PLEASE HELP!
At the given point fine the line that is normal to the curve. Y^4+x^3=y^2+10x, normal at (0,1). I have the answer but I need the steps!

Question

PLEASE HELP! 
At the given point fine the line that is normal to the curve. Y^4+x^3=y^2+10x, normal at (0,1). I have the answer but I need the steps!

anonymous · Answer

find*

zepdrix · Answer

Still stuck on this one or no?

anonymous · Answer

No, I got this one!

anonymous · Answer

Wait I need this one, sorry!

zepdrix · Answer

lol

anonymous · Answer

I thought it was the same one haha

zepdrix · Answer

$$\large y^4+x^3=y^2+10x$$

$$\large y'(x,y)=m$$
M gives us the slope of the line `tangent` to our function.
Do you know the relationship between the `tangent` and `normal` lines?

anonymous · Answer

Sort of, like its a perpendicular line? Or am I totally off?

zepdrix · Answer

Good, yes :) their perpendicular to one another.
To get the slope of the normal line, we'll take the `negative reciprocal` of our tangent slope.

So let's start by finding $\large y'$.

anonymous · Answer

Okay, so where I was stuck on was the powers, how do I get it in y equals form since there are like 3 different powers used

zepdrix · Answer

Each time we differentiate a y term, a y' will pop out right?
So you stuck at this point?

$$\large 4y^3y'+3x^2=2yy'+10$$

anonymous · Answer

yes!

zepdrix · Answer

We want to solve for y'.
Let's start by getting all the (y')'s to one side.
We'll subtract 2yy' from each side,
and subtract 3x^2 from each side,$$\large 4y^3y'-2yy'=10-3x^2$$
From here we can `factor` y' out of each term on the left,$$\large y'(4y^3-2y)=10-3x^2$$Then to solve for y', we divide both sides by that big mess,

anonymous · Answer

Okay, so I understand up to here

anonymous · Answer

That would be y'=10-3x^2/4y^3-2y

zepdrix · Answer

Ok good :)
Since y' is a function of `both` x and y, let's write it like this,
$$\large y'(x,y)=\frac{10-3x^2}{4y^3-2y}$$

The slope of the line tangent to our function at a particular point will be given by,$$\large y'(0,1)$$

So plug that in and see what you get.

anonymous · Answer

5

anonymous · Answer

So the equation I get is y=5x+1?

zepdrix · Answer

So if we were asked to find an equation for the line `tangent to the curve`, we would say, "ok so 5 is the slope of our line"

And we would write it in this form,$$\large y=mx+b$$

But in this problem we want an equation for the line `normal to the curve`. So we'll need to take the negative reciprocal of the slope value we found.

That will be the M in our normal line equation.

anonymous · Answer

So the negative reciprocal is -1/5? So our final answer is y=-1/5x+1?

zepdrix · Answer

You were able to find the y-intercept with no problem? Ok cool.

Yah I think that's right :)
Lemme check on wolfram real quick to make sure I didn't make a silly mistake somewhere.

zepdrix · Answer

Blah the function is too complicated to input XD lol
I'm pretty sure we did that right anyway :D

zepdrix · Answer

Oh do you have an answer key for this one? :3

anonymous · Answer

No if you say that's right then it is, cause I have the correct answer! I just didn't know the process but I followed what you did and I came out with the same answer! You are so helpful! :) Do you think you could help me on a couple more?

zepdrix · Answer

Turned out correct? Ah good!
Sure.

anonymous · Answer

You are a lifesaver!

zepdrix · Answer

The best way would be to umm, go to your questions and type @zepdrix in the description somewhere.

It sends me a notification, making it easier to find the problem.

anonymous · Answer

Okay I'll do that! so I just type it in the ask a question box?

zepdrix · Answer

Yes that's fine.
Or you can Ask the question, and then type it in the posts area.