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# C1 Soloman Differentiation questions help puhleasee :3 watch

1. There is a few questions lol

3) The diagram shows the curve y=x^2 + x - 2. The curve crosses the x-axi sat the points A(a,0) and B(b,0) where a<b.

a) Find the values of a and b.

I got A(-2,0) and B(1,0)

b) Show that the normal to the curve at A has the equation x- 3y + 2= 0

so dy/dx = 2x + 1
when x = -2
dy/dx= 2(-2) + 1
dy/dx= -3
y=-3(x+2)
y= -3x -6 (tangent)

The gradient of the normal will be 1/3

Hence, y=1/3(x+2)
3y = x + 2
x - 3y + 2 = 0

c) The tangent to the curve at B meets the normal to the curve at A at the point C.
Find the exact coordinates of C.

(Yeah I don't know)

Thanks and I have more questions so pls don't run away
2. You didn't get minus 2, look at the line above. You just wrote the sign wrong.
3. 4) Given that y= (x^2 - 6x - 3)/(3x^1/2), show that dy/dx can be expressed in the form (x+a)^2/(bx^3/2), where a and b are integers to be found.

Tried, but no lol no
4. (Original post by Aph)
You didn't get minus 2, look at the line above. You just wrote the sign wrong.

Oops

Thanks lololol
5. How about part c to that question?
6. (Original post by Dinaa)
4) Given that y= (x^2 - 6x - 3)/(3x^1/2), show that dy/dx can be expressed in the form (x+a)^2/(bx^3/2), where a and b are integers to be found.

Tried, but no lol no
That's quite easy too, just bring your x^1/2 to the numerator.
7. (Original post by Dinaa)
How about part c to that question?
Find the equasion for the tangent at B and use simultaneous equations.
8. Um
9. (Original post by Dinaa)
Um
You can do this... It's really not all the difficult.
10. (Original post by Aph)
You can do this... It's really not all the difficult.
Saying that really doesn't help lol

I don't know which equations to use, for simultaneous equations.
11. (Original post by Dinaa)
Saying that really doesn't help lol

I don't know which equations to use, for simultaneous equations.
You use the tangent at B, doing what you did to calculate the tangent at A and the normal at A which is what you proved in part b)
12. (Original post by Aph)
You use the tangent at B, doing what you did to calculate the tangent at A and the normal at A which is what you proved in part b)
But I don't have coordinates to find the equation of the tangent at B
13. (Original post by Dinaa)
But I don't have coordinates to find the equation of the tangent at B
14. (Original post by aph)
o m g
u
s t a r

15. (Original post by Dinaa)
o m g
u
s t a r

All I did was point out the obvious, you're doing all the hard work...
16. (Original post by Aph)
All I did was point out the obvious, you're doing all the hard work...
Com'on it was not that obvious

I got y=3x -3 for the tangent at B

so then..

x - 3(3x-3) + 2 = 0
x - 9x+9 + 2 = 0
-8x + 11 = 0
8x = 11
x = 11/8

when x = 11/8
y= 3(11/8) - 3
y= 9/8

so (11/8, 9/8)

looks kind of wrong
17. (Original post by Dinaa)
Com'on it was not that obvious

I got y=3x -3 for the tangent at B

so then..

x - 3(3x-3) + 2 = 0
x - 9x+9 + 2 = 0
-8x + 11 = 0
8x = 11
x = 11/8

when x = 11/8
y= 3(11/8) - 3
y= 9/8

so (11/8, 9/8)

looks kind of wrong
No, that's right don't always expect to get nice numbers but I'd be happy with that answer.
18. (Original post by Aph)
No, that's right don't always expect to get nice numbers but I'd be happy with that answer.
(Original post by Aph)
That's quite easy too, just bring your x^1/2 to the numerator.
(Original post by Dinaa)
4) Given that y= (x^2 - 6x - 3)/(3x^1/2), show that dy/dx can be expressed in the form (x+a)^2/(bx^3/2), where a and b are integers to be found.

Tried, but no lol no
Thank you

So for this other question, I got y= 1/2x^1/2 - x^-1/2 + x^-3/2

when... I tried.. to simplify?

EDIT: wait wat did I just type
19. (Original post by Dinaa)
Thank you

So for this other question, I got y= 1/2x^1/2 - x^-1/2 + x^-3/2

when... I tried.. to simplify?
Is that after differentiation?
20. (Original post by Aph)
Is that after differentiation?
Oops lol

When I simplified, I got y = 1/3x^3/2 - 2x^1/2 - x^-1/2

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