i can find the tangent and the norm with some easy when the equation to start is something like
3x^2+8x+4 when x=1
but i have a new equation
3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
i can differentiate to get
6x+2x2y+2y^2=6y^2
but dont get how i use y=mx+c
if x=2 and y=3 this means
3=m2+c
all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you

2. Originally Posted by alpharebel10
i can find the tangent and the norm with some easy when the equation to start is something like
3x^2+8x+4 when x=1
but i have a new equation
3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
i can differentiate to get
6x+2x2y+2y^2=6y^2
but dont get how i use y=mx+c
if x=2 and y=3 this means
3=m2+c
all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you
You need to be careful when differentiating implicitly. For your function

$3x^2+2xy^2+6=2y^3$

you have $6x + 2y^2 + 4xy y' = 6y^2 y'$ which solving for y' gives

$
y' =\frac{3x + y^2}{3y^2-2xy}
$
and at (2,3) this becomes $y' = 1$, which is the slope of the tangent. Now the equation of the tangent is $y - 3 = 1(x-2)$ or $y = x + 1$. For the normal, use $y - 3 = -1(x-2)$

3. Originally Posted by alpharebel10
i can find the tangent and the norm with some easy when the equation to start is something like
That's "normal", not "norm", which has quite a different meaning in mathmathics. Also it is not clear whether you want the tangent and normal lines or vectors. I will asume you mean lines.

3x^2+8x+4 when x=1
but i have a new equation
3x^2+2xy^2+6=2y^3 and have to find the tangent and norm at point (2,3)
i can differentiate to get
6x+2x2y+2y^2=6y^2
No, you don't. Since y is a function of x and you are differentiating with respect to x, you have to use the "chain rule". The derivative of 2xy^3, with respect to x, is, by the product rule 2y^2+ 2x(y^2)'= 2y^2+ 4xyy' and the derivative of 2y^3, with respect to x is 6y y'.
The derivative of 3x^2+ 2xy^2+ 6= 2y^3 is 6x+ 2y^2+ 4xyy'= 6y^2y'. Solve that for y' (when x= 2 and y= 3) and use that for your m.

but dont get how i use y=mx+c
if x=2 and y=3 this means
3=m2+c
all i get 4 m is 54 and i dont think this is right can u please put me on the right line thank you