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?

Question

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?

anonymous · Answer

The normal is a line perpendicular to the tangent line

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

myininaya · Answer

thats what i was fixing to ask lol

anonymous · Answer

(1,2)

anonymous · Answer

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

myininaya · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

correct?

anonymous · Answer

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

anonymous · Answer

so then y=3

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

close... -1/3

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

perpendicular lines are negative reciprocals

anonymous · Answer

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

anonymous · Answer

yes

anonymous · Answer

and m=$$-1/3$$

anonymous · Answer

yes

anonymous · Answer

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

anonymous · Answer

wich solves to 1=0

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

and solve for b?

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

yes you could

anonymous · Answer

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

anonymous · Answer

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

anonymous · Answer

Cheers , hope I didin't confuse you

anonymous · Answer

no you did fine.  Have to go now bye