- #1
ngm01
- 8
- 0
Greetings, I faced with a problem that states - Find slope of the tangent line to 〖 x〗^4 –
xy^2+ 4xy^2 = 20,at (1,2)
First I set F= x^4 – xy^2+ 4xy^2-20=0
I found dF/dx = 4x^3 + y^2+ 4y^2= 24 ,at (1,2)
then dF/dy = 2xy+8xy,=20 at (1,2) …. Armed with this I then made the assumption that
dy/dx = dF/dX multiplied dy/dF where dy/dF is simply the inverse of dF/dy which led
to dy/dx = 24/20 = 1.2…. however that’s not the answer. The answer is showed as – 1.2,
explained as the perpendicular slope is 20/24 and the tangent slope as - 24/20
Can you tell me what I’m missing here?
Thanks
xy^2+ 4xy^2 = 20,at (1,2)
First I set F= x^4 – xy^2+ 4xy^2-20=0
I found dF/dx = 4x^3 + y^2+ 4y^2= 24 ,at (1,2)
then dF/dy = 2xy+8xy,=20 at (1,2) …. Armed with this I then made the assumption that
dy/dx = dF/dX multiplied dy/dF where dy/dF is simply the inverse of dF/dy which led
to dy/dx = 24/20 = 1.2…. however that’s not the answer. The answer is showed as – 1.2,
explained as the perpendicular slope is 20/24 and the tangent slope as - 24/20
Can you tell me what I’m missing here?
Thanks