anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
The normal is a line perpendicular to the tangent line
anonymous
  • anonymous
Could you explain how to do this problem? It has been a while since I have done this kind of problem.
anonymous
  • anonymous
for what point are you trying to find the normal to?

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anonymous
  • anonymous
Also if the way I have put in the equation is confusing it should be: \[y=x^3+1\]
myininaya
  • myininaya
thats what i was fixing to ask lol
anonymous
  • anonymous
(1,2)
anonymous
  • 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
myininaya
  • 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
anonymous
  • anonymous
for the derivative I got \[y=3x^2\]
anonymous
  • anonymous
and now I put in 2 for y and 1 for x, such that \[2=3(1)^2\]
anonymous
  • anonymous
correct?
anonymous
  • anonymous
no..... sorry just 1 for x, bc y' is the slope
anonymous
  • anonymous
so then y=3
anonymous
  • anonymous
and to get x i plug in 3 for y in the derivative?
anonymous
  • anonymous
yeah thats your slop for the tanget, now you have to find the slope for the normal
anonymous
  • 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
anonymous
  • anonymous
So if the sope of the tangent is 3, then the slope of the normal is 1/3, correct?
anonymous
  • anonymous
close... -1/3
anonymous
  • anonymous
If the reciprocal is the number written as a fraction then flipped why did the sign change?
anonymous
  • anonymous
sorry it is 1/3, I'm being retarded
anonymous
  • anonymous
and then thats it? the answer is 1/3?
anonymous
  • anonymous
now use the point slope formula to get the equation of the normal
anonymous
  • anonymous
refresh my memory, what is the point slope formula? Or can i just use y=mx+b?
anonymous
  • anonymous
wait the slope is -1/3 bc perpendicular slope have opposite signs and the point slope formula is \[y-y_1=m(x-x_1)\]
anonymous
  • anonymous
perpendicular lines are negative reciprocals
anonymous
  • anonymous
ok. y1 and x1 are from the points given at the start, right?
anonymous
  • anonymous
yes
anonymous
  • anonymous
and m=\[-1/3\]
anonymous
  • anonymous
yes
anonymous
  • anonymous
so plugging everything in gets\[3-2=-1/3(1-1)\]
anonymous
  • anonymous
wich solves to 1=0
anonymous
  • anonymous
no.... you are looking y, the values you have are y1, x1, and m
anonymous
  • anonymous
y1=2, x1=1 and m=-1/3
anonymous
  • anonymous
so then it should be \[y-2=-1/3(x-1)\]
anonymous
  • anonymous
yeah... now put it in terms of y=mx+b
anonymous
  • anonymous
\[2=-1/3(1)+b\]
anonymous
  • anonymous
and solve for b?
anonymous
  • 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
anonymous
  • 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
anonymous
  • anonymous
then move -x+1 over to give x+3y-7=0
anonymous
  • anonymous
yes you could
anonymous
  • anonymous
ok thank you for the help. yo just got a new fan!
anonymous
  • anonymous
also now i'll use this site for help with math if the teacher is unavailable. Thanks again!
anonymous
  • anonymous
Cheers , hope I didin't confuse you
anonymous
  • anonymous
no you did fine. Have to go now bye

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