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Give the equation of the normal line to the graph of $$y = 2x \sqrt{x^2+8} + 2$$ at the point $(0,2)$

What I've done so far is:

Taken the derivative and got

$$(2x^2)/\sqrt{ x^2+8} + 2\sqrt{x^2+8}$$

I have no idea if this is right, it was pretty hard to get the derivative of that. Before I go on any further, is this derivative right?

Source: http://online.math.uh.edu/apcalculus/exams/AP_AB_version1_1.htm - #9

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  • $\begingroup$ Check whether the above edited function is correct, @Hello. $\endgroup$
    – Timbuc
    Aug 27, 2014 at 23:44
  • $\begingroup$ That's a lot of tedious algebra work for one multiple choice question... $\endgroup$
    – IAmNoOne
    Aug 27, 2014 at 23:45
  • $\begingroup$ @Timbuc yes thank you. $\endgroup$
    – Hello
    Aug 27, 2014 at 23:45
  • $\begingroup$ @Nameless yeah i know haha $\endgroup$
    – Hello
    Aug 27, 2014 at 23:45
  • $\begingroup$ Ok @Hello, now you edit your post and make it correct copying what I edited. $\endgroup$
    – Timbuc
    Aug 27, 2014 at 23:47

2 Answers 2

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$$f(x)=2x \sqrt{x^2+8}+2 \Rightarrow f'(x)=2\sqrt{x^2+8}+\frac{x}{\sqrt{x^2+8}}2x=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{2(x^2+8)+2x^2}{\sqrt{x^2}8}=\frac{4x^2+16}{\sqrt{x^2+8}}$$

At the point $(0,2)$, $f'(0)=\frac{16}{\sqrt{8}}=\frac{16}{2 \sqrt{2}}=\frac{8}{\sqrt{2}}$.

The slope of the normal line is $\frac{-1}{f'(0)}=-\frac{\sqrt{2}}{8}$.

Therefore, the equation of the normal line at the point $(0,2)$ is the following:

$$y-y_1=m(x-x_1) \Rightarrow y-2=-\frac{\sqrt{2}}{8}(x-0) \Rightarrow 8y-16=-\sqrt{2}x \Rightarrow 8y+\sqrt{2}x=16$$

$$8y+\sqrt{2}x=16 \Rightarrow 8 \sqrt{2}y+2x=16 \sqrt{2} \Rightarrow 4 \sqrt{2}y+x=8 \sqrt{2}$$

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  • $\begingroup$ Shouldn't the slope be negative when you said "The slope of the normal line is..." you wrote -1/f'(0) but never actually made it negative $\endgroup$
    – Hello
    Aug 27, 2014 at 23:57
  • $\begingroup$ Oh, sorry! I corrected it! $\endgroup$
    – Mary Star
    Aug 28, 2014 at 0:00
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    $\begingroup$ I understand how you got that and I got it too. But it's a multiple choice question (check the OP where I posted the link (#9)). Any ideas how to turn that into that? $\endgroup$
    – Hello
    Aug 28, 2014 at 0:01
  • $\begingroup$ I edited my answer! Therefore, it is $b$. $\endgroup$
    – Mary Star
    Aug 28, 2014 at 0:06
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$$f'(x)=\left(2x\sqrt{x^2+8}+2\right)'=2\sqrt{x^2+8}+\frac{2x^2}{\sqrt{x^2+8}}=\frac{4x^2+16}{\sqrt{x^2+8}}$$

So

$$f'(0)=\frac{16}{\sqrt8}=\frac{8}{\sqrt2}=4\sqrt2\implies-\frac1{f'(0)}=-\frac1{4\sqrt{2}}$$

Try now to take it from here.

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  • $\begingroup$ Wait before I continue, I had the right numbers? lol $\endgroup$
    – Hello
    Aug 27, 2014 at 23:47
  • $\begingroup$ I can't be sure until you first edit your question and make it clear... $\endgroup$
    – Timbuc
    Aug 27, 2014 at 23:48
  • $\begingroup$ Yeah I actually got that far already. I got $-1/(2\sqrt{8})$ which is the same thing $\endgroup$
    – Hello
    Aug 27, 2014 at 23:51

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