Finding an equation of the normal to curves

[Monkeyseat]

Ok, I would appreciate if you could check these 2 questions I have done. I believe I have done them correctly but the answers say otherwise.

Question 1:

Find an equation of the normal to each of the following curves at the point indicated:

y = x^3 + 5x^2 - 7x + 3 at the pont (1,2)

My working

dy/dx = 3x^2 + 10x - 7

When x = 1, dy/dx = (3 * 1) + (10 * 1) - 7 = 6

Gradient of normal:

m1 * m2 = -1
6 * (-1/6) = -1

Therefore the gradient of the normal = -1/6

Equation of normal:

y - y1 = m(x - x1)
y - 2 = -1/6(x - 1)
6y - 12 = -x + 1
6y + x - 11 = 0

For Question 1, the book says the answer is 6y + x = 18, who is correct?

Question 2:

Find an equation of the normal to each of the following curves at the point indicated:

y = 6x - x^4 at the point (2, -4)

My working

dy/dx = 6 - 4x^3

When x=2, dy/dx = 6 - (4 * 8) = -26

Gradient of normal:

m1 * m2 = -1
-26 * (1/26) = -1

Therefore the gradient of the normal = 1/26

Equation of normal:

y - y1 = m(x - x1)
y + 4 = 1/26(x-2)
26(y+4) = x - 2
26y + 104 = x - 2
26y - x + 106 = 0

For Question 2, the book says the answer is 26y - x + 107 = 0, who is correct?

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I have done a lot of these questions and these are the only 2 I have done wrong apparently. If you could check them and tell me where I'm going wrong it would be appreciated (preferably using the methods I have learnt) as I have checked many times and don't know what error I have made. The book makes quite a few mistakes however, so I wasn't sure.

Thanks!

EDIT:

Typos corrected, any more let me know.

 

[stapel]

Question 1:....

y - y1 = m(x - x1)
y - 2 = -1/6(x - 1)

I agree with you to this point. But then you appear to lose a "minus" sign somewhere. I think you should get the following:

. . . . .6[y - 2] = 6[-1/6(x - 1)]

. . . . .6y - 12 = -(x - 1)

. . . . .6y - 12 = -x + 1

. . . . .6y - 12 + x - 1 = 0

. . . . .x + 6y - 13 = 0

So I think neither answer was correct, but you were a lot closer. :wink:

Question 2:....

I agree with your answer, other than signs. (The "standard" way of formatting is usually to have the x-term be positive, so the equation "should" be "x - 26y - 106 = 0". I would accept your answer either way, but I'm not doing your grading.)

Eliz.

 

[Monkeyseat]

Question 1:....

y - y1 = m(x - x1)
y - 2 = -1/6(x - 1)

I agree with you to this point. But then you appear to lose a "minus" sign somewhere. I think you should get the following:

. . . . .6[y - 2] = 6[-1/6(x - 1)]

. . . . .6y - 12 = -(x - 1)

. . . . .6y - 12 = -x + 1

. . . . .6y - 12 + x - 1 = 0

. . . . .x + 6y - 13 = 0

So I think neither answer was correct, but you were a lot closer. :wink:

Question 2:....

I agree with your answer, other than signs. (The "standard" way of formatting is usually to have the x-term be positive, so the equation "should" be "x - 26y - 106 = 0". I would accept your answer either way, but I'm not doing your grading.)

Eliz.

Thanks for the reply. I agree with this (I was a tiny bit confused in your first line but I presumed you were multiplying both sides by 6 to get rid of the fraction?):

x + 6y - 13 = 0

I added the 1 instead of taking it off when moving it over by accident so that's why I had - 11.

Could it also be 0= - x - 6y + 13 (although this would not be correct notation)? I believe I must have been doing it like this (and got -11 because I probably did 1 - 12 or -12 + 1) but still messed it up! :lol: Not a typo, just a stupid mistake lol. I actually got the answer you gave on a seperate piece of paper but I just didn't realise the difference lol, I must've thought I wrote -13. I need to go a bit slower I suppose.

Thanks for the fast response, so could it also be - x - 6y + 13 in terms of changing the signs like you did in Question 2 despite it being wrong? I don't think my tutor minds but I suppose it is best, it's just annoying with the wrong answers in the book, I wish they'd check them before publishing.

Cheers!

Hopefully this is typo free...