IsTim
  • IsTim
Determine the equation of the tangent to the curve at a given point. x^2+9y^2=37 (1,2)
Mathematics
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IsTim
  • IsTim
Determine the equation of the tangent to the curve at a given point. x^2+9y^2=37 (1,2)
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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IsTim
  • IsTim
I've gotten the slope of -1/18, but I don't know if that's correct. And if so, how do I continue?
IsTim
  • IsTim
So, first, dy/dx=-x/dy. I then subbed in (1,2) and got -1/18.
anonymous
  • anonymous
Lol uhh so I cant give answers so ill just leave now but what youv done so far is correct :D

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IsTim
  • IsTim
I don't need the answer. I need to know what my process afterwards should be.
inkyvoyd
  • inkyvoyd
What's the derivative?
inkyvoyd
  • inkyvoyd
And Chad is in 11th grade, he won't be able to help you
IsTim
  • IsTim
1. Derivative is 2x+18y(dy/dx)=0. Right? 2. I figured that out already...
anonymous
  • anonymous
lol ink I can give him an answer ;D
inkyvoyd
  • inkyvoyd
x^2+9y^2=37 dy/dx 37=dy/dx x^2+dy/dx 9y^2 0=2x+18y dy/dx
inkyvoyd
  • inkyvoyd
No, you won't give him the answer he's looking for. THe answer he's looking for is an explanation, because, as everyone knows, it's the explanantion that matters in calc
IsTim
  • IsTim
I already have the answer in my textbook and potentially Wolfram Alpha. I need to know how to GET there though.
inkyvoyd
  • inkyvoyd
Normally at this point I try to isolate dy/dx for convenience
anonymous
  • anonymous
from conic section?
inkyvoyd
  • inkyvoyd
0=2x+18y dy/dx (-2x)/(18y)=dy/dx x/(-9y)=dy/dx
IsTim
  • IsTim
Yeah, I think I did that.It's -x/dy right? @ Shruti: I don't think so...
inkyvoyd
  • inkyvoyd
And, my answer agrees with your answer here.
inkyvoyd
  • inkyvoyd
Although my work here is moot, because i'm in 9th grade.
IsTim
  • IsTim
Ok. Thanks for the help so far.
IsTim
  • IsTim
Ok. How would I get to the answer of x+18y-37=0 from knowing that the slope is -1/18?
inkyvoyd
  • inkyvoyd
eh?
inkyvoyd
  • inkyvoyd
If you had the derivative and wanted to get the original equation?
IsTim
  • IsTim
I guess?
IsTim
  • IsTim
@shruti By conics did you mean http://www.purplemath.com/modules/circle3.htm
amistre64
  • amistre64
since this is an equation of an ellipse, y can be considered an implicit function of x right?
IsTim
  • IsTim
I guess. My notes only reveal the steps. This work is not from a textbook.
IsTim
  • IsTim
In fact, it's not even addressed in my textbook. That's strange.
amistre64
  • amistre64
it can be .... to derive implicitly, ignore the form of the variable and treat everything as tho it acted like you normally think of deriving x^2+9y^2=37 ; just derive it all, but dont toss out the derived bits 2x x' + 18y y' = 0 now lets consider this when we derive it with respect to x x' = dx/dx = 1 y' = dy/dx like normal so, lets solve it for y'
inkyvoyd
  • inkyvoyd
If you have the derivative, and want to get the original equation, you take the anti-derivative of it
inkyvoyd
  • inkyvoyd
If you're asking about taking anti-derivatives (indefinite integrals), and have not yet learned about rieman sums, don't worry about the question
inkyvoyd
  • inkyvoyd
Otherwise, I might have to look into it.
IsTim
  • IsTim
Y' is -x/dy, which, when we sub in (1,2) is -1/18 right?
anonymous
  • anonymous
yes tim, -1/18 is right
IsTim
  • IsTim
What would I do next then?
inkyvoyd
  • inkyvoyd
Well, you have a point, and a slope.
inkyvoyd
  • inkyvoyd
(1,2) -> point -1/18 -> slope
inkyvoyd
  • inkyvoyd
remember, y=mx+b
IsTim
  • IsTim
Would I sub those into y=mx+b?
anonymous
  • anonymous
you have slope, you have a point, put it in point slope form...
inkyvoyd
  • inkyvoyd
put in the point and the slope, and then you have b
inkyvoyd
  • inkyvoyd
then, put in y=mx+b, cept don't put in that point
inkyvoyd
  • inkyvoyd
This isn't a calculus problem, this is an algebra problem, :D
IsTim
  • IsTim
Sorry, what is point slope form? It's algebra? I must really have been "dumbing" down these years...
inkyvoyd
  • inkyvoyd
Ok.
inkyvoyd
  • inkyvoyd
There's multiple ways of writing the equation for a line.
inkyvoyd
  • inkyvoyd
This is point slope: y-y1=m(x-x1)
inkyvoyd
  • inkyvoyd
this is slope intecept: y=mx+b
inkyvoyd
  • inkyvoyd
This is standard form: ax+by=c
inkyvoyd
  • inkyvoyd
I normally work with slope intercept, because i'm more used to it.
inkyvoyd
  • inkyvoyd
However, point slope is the fastest way to go here.
inkyvoyd
  • inkyvoyd
y-y1=m(x-x1) (x1,y1)=(1,2) m=-1/18
IsTim
  • IsTim
The answer is in the form of ax+by=c I guess, but I am also more familiar with slope intercept.
IsTim
  • IsTim
Wait, I already have the slope... I know it is -1/18. But how do I get to x+18y-37?
inkyvoyd
  • inkyvoyd
y-1=(-1/18)(x-2) now, just isolate y, and distribute x
inkyvoyd
  • inkyvoyd
They have given you the point that the line must pass through, and you have found the slope,
IsTim
  • IsTim
Oh. I finally understand. Thank you very much. I will call out again if I need help.
inkyvoyd
  • inkyvoyd
Ok. :D
IsTim
  • IsTim
Wait, 1 and 2 in y-1=(-1/18)(x-2)/ Shouldn't it be y-2=(-1/18)(x-1)? Because (1,2)?
inkyvoyd
  • inkyvoyd
Yes
inkyvoyd
  • inkyvoyd
I got it confused myself. xD
IsTim
  • IsTim
Why doesn't my answer look like that provided on the sheet, which is x+18y-37?
IsTim
  • IsTim
Oh. I figured that out. Sorry.
IsTim
  • IsTim
Thank you all for putting forth the effort to assist me.
IsTim
  • IsTim
Or at the very least, looking over the question.
inkyvoyd
  • inkyvoyd
No problem :)
inkyvoyd
  • inkyvoyd
This is good premptive practice for me ;)

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