Question

what is the slope of the line normal to the graph y= x^2/ (3 in the radical thing, not sure what its called) √(3x^2+1) at x =1

Answer 1

\[\Large\bf\sf y=\frac{x^2}{\sqrt[3]{3x^2+1}},\qquad\qquad\text{at } x=1\]

Answer 2

That look right?

Answer 3

yes :)

Answer 4

We only need the `slope` of the normal line?
We don't need to setup an equation for it?
Oh that's nice :)

So we'll start by finding the slope of the line `tangent` to the curve at x=1.
How do we find tangent lines? Remember? :X

Answer 5

we derive to get slope? sorry;) and then y =mx +b

Answer 6

Since they didn't ask for the `equation` of the line normal to the curve, we don't have to worry about the y=mx+b part!

We'll simply find the m, (which as you said, we'll derive to find it) and then we'll do some magic to find the slope of the normal line.

Answer 7

lol

Answer 8

Have you tried taking a derivative yet?
This one is a bit of a doozy.

Make sure you don't try to simplify.

Answer 9

no so we use quotient rule

Answer 10

sounds good.

Answer 11

ok ill do

Answer 12

\[\Large\bf\sf y'=\frac{\color{orangered}{(x^2)'}\sqrt[3]{3x^2+1}-x^2\color{orangered}{(\sqrt[3]{3x^2+1})'}}{\left(\sqrt[3]{3x^2+1}\right)^2}\]

There's our quotient rule setup,
Gotta differentiate the orange parts.
Get to it girl! :O

Answer 13

i finished lol

Answer 14

Hmm kinda hard to write it all out on here isn't it? :\

Answer 15

jk still working on it wait im simplifying

Answer 16

\[\Large\bf\sf y'=\frac{\color{royalblue}{(2x)}\sqrt[3]{3x^2+1}-x^2\cdot\color{royalblue}{6x(3x^2+1)^{-2/3}}}{\left(\sqrt[3]{3x^2+1}\right)^2}\]

Answer 17

Nooo don't simplify :O that's a waste of time.

Answer 18

Did you get something like that?

Answer 19

lol yea i got that

Answer 20

Woops I forgot my 1/3 in front of the second term. my bad.

Answer 21

From the power rule on the cube root term.

Answer 22

\[\Large\bf\sf y'=\frac{\color{royalblue}{(2x)}\sqrt[3]{3x^2+1}-x^2\cdot\color{royalblue}{2x(3x^2+1)^{-2/3}}}{\left(\sqrt[3]{3x^2+1}\right)^2}\]

Answer 23

Now we'll evaluate it at x=1.

This is why we don't want to simplify ahead of time :U
Plugging in 1 should shrink it down pretty quickly.

Answer 24

hold pls i messed up whit the exponent s ill tell u

Answer 25

ok im their

Answer 26

so now we just plug in 1 to F(x)

Answer 27

y'x

Answer 28

ya find y'(1).

Answer 29

31/4??

Answer 30

wait forgot the bottom

Answer 31

sorry im confusing myself

Answer 32

Hmm this one is a doozy :\

Answer 33

yea help me pls

Answer 34

Mmm let's assume we did our derivative correctly and just see what happens.
\[\Large\bf\sf y'=\frac{(2)4^{1/3}-2(4)^{-2/3}}{(4)^{2/3}}\]

Answer 35

So that's what I got when I plugged 1 in.
Were you able to get to that step or so? :o

Answer 36

i forgot how to convert the square roots to fractions but i got it now

Answer 37

thank you

Answer 38

I hope we did that correctly D:
Imma try the derivative again.. looks weird..

Answer 39

Ya if you use your calculator to simplify it down further, you'll end up with,\[\Large\bf\sf y'(1)\approx 0.9449408\]Which is good. That's correct, so our derivative looks good.
That's the slope of the line `tangent` to our curve at x=1.

Remember how a normal line relates to a tangent line?

Answer 40

If your teacher wants the answer in exact form, then maybe we should avoid the decimal :d it's up to you.

Answer 41

i honestly forgot but something simple

Answer 42

ya simple :d
you flip and negate.

The slope of the normal line will be the `negative reciprocal` of the slope of the tangent line.

Answer 43

\[\Large\bf\sf m_{\tan}\quad=\quad -\frac{1}{m_{norm}}\]

Answer 44

so is .945 the slope of the normal line b/c its one of my answers?

Answer 45

It's the slope of the tangent line. Or it should be at least :x

Answer 46

For the slope of the normal line, we need to flip it, and negate it.\[\Large\bf\sf m_{\tan}\approx 0.9449408\]

The answer we're looking for, the slope of the normal line,\[\Large\bf\sf m_{norm}\approx -\frac{1}{0.9449408}\]

Answer 47

Hopefully this will give us one of your answer choices also ^^ lol

Answer 48

Do you have a fancy calculator you can do that step with? :D

Answer 49

yea i fgot the ti inspire thank you

Answer 50

Thank You Zepdrik Goodnight <3

Answer 51

get the right answer? :O go to bed!