Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find two unit vectors that are normal to $\sqrt{\frac{x+z}{y-1}}=z^2$ at $P(3,5,1)$.

Attempt: I have $f(x,y,z)=\sqrt{\frac{x+z}{y-1}}-z^2$. First I found the gradient : $$\nabla f=<f_x,f_y,f_z>=<\frac{1}{2 (y-1) \sqrt{\frac{x+z}{y-1}}};-\frac{x+z}{2 (y-1)^2 \sqrt{\frac{x+z}{y-1}}};\frac{1}{2 (y-1) \sqrt{\frac{x+z}{y-1}}}-2 z>$$ $$\nabla f(3,5,1)=<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>$$

So the two unit vectors would be the positive and negative normalized gradients: $$\vec{u_1}=\frac{<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>}{||<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>||}=\frac{8}{\sqrt{227}}<\frac{1}{8}, \frac{-1}{8}, \frac{-15}{8}>=\frac{1}{\sqrt{227}}<1,-1,-15>$$


However, in the answer key the answer is $(\pm\frac{1}{\sqrt{365}}<1,-1,-19>$). I checked the derivatives multiple times and the gradient as well and I have no idea why my answer does not match the answer key. I do not think it is a typo, so help please.

share|cite|improve this question
The answer key answers aren't even unit vectors, unless that was suppposed to be $\sqrt{363}$. Your solution appears to be correct. – Jonas Meyer May 24 '12 at 5:20
Wow. Finally, it is a typo and not me making a mistake. On Thursday I will confirm again with my professor who has the solution manual. – Koba May 24 '12 at 6:27
Yes it was a typo in the textbook. – Koba May 25 '12 at 4:59

here shows the reason. T'(t) N(t) = _____ ||T'(t)||

your answer is like N(t)=f'(t)/(||f'(t)||)

share|cite|improve this answer
This question was already answered by the looks of the comments above. And this was more than a year ago. How is your answer helpful at all? I can't see it. – Daniel R Nov 1 '13 at 13:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.