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# Calculus Maths Question watch

1. A cubic curve has equation y=x^3 - 3x^2 +1

i)Show that the tangent to the curve at the point where x=-1 has gradient 9.

ii)Find the coordinates of the other point P, on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P.

iii)Show that the area of the triangle bounded by the normal at P and the x and y axes is 8 square units.

For the first part, I've worked out dy/dx= 3x^2 -6x

So if x=-1, dy/dx=9

I'm not sure about how to continue though..

I'd really appreciate some help with these questions. Thank you
2. (Original post by x-Sophie-x)
ii)Find the coordinates of the other point P, on the curve at which the tangent has gradient 9 and find the equation of the normal to the curve at P.

iii)Show that the area of the triangle bounded by the normal at P and the x and y axes is 8 square units.
If you want to find the other point P where the tangent has a gradient of 9, set dy/dx = 9 and solve the quadratic - you want the root which isn't equal to -1 as you already have that
3. ii) solve the quadratic for dy/dx =9

iii) plug answer for 'ii' into original function to get co-ordinate of P. Then solve geometrically.

Posted from TSR Mobile
4. i) like you did, put in -1 to get 9
ii) make dy/dx = 9 then u should get a quadratic to solve and get two x values for which one of them would be -1 and use the other to sub into dy/dx and make it -1 over dy/dx since its the normal then find y1 by subbing in that value into the cubic to find your equation at P
iii) draw your equation in ii) to make it easier to see
5. (Original post by Felix Felicis)
If you want to find the other point P where the tangent has a gradient of 9, set dy/dx = 9 and solve the quadratic - you want the root which isn't equal to -1 as you already have that
Thanks, how did I not realise to do that! So P is (3,1).

And thanks a lot for the quick reply (:
6. (Original post by x-Sophie-x)
Thanks, how did I not realise to do that! So P is (3,1).

And thanks a lot for the quick reply (:
Yep and no problem
7. (Original post by Skilled)
i) like you did, put in -1 to get 9
ii) make dy/dx = 9 then u should get a quadratic to solve and get two x values for which one of them would be -1 and use the other to sub into dy/dx and make it -1 over dy/dx since its the normal then find y1 by subbing in that value into the cubic to find your equation at P
iii) draw your equation in ii) to make it easier to see
Thanks a lot.

I get P as (3, 1) and the equation of the normal is y=-1/3x + 4/3 ?

Also, which two values would I integrate between to find the area?
Because when I draw out the curve, there are several different sections, and surely it's not possible to work out the area as one integration?
Or am I being stupid again.
8. (Original post by x-Sophie-x)
Thanks a lot.

I get P as (3, 1) and the equation of the normal is y=-1/3x + 4/3 ?

Also, which two values would I integrate between to find the area?
Because when I draw out the curve, there are several different sections, and surely it's not possible to work out the area as one integration?
Or am I being stupid again.
I got the equation to be y = -1/9 x + 4/3

Also, it says between the y-axis and the x-axis - presumably this means between x = 0 and the x-intercept of the line
9. (Original post by Felix Felicis)
I got the equation to be y = -1/9 x + 4/3

Also, it says between the y-axis and the x-axis - presumably this means between x = 0 and the x-intercept of the line
Ah yeah, that's what I meant. Typing error, sorry!

Ah, okay. I'll try and work this out, thanks.
10. (Original post by x-Sophie-x)
Thanks a lot.

I get P as (3, 1) and the equation of the normal is y=-1/3x + 4/3 ?

Also, which two values would I integrate between to find the area?
Because when I draw out the curve, there are several different sections, and surely it's not possible to work out the area as one integration?
Or am I being stupid again.
Sorry for not getting back to you, what the other person has said, makes sense
11. (Original post by Skilled)
Sorry for not getting back to you, what the other person has said, makes sense
That's fine.

Thanks very much for your help (:
12. (Original post by x-Sophie-x)
That's fine.

Thanks very much for your help (:
Np
Ask me any other maths help and I will try to answer, Im doing A2 edexcel math right now
13. (Original post by Skilled)
Np
Ask me any other maths help and I will try to answer, Im doing A2 edexcel math right now
Haha, I have a maths questions on the series and sequences topic in core 2, but I'm not even going to attempt looking at maths at this kind of time!

Much appreciated though

Posted from TSR Mobile
14. Just in case it asks for the equation of a normal in a simplified form on a later question, I'd probably multiply through by 9 to get 9y = -x + 12, then rearrange so that x + 9y -12 = 0.

Nothing necessary, just tidies it up a bit and stops short-sighted examiners from dropping marks.

Posted from TSR Mobile
15. (Original post by Snakefingers13)
Just in case it asks for the equation of a normal in a simplified form on a later question, I'd probably multiply through by 9 to get 9y = -x + 12, then rearrange so that x + 9y -12 = 0.

Nothing necessary, just tidies it up a bit and stops short-sighted examiners from dropping marks.

Posted from TSR Mobile
Thanks for the advice I think I had to work that out for part 5 of the question anyway so I could work out something else.

Posted from TSR Mobile

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