You are Here: Home >< Maths

# Just a quick question (differentiation)

Announcements Posted on
TSR's new app is coming! Sign up here to try it first >> 17-10-2016
1. (Original post by Zacken)
Sounds like whatever answers you're looking at are ****ed to hell.
So I'm not going crazy Well I'll get cracking with part (b) I'll update you in a second...(or a more like 5 minutes)
2. (Original post by homeland.lsw)
K well this is another one...yuck

The sketch shows the curve
and the normal at

a) Find the equation of the normal at the point

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?
3. (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...

Zacken (in case you were curious)

Edit sorry for the upside downedness...
4. (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.

Can you post a picture of the question, homeland?
5. (Original post by homeland.lsw)
Yup this is the question direct from the book...

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.

The actual question should be .
6. for the benefit of everyone's neck...

7. (Original post by Student403)
for the benefit of everyone's neck...
Thank you .
8. (Original post by Zacken)
This book seems very incompotent, the sketch given doesn't even look like x^2 - 5 at all.

The actual question should be .
I was thinking that because surely at x=0 , y=-5 so would be no where near the origin...

(Original post by Student403)
for the benefit of everyone's neck...
Like I said, 11 gemmers know what they are doing
9. Zacken so that means my part (a) is wrong?
10. (Original post by homeland.lsw)
Zacken so that means my part (a) is wrong?
According to the way the question is written, it's correct. Because the question is wrong.

Do it again with though.
11. (Original post by Zacken)
According to the way the question is written, it's correct. Because the question is wrong.

Do it again with though.
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
12. (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.

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

So gradient of normal is 1.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: March 31, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### How does exam reform affect you?

From GCSE to A level, it's all changing

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams