How to Find Slope of Normal to Curve

In the application of derivatives, tangents and normals are important concepts. Each normal line is perpendicular to the tangent line drawn at the point where the normal meets the curve. So the slope of each normal line is the opposite reciprocal of the slope of the corresponding tangent line, which can be derived by the derivative. In this section, aspirants will learn how to find the slope of normal to a curve with the help of solved examples.

By knowing the equation of tangent, the equation of normal can be written as yy0=1f(x0)[xx0]y – y_{0} = \frac{-1}{f'(x_{0})}[x-x_{0}]

How do you find the slope of a line normal to a curve?

In analytic geometry, the equation of a line, say y = mx + b, wherever m is slope and b is the y-intercept. The slope of any line perpendicular to a line with slope m is the negative reciprocal, that is -1/m. Currently, we tend to don’t continually y-intercept, therefore a rather totally different variety of line equations is commonly useful: yy1=m(xx1)y-y_{1}=m(x-x_{1}), wherever m is our slope, and x1 and y1 are the coordinates of some extent.

Solved Examples

Example 1: Find the slope of the curve y = (1 + x)sin x at x = π/4.

Solution: Given equation is y = (1 + x)sin x

Apply derivative on above equation

dy/dx = (1 + x) cos x + sin x

[Using product rule]

Now, slope at x = π/4

dy/dx = (1 + π/4) cos (π/4) + sin(π/4)

= 1/√2 . (2 + π/4) or (2 + π/4)/√2

Which is the slope of the given curve.

Example 2: Find the equation of a normal line on a curve y = x3 – 2x2 + 2x – 3 at x = 1.

Solution:

y = x3 – 2x2 + 2x – 3 …(1)

Derivative given equation, we have

y’ = 3x2 – 4x + 2

we tend to judge this spinoff at x = 1, which is y’ = 1

Or slope, m, from this performs at x = 1 is m = 1.

Slope of normal line, m’ = 1/m = -1

Find the value of y, by substituting x = 1 in equation (1)

y = -3

Now, equation of a normal line at point (1, -3) and with slope -1 is

y – (-3) = -1(x – 1)

Or y + x + 2 = 0

Example 3: Find the slope of the normal to the curve y = 2x^2+ 3 sin x at x = 0.

Solution:

The slope of normal to a curve is given as,

m = −1 / [dy/ dx]

Here, the equation of the curve is, y = 2x^2 + 3 sinx

⇒ dy/ dx = 4x + 3 cosx

∴dy/ dx at x = 0 = 4 (0) + 3 cos 0 = 3

∴ The slope of the normal = −1 / 3.

Example 4: Find the equation of the normal line to the curve y=x3+exat x0=0.y = {x^3} + {e^x} \text at\ {x_0} = 0.

Solution: 

Determine the value of the function at x0=0.{x_0} = 0. y0=y(0)=03+e0=1.{y_0} = y\left( 0 \right) = {0^3} + {e^0} = 1.

At the point x0=0,{x_0} = 0, it equals

y(0)=302+e0=1.y^\prime\left( 0 \right) = 3 \cdot {0^2} + {e^0} = 1.

Thus, the equation of the normal is written as follows:

yy0=1y(x0)(xx0),y1=11(x0),y=x+1.y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\\ y – 1 = – \frac{1}{1}\left( {x – 0} \right),\\ y = – x + 1.\\

Example 5: Find the equations of the tangent line and normal line to the parabola y=2x2y = 2{x^2} at the point (2,8)\left( {2,8} \right).

Solution: 

Find the derivative of the function

y=(2x2)=4x,    y(2)=8.{y^\prime = \left( {2{x^2}} \right)^\prime = 4x,}\;\; \Rightarrow {y^\prime\left( 2 \right) = 8.}

The equation of tangent line is

yy0=y(x0)(xx0),y8=8(x2),y8=8x16,8xy8=0.y – {y_0} = y^\prime\left( {{x_0}} \right)\left( {x – {x_0}} \right),\\ y – 8 = 8\left( {x – 2} \right),\\ y – 8 = 8x – 16,\\ 8x – y – 8 = 0.\\

The equation of normal line is

yy0=1y(x0)(xx0),y8=18(x2),y8=x8+14,8y64=x+2,x+8y66=0.y – {y_0} = – \frac{1}{{y^\prime\left( {{x_0}} \right)}}\left( {x – {x_0}} \right),\\ y – 8 = – \frac{1}{8}\left( {x – 2} \right),\\ y – 8 = – \frac{x}{8} + \frac{1}{4},\\ 8y – 64 = – x + 2,\\ x + 8y – 66 = 0.\\

Hence, 8x – y – 8 = 0 and x + 8y – 66 = 0.

Example 6: Write an equation of the normal to the ellipse x24+y21=1\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1} = 1 at the point (1,32)\left( {1,\large\frac{{\sqrt 3 }}{2}\normalsize} \right).

Solution:

Find the derivative of the function

(x24+y21)=1,    2x4+2yy=0,    4yy=x,    y=x4y.{{\left( {\frac{{{x^2}}}{4} + \frac{{{y^2}}}{1}} \right)^\prime } = 1′,\;\;}\Rightarrow\\ {\frac{{2x}}{4} + 2yy’ = 0,\;\;}\Rightarrow\\ {4yy’ = – x,\;\;}\Rightarrow\\ {y = – \frac{x}{{4y}}.}\\

The derivative is the point of tangency is equal to

y(x0,y0)=y(1,32)=1432=123.{y’\left( {{x_0},{y_0}} \right) = y’\left( {1,\frac{{\sqrt 3 }}{2}} \right) }\\ = { – \frac{1}{{\frac{{4\sqrt 3 }}{2}}} }\\ = { – \frac{1}{{2\sqrt 3 }}.}\\

Then the equation of the normal is written as

yy0=1y(x0,y0)(xx0),    y32=1(123)(x1),    y32=23x23,    y=23x23+32,    y=23x3323,46x2,60.{y – {y_0} = – \frac{1}{{y’\left( {{x_0},{y_0}} \right)}}\left( {x – {x_0}} \right),\;\;}\Rightarrow\\ { y – \frac{{\sqrt 3 }}{2} = – \frac{1}{{\left( { – \frac{1}{{2\sqrt 3 }}} \right)}}\left( {x – 1} \right),\;\;}\Rightarrow\\ {y – \frac{{\sqrt 3 }}{2} = 2\sqrt 3 x – 2\sqrt 3 ,\;\;}\Rightarrow\\ {y = 2\sqrt 3 x – 2\sqrt 3 + \frac{{\sqrt 3 }}{2},\;\;}\Rightarrow\\ {y = 2\sqrt 3 x – \frac{{3\sqrt 3 }}{2}} \approx {3,46x – 2,60.}\\