Calculus 1-->Curve of Serpentine

Well for the Slope of the Tangent..
dy/dx = [(1 + x^2) - (x)(2x)]/(1 + x^2)^2 = m (Slope)
Therefore m = (1 - x^2)/(1 + x^2)^2
Slot in for x and M = (1 - 9)/[(1 + 9)^2)
m = 8/100 = 2/25

Then y - y' = m(x - x')
y - 0.3 = 2/25(x - 3)
25y - 7.5 = 2x - 6
2x - 25y + 1.5 = 0 is a tangent to that curve at (3 , 0.3)

ok
Excellent notes. I thank you very much.

anything.. your welcome..

Curve of Serpentine
Find the equations of the tangent line and normal line to the given curve at the specified point.

y=sqrt(x)/x+2 , (1,1/3)

Tangent y=

Normal y =

Curve of Serpentine
The curve below is called a serpentine. Find an equation of the tangent line to this curve at the point (-3, -0.3).

y=x/1+x^2

y=

Re: Curve of Serpentine


Same thing

you get the slope of a line by finding the derivative (dy/dx), plug in the x0 to find m0 and use y0-m0x0=b. The tangent line then is y=m0x+b

The normal to that line is determined using the condition that the product of the slope of a line and the slope of a perpendicular line (normal) is -1. So m0*mn0=-1 or mn0=-1/m0.

Rap

Thank you

wouldn't it be negative 2/25? cause 1-9 = -8 and then dividing it by 100 you're gonna get negative. -8/100 = -2/25...