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# Just a quick question (differentiation)

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1. a) Find the rate of change of with respect to when
b) Find the coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this?

Zacken or anyone else?
2. (Original post by homeland.lsw)

a) Find the rate of change of with respect to when
b) Find the coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this?

Zacken or anyone else?
So, the rate of change at any given point is given by , the rate of change at is given by plugging this into the above equation.

We notate this as:
3. (Original post by Zacken)
So, the rate of change at any given point is given by , the rate of change at is given by plugging this into the above equation.

We notate this as:
So in simple terms
???
therefor

4. (Original post by homeland.lsw)

a) Find the rate of change of with respect to when
b) Find the coordinates of the two points where the gradient of the curve is 1

Can anyone help me with this?

Zacken or anyone else?
That's part (a) dealt with, not part (b).

The gradient of the curve at a general point is given by , so the gradient is precisely when , solving this for gives you the x co-ordinate.

Now we want the y-coordinate, but all we have is so what we do is integrate! .

Can you take it from here?
5. (Original post by homeland.lsw)
So in simple terms
???
therefor

Nopes, you replace with 3, and then . 6 is the gradient of the curve when .
6. (Original post by Zacken)
Nopes, you replace with 3, and then . 6 is the gradient of the curve when .
The answer is -9 but why??
7. (Original post by homeland.lsw)
The answer is -9 but why??
8. (Original post by Zacken)
I dunno, I checked the answers to see if that gave me an idea and it said -9...
9. (Original post by homeland.lsw)
I dunno, I checked the answers to see if that gave me an idea and it said -9...
Have you copied the question down right? The correct answer is 6.
10. (Original post by Zacken)
Have you copied the question down right? The correct answer is 6.
yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?
11. (Original post by homeland.lsw)
yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?
Probably just the answers being wrong as usual, anyway. When x=3, dy/dx is 6. So that's part (a).
12. (Original post by homeland.lsw)
yup dy/dx=9-x

find the rate of change of y with respect to x when x=3...

maybe a misprint?
Are you sure it is a 3 and not 0?
13. (Original post by zetamcfc)
Are you sure it is a 3 and not 0?
Even then, the answers say -9
14. (Original post by Zacken)
Even then, the answers say -9
lol,such an idiot ! am, must be a typo in the question
15. (Original post by Zacken)
Probably just the answers being wrong as usual, anyway. When x=3, dy/dx is 6. So that's part (a).
wait so that's it? That's the answer?
16. (Original post by homeland.lsw)
wait so that's it? That's the answer?
Yeah.
17. 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
18. (Original post by homeland.lsw)
y= -1/2x + 11/2 for part a?

Zacken
Is the bolded bit a typo? If so, looks good other than that! Well done.
19. (Original post by Zacken)
Is the bolded bit a typo? If so, looks good other than that! Well done.
Ah yes it should be -1/4 x for the gradient...

But then why does the answer say
y=x-8...that's really confusing me
20. (Original post by homeland.lsw)
Ah yes it should be -1/4 x for the gradient...

But then why does the answer say
y=x-8...that's really confusing me
Sounds like whatever answers you're looking at are ****ed to hell.

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