C1 Soloman Differentiation questions help puhleasee :3

There is a few questions lol :blushing:

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 :mmm:

(edited 8 years ago)

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Reply 1

Aph

You didn't get minus 2, look at the line above. You just wrote the sign wrong.

Reply 2

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 :rofl:

Reply 3

Dinaa

Original post by Aph

You didn't get minus 2, look at the line above. You just wrote the sign wrong.

Oops

Thanks lololol

Reply 4

Dinaa

How about part c to that question? :s-smilie:

Reply 5

Aph

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 :rofl:

That's quite easy too, just bring your x^1/2 to the numerator.

Reply 6

Aph

Original post by Dinaa

How about part c to that question? :s-smilie:

Find the equasion for the tangent at B and use simultaneous equations.

Original post by Dinaa

You can do this... It's really not all the difficult.

Reply 9

Dinaa

Original post by Aph

You can do this... It's really not all the difficult.

Saying that really doesn't help lol :colonhash:

I don't know which equations to use, for simultaneous equations.

Reply 10

Aph

Original post by Dinaa

Saying that really doesn't help lol :colonhash:

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)

Reply 11

Dinaa

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 :colondollar:

Reply 12

Aph

Original post by Dinaa

But I don't have coordinates to find the equation of the tangent at B :colondollar:

Yes you do, look at your answer to part a)

Reply 13

Dinaa

Original post by aph

yes you do, look at your answer to part a)

o m g
u
s t a r

:colondollar:

Reply 14

Aph

Original post by Dinaa

o m g
u
s t a r

:colondollar:

All I did was point out the obvious, you're doing all the hard work...

Reply 15

Dinaa

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 :blushing:

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 :K:

Reply 16

Aph

Original post by Dinaa

Com'on it was not that obvious :blushing:

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 :K:

No, that's right :h:

don't always expect to get nice numbers but I'd be happy with that answer.

Reply 17

Dinaa

Original post by Aph

No, that's right :h:

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 :rofl:

Thank you