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C3, Somebody Save Me... (equation of normal etc etc...) watch

1. This question is doing my nut in, think some c1 is required here but ive forgotten all about it :/

Anyway "Find the equation of the normal to the curve y = 5 (Over) 2(3x-1)^4 at the pint at which x=0, giving your in the form ax + by + c = 0 where a, b and c are integers..." ARGH!!!!
2. yea?
3. What have you done so far?
4. Use the quotient rule to differentiate it with regards to x.

Can you see where you go from here?

next time I'll actually read the question :/
5. (Original post by JamesyB)
yea?
yeayah how did you do that!?
6. Firstly, put in x = 0 so you have a set of co-ordinates; (x,y).

Now differentiate to get the gradient of the tangent.

Re-arrange to find gradient of normal.

Equation of a line (normal or tangent) is in the form y - y1 = m(x - x1); (x,y) from above are x1 and y1, gradient is m.

Done!
7. (Original post by Greeze Geez)
yeayah how did you do that!?
http://www.thestudentroom.co.uk/wiki/LaTex
8. (Original post by Keckers)
Use the quotient rule to differentiate it with regards to x.

Can you see where you go from here?
9. (Original post by Greeze Geez)
how did you do that!?
See http://www.thestudentroom.co.uk/wiki/LaTex

Edit: Too slow
10. (Original post by ghostwalker)
See http://www.thestudentroom.co.uk/wiki/LaTex

Edit: Too slow
That can never be true
11. (Original post by boromir9111)
That can never be true
Ha! If only.

Anyhow, enough off topic
12. thanks every1! i clocked just now that i have to differentiate, but im having trouble applying the rule, the 5 on the top is doing my nut in, we havent been taight this yet :/ could anybosy shw me the differentiation process or at least the result... Thanks people, truly
13. (Original post by Greeze Geez)
thanks every1! i clocked just now that i have to differentiate, but im having trouble applying the rule, the 5 on the top is doing my nut in, we havent been taight this yet :/ could anybosy shw me the differentiation process or at least the result... Thanks people, truly
Have you differentiated (ax+b)^n using the chain rule?

This is the same thing with a 2.5 (5/2) infront.

y = k(ax+b)^n

dy/dx = ...

Have you differentiated (ax+b)^n using the chain rule?

This is the same thing with a 2.5 (5/2) infront.

y = k(ax+b)^n

dy/dx = ...

wait so is dy/dx 2.5(3x-1)^3 or 10(3x-1)^3???

sorry by the way people... this is what can only be desciribed as a long ting..
15. (Original post by Greeze Geez)
yeayah how did you do that!?
LaTeX, it's a system for typing mathematical expressions. Very, VERY useful See here for more info.

Back to the question :P How could you re-write this so that it would help make it into something easier?

16. (Original post by JamesyB)
LaTeX, it's a system for typing mathematical expressions. Very, VERY useful See here for more info.

Back to the question :P How could you re-write this so that it would help make it into something easier?

Fam, is this is this right>> dy/dx = -40(3x-10)^-5??
17. It is worth just knowing how to write down the answer for the derivative of a linear function raised to a power.

18. (Original post by Mr M)
It is worth just knowing how to write down the answer for the derivative of a linear function raised to a power.

Oh i see, yh it was the k that was doing me in, we hadnt worked on examples with "k" involved, so i take dy/dx is now -30(3x-1)^-5??
19. (Original post by Greeze Geez)
Oh i see, yh it was the k that was doing me in, we hadnt worked on examples with "k" involved, so i take dy/dx is now -30(3x-1)^-5??
Yes
20. (Original post by Greeze Geez)
Oh i see, yh it was the k that was doing me in, we hadnt worked on examples with "k" involved, so i take dy/dx is now -30(3x-1)^-5??
Yup!

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