Equation of the Normal Line...4

[nycmathguy]

Find the equation of the normal line to the curve f(x) = x^3 + 2 at the point P (1, 4).

dy/dx = 3x

Let x = 1.

3(1) = 3 = slope of tangent line.

The slope of the normal line in this case is m = -1/3.

Use y = mx + b to find b.

4 = (-1/3)(1) + b

4 = (-1/3) + b

4 + (1/3) = b

13/3 = b

The equation of the normal line to the given curve at the point (1, 4) is y = (-1/3)(x) + (13/3).

You say?

 

[MathLover1]

correction: dy/dx = 3x^2

(dy/dx) f(1) =3*1^2 =3

the slope of tangent line at x=1 is m=3 and passing through (1, 3)
y-3=3(x-1)
y=3x-3+3
y=3x ->tangent line

=> the slope of normal line will be m= -1/3 and the point (1, 3), so the equation of a normal line will be:
y-y1=m(x-x1)
y-3=-(1/3)(x-1)
y-4=-(1/3)x+1/3
y=-(1/3)x+1/3+3
y=-(1/3)x+10/3 ->normal line

 

[nycmathguy]

correction: dy/dx = 3x^2

(dy/dx) f(1) =3*1^2 =3

the slope of tangent line at x=1 is m=3 and passing through (1, 3)
y-3=3(x-1)
y=3x-3+3
y=3x ->tangent line

=> the slope of normal line will be m= -1/3 and the point (1, 3), so the equation of a normal line will be:
y-y1=m(x-x1)
y-3=-(1/3)(x-1)
y-4=-(1/3)x+1/3
y=-(1/3)x+1/3+3
y=-(1/3)x+10/3 ->normal line

I see my error. Yes, dy/dx is 3x^2. I worked out all of these problems without a single hour of sleep. However, I'm not making excuses. I am simply being honest. The important thing here is that I understand how to find the normal line.