I don't understand how you did part d

Also, for part c, is my answer and what you differentiated the same thing? Did you pick that one because it's easier?

@Hello: For your first comment: Since $(a,-2a)$ is a point on the curve, you can put $x=a,y=-2a$ in $x^2+3y^2+xy=3$, so you can find $a$. For the second comment: part c? maybe you mean part a or b?

no i mean part c when you found the second derivative

@Hello: Well, I cannot find any answer for $c$, so maybe you mean the answers for part a,b: Your answer for part a is correct, but your answer for part b is not correct. The correct answer is $y-1=((6\cdot 1+0)/(2\cdot 0+1))x\iff y-1=6x$. (This comes from $y-1=((6Y+X)/(2X+Y))x$ but you need put $(X,Y)=(1,0)$ in it.)

Duh, thanks for the help on part b. For part d, what does "the nature of the line" mean?

@Hello: First, please check if you have no typo in the question d. If you have no typo, then we can get two distinct tangent lines. So, this means that we can get the two concrete lines. Hence, all we can do is to show their slopes and y-intercept of the line, I guess.

Nope no mistake. The question can be found here (#5): online.math.uh.edu/apcalculus/exams/AB_SECTION_II_version_1.pdf

@Hello: OK. My understanding is that we are asked to show their equations. If the question asks the other things, I think the question asks badly.

So I have a and (now) b correct. c you said you don't know? and d, what do you mean by "other things"?

@mathlove Hi

The answer for c is $-17/36$. For d, I don't mean anything by "other things". I'm just saying that my understanding is that we are asked to show their equations of the tangent lines.

how did you get that for c?

Since y'=(-2x-y)/(6y+x), putting x=0,y=1 gives you y'=-1/6. Now putting x=0,y=1,y'=-1/6 in y''=(-2-2y'-6(y')^2)/(6y+x) gives you the answer.

i don't understand how to get the second derivative

OK, first, do you understand how to get 2x+6yy'+y+xy'=0?

the yy' is confusing me

is there another way to write that

yy' represents that y multiplied by y'. Another way is y*y'...

yes i understand how you got that. i just wrote dy/dx instead of y'

where do we go from there?

Yeah, you can write dy/dx, but for me y' is easy to write and understand because it is not in the form of fraction. After getting 2x+6yy'+y+xy'=0, can you differentiate the both sides?

when you differentiate that, does that mean you'll get 6(dy/dx)(dy/dx)? 2 of them?

wait

i get it

Fine. any problem?

so far i have 2+6(d2y/dx2). not sure what to do with x(dy/dx). is it just 1? where does the dy/dx go?

You need to be careful about (f(x)g(x))'=f'(x)g(x)+f(x)g'(x). So, for example, (yy')'=y'y'+yy'', or d(y(dy/dx))/dx=(dy/dx)(dy/dx)+y(d^2y/dx^2).

this is really hard to read in text D:

Ok, I will write it in my answer. Wait a minute.

I wrote it. Take a look.

what level of math would you say this question is?

I have to say this is a fundamental level.

precalc?

not precalculus, I think.

oh. i just finished precalc and this is on my summer assignment for ap

I see. I really would like you to understand how to solve this question. Is there anything unclear?

it's just taking the second derivative. i've never had to deal with an implicit second derivative

I see. By the way, I think you may want to learn how to use y' because using y' makes things easier, I think.

idk my teacher uses dy/dx (he was my precalc teacher and will be my ap calc teacher). so it'll probably be best to stick with what he uses

Fine. No problem. Do you understand what I wrote in my answer, then?

not really

You know (f(x)g(x))'=f'(x)g(x)+f(x)g'(x), don't you?

yeah the product rule

I used it for y(dy/dx).

i just realized, isn't there some way to find it out graphically?

graphically? I have no idea.

like on the calculator perhaps

You mean like this? wolframalpha.com/input/?i=d%28y%28dy%2Fdx%29%29%2Fdx

nevermind. the instructions say calculators aren't permitted

Ok, then, please ask something if you have. If you don't have anything unclear, I would like you to check my answer and finish talking.

i don't understand your answer

Ok, then, which part.

imagizer.imageshack.us/v2/240x320q90/913/eRTCV9.jpg good so far?

Yes. Sorry the correct answer is -11/36 though I said it is -17/36.

so can i just replace the dy/dx there and plug in? or should i simplify then plug in

whichever you like. since x=0, y=1 are very simple, you can plug them in now.

how can i simplify it?

Just expand the numerator.

too annoying. i'll just plug in now

yeah:)

finally. part d time!

Well done!

what did you do for part d in the answer?

i see you plugged it in, but why the original?

Because a point (a,-2a) is on the curve x^2+3y^2+xy=3. This means x=a,y=-2a satisfy x^2+3y^2+xy=3.

oh. i wasn't sure because it said determine the nature of the line tangent to the curve

I see what you mean. Then, we can find the value of a.

is ±sqrt(3/11) the anser?

answer*

Exactly.

even if it says determine the nature of the line?

Yes. but the final answer is not the value of a. You need to find the equation of the two tangent lines.

so i'm not finished?

Yes. You got two values of a. This means that you need to find the equation of the tangent line at (sqrt(3/11),-2sqrt(3/11)) and (-sqrt(3/11),2sqrt(3/11)).

Sorry, I meant no. you are not finished.

so is the final answer going to be an equation?

Yes. Two equations.

doesn;t the nature mean like min/max

what do you mean min/max of a line?

idk what i'm talking about lol. i just dontk now if its supposed to be an equation

The best way would be to ask your teacher if you can.

if this is an assignment.

well it's a summer assignment so i can't

i don't think he's grading it anyway since i haven't actually learned all of ap calc

Ok, all I can say is that my understanding is that we are asked to find the equations of the tangent lines. so?

Don't delete the question. It's so rude.

Oh i thought since it was solved it was supposed to be. im new

by the way we had to plug into the derivative not the original :(

no it is not supposed to do so. yeah, plugging into the derivative wins, but as a result, we lead the same conclusion, anyway if we find the equations, haha.