't even look like x^2 - 5 at all. :laugh:

The actual question should be y=x25xy = x^2 - 5x. :tongue:


I was thinking that because surely at x=0 , y=-5 so would be no where near the origin...:dontknow:


Original post by Student403
for the benefit of everyone's neck...



Like I said, 11 gemmers know what they are doing
@Zacken so that means my part (a) is wrong? :getmecoat:
Reply 29
Original post by homeland.lsw
@Zacken so that means my part (a) is wrong? :getmecoat:


According to the way the question is written, it's correct. Because the question is wrong.

Do it again with y=x25xy = x^2 - 5x though. :smile:
Original post by Zacken
According to the way the question is written, it's correct. Because the question is wrong.

Do it again with y=x25xy = x^2 - 5x though. :smile:


dy/dx = 2x - 5
Then?
2x-5=0
2x=5
x=5/2

or
2x-5=0
2(2)-5
-1?

I'm sorry, I'm confused af
Reply 31
Original post by homeland.lsw

2x-5=0
2(2)-5
-1?

I'm sorry, I'm confused af


Should be this 2(2) - 5, but you shouldn't say =0. :tongue:

Just: gradient of tangent is 2(2) - 5 = 4-5 = -1.

So gradient of normal is 1.

Quick Reply

Latest