umm i don't remember how to find the slope for this problem.
use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)
e^y + x^2 - 3y^2=5
so i know the tangent formula is (y-y1) = m (x-x1)
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umm i don't remember how to find the slope for this problem.
use implicit differentiation to find and equation of the line tangent to the graph at the point (2,0)
e^y + x^2 - 3y^2=5
so i know the tangent formula is (y-y1) = m (x-x1)
Well,Quote:
Originally Posted by FinalFantasy9291
you have x1 = 2 , y1 = 0 ..
By differentiating both sides w.r.t. x, we got:
solve for y' , then evaluate it at (2,0), i.e. when x=2 and y=0 to get the value of m ..
then substitute x1,y1 and m the tangent line's formula ..
Please, note that its called "tangent line's formula" not "tangent formula" ..
..
thank you for that correction haha.
so from what you gave me i got (y-0)=-4(x-2) is that correct?
Let me see your work ..Quote:
Originally Posted by FinalFantasy9291
mayba you have mistakes ..
If you got the right final answer, that does not mean you've solved the problem correctly ..
okay so..
y'e^y+2x-6yy' = 0
btw is there a way i can type this out without having to use ^ so it's not as confusing for you or me?
y'e^0+(2)-6(0)y'=0
y' + 4= 0
y'=-4
so... (y-0)=-4(x-2)
Yes. Use the latex.Quote:
Originally Posted by FinalFantasy9291
I think that 2 is just a typo .. Right?Quote:
Originally Posted by FinalFantasy9291
It should be 4..
Correct.Quote:
Originally Posted by FinalFantasy9291
hehe. how do i use the latex?
tyvm
http://www.mathhelpforum.com/math-help/latex-help/Quote:
Originally Posted by FinalFantasy9291