Equation of the Normal Line...2

[nycmathguy]

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

 

[MathLover1]

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

We are expected to find the equation of a line normal to the curve f(x)=x^2
at the point P (2, 4). A normal line is a line perpendicular to the tangent line, so we will take the derivative of f(x) to find the slope of the tangent line, and then take the negative reciprocal of this slope, to find the slope of the normal line.
(d/dx)f(x)=2x
then for x=2
(d/dx)f(2)=2*2=4 ->the slope of the tangent line is 4

and the slope of the normal line will be m=-1/4

The equation of a normal line will have the form y=mx+b and we know a slope and the point P (2, 4), so use it to calculate b

4=(-1/4)*2+b

b=9/2

then, equation of the normal line is:
y=-(1/4)x+9/2

 

[nycmathguy]

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

We are expected to find the equation of a line normal to the curve f(x)=x^2
at the point P (2, 4). A normal line is a line perpendicular to the tangent line, so we will take the derivative of f(x) to find the slope of the tangent line, and then take the negative reciprocal of this slope, to find the slope of the normal line.
(d/dx)f(x)=2x
then for x=2
(d/dx)f(2)=2*2=4 ->the slope of the tangent line is 4

and the slope of the normal line will be m=-1/4

The equation of a normal line will have the form y=mx+b and we know a slope and the point P (2, 4), so use it to calculate b

4=(-1/4)*2+b

b=9/2

then, equation of the normal line is:
y=-(1/4)x+9/2

View attachment 2305

Not too bad at all. I will post and solve several similar equations when time allows. I might take the new room, which means busy, busy, busy.