## anonymous 5 years ago This is a question on my math homework, but I can't make any sense of it. Here it is: find the equation of the normal to the curve with equation y=(x^3)+1. Is there a typo or am I just missing something?

1. anonymous

The normal is a line perpendicular to the tangent line

2. anonymous

Could you explain how to do this problem? It has been a while since I have done this kind of problem.

3. anonymous

for what point are you trying to find the normal to?

4. anonymous

Also if the way I have put in the equation is confusing it should be: $y=x^3+1$

5. myininaya

thats what i was fixing to ask lol

6. anonymous

(1,2)

7. anonymous

k... now take the derivative of you function.... plug in the values (1,2) this will give you the slope, Let me know what you get

8. myininaya

find f'(x) and evaluate it at x=1 then take the negative reciprocal of that this gives you the slope of the normal line

9. anonymous

for the derivative I got $y=3x^2$

10. anonymous

and now I put in 2 for y and 1 for x, such that $2=3(1)^2$

11. anonymous

correct?

12. anonymous

no..... sorry just 1 for x, bc y' is the slope

13. anonymous

so then y=3

14. anonymous

and to get x i plug in 3 for y in the derivative?

15. anonymous

yeah thats your slop for the tanget, now you have to find the slope for the normal

16. anonymous

since the normal is perpendicular to the tangent, take the reciprocal of the slope of the tangent to get the slope of the normal

17. anonymous

So if the sope of the tangent is 3, then the slope of the normal is 1/3, correct?

18. anonymous

close... -1/3

19. anonymous

If the reciprocal is the number written as a fraction then flipped why did the sign change?

20. anonymous

sorry it is 1/3, I'm being retarded

21. anonymous

and then thats it? the answer is 1/3?

22. anonymous

now use the point slope formula to get the equation of the normal

23. anonymous

refresh my memory, what is the point slope formula? Or can i just use y=mx+b?

24. anonymous

wait the slope is -1/3 bc perpendicular slope have opposite signs and the point slope formula is $y-y_1=m(x-x_1)$

25. anonymous

perpendicular lines are negative reciprocals

26. anonymous

ok. y1 and x1 are from the points given at the start, right?

27. anonymous

yes

28. anonymous

and m=$-1/3$

29. anonymous

yes

30. anonymous

so plugging everything in gets$3-2=-1/3(1-1)$

31. anonymous

wich solves to 1=0

32. anonymous

no.... you are looking y, the values you have are y1, x1, and m

33. anonymous

y1=2, x1=1 and m=-1/3

34. anonymous

so then it should be $y-2=-1/3(x-1)$

35. anonymous

yeah... now put it in terms of y=mx+b

36. anonymous

$2=-1/3(1)+b$

37. anonymous

and solve for b?

38. anonymous

no... so you have $y-2=-\frac{1}{3}(x-1)$ Expand this and you get $y-2=-\frac{1}{3}x+\frac{1}{3}$ now combine like terms $y=-\frac{1}{3}x+\frac{1}{3}+2$ $y=-\frac{1}{3}x+\frac{7}{3}$ which is in the form of y=mx+b

39. anonymous

could you also distribute the negative and move the 1/3 from $y-2=-1/3(x-1)$ to give 3y-6=-x+1

40. anonymous

then move -x+1 over to give x+3y-7=0

41. anonymous

yes you could

42. anonymous

ok thank you for the help. yo just got a new fan!

43. anonymous

also now i'll use this site for help with math if the teacher is unavailable. Thanks again!

44. anonymous

Cheers , hope I didin't confuse you

45. anonymous

no you did fine. Have to go now bye