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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?
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?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The normal is a line perpendicular to the tangent line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Could you explain how to do this problem? It has been a while since I have done this kind of problem.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for what point are you trying to find the normal to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Also if the way I have put in the equation is confusing it should be: \[y=x^3+1\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0thats what i was fixing to ask lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0k... 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
 5 years ago
Best ResponseYou've already chosen the best response.0find 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
 5 years ago
Best ResponseYou've already chosen the best response.0for the derivative I got \[y=3x^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and now I put in 2 for y and 1 for x, such that \[2=3(1)^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no..... sorry just 1 for x, bc y' is the slope

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and to get x i plug in 3 for y in the derivative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah thats your slop for the tanget, now you have to find the slope for the normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the normal is perpendicular to the tangent, take the reciprocal of the slope of the tangent to get the slope of the normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So if the sope of the tangent is 3, then the slope of the normal is 1/3, correct?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If the reciprocal is the number written as a fraction then flipped why did the sign change?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry it is 1/3, I'm being retarded

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then thats it? the answer is 1/3?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now use the point slope formula to get the equation of the normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0refresh my memory, what is the point slope formula? Or can i just use y=mx+b?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait the slope is 1/3 bc perpendicular slope have opposite signs and the point slope formula is \[yy_1=m(xx_1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0perpendicular lines are negative reciprocals

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok. y1 and x1 are from the points given at the start, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so plugging everything in gets\[32=1/3(11)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no.... you are looking y, the values you have are y1, x1, and m

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0y1=2, x1=1 and m=1/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then it should be \[y2=1/3(x1)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah... now put it in terms of y=mx+b

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no... so you have \[y2=\frac{1}{3}(x1)\] Expand this and you get \[y2=\frac{1}{3}x+\frac{1}{3}\] 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
 5 years ago
Best ResponseYou've already chosen the best response.0could you also distribute the negative and move the 1/3 from \[y2=1/3(x1)\] to give 3y6=x+1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then move x+1 over to give x+3y7=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thank you for the help. yo just got a new fan!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also now i'll use this site for help with math if the teacher is unavailable. Thanks again!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Cheers , hope I didin't confuse you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no you did fine. Have to go now bye
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