## Sunday, February 5, 2017

### High School Math Solutions – Derivative Applications Calculator, Normal Lines

Last blog post, we talk about using derivatives to compute the tangent lines of functions at certain points. Another application of derivatives is finding the normal line of a function at a certain point.

A normal line is a line that is perpendicular to a tangent line. The slope of the normal line is the negative reciprocal of the slope of the tangent line.

The steps for finding the equation of a normal line is pretty simple, as long as you’ve mastered finding the equation of a tangent line.

Steps for finding the equation of a normal line:

1.  Find the normal point
• The normal point is the same as the tangent point
2.  Compute the slope of the function at the x coordinate
• Compute the derivative of the function at the x coordinate
• This is the slope of the tangent line
3.  Compute the slope of the perpendicular line at the x coordinate
• The slope (m) is the negative reciprocal of the slope (m_1) of the tangent line
• m=\frac{-1}{m_1}
4.  Use the point-slope formula to find the equation of the normal line
• y-y_1=m(x-x_1)

Let’s see some examples.

Find the normal line of y=x^2-x-1 at x=2

1. Find the normal point

y=(2)^2-2-1=1

(2,1)

2. Compute the slope of the function at the x coordinate

y^'=2x-1

y^'=2(2)-1=3

m_1=3

3. Compute the slope of the perpendicular line at the x coordinate

m=\frac{-1}{m_1}
m=\frac{-1}{3}

4. Use the point-slope formula to find the equation of the normal line

(2,1)       m=\frac{-1}{3}

y-y_1=m(x-x_1)

y-1=\frac{-1}{3}(x-2)

y=\frac{-1}{3} x+\frac{5}{3}

Find the normal line of \f(x)=x^4+2e^x  at (0,2)

1. Find the normal point

2. Compute the slope of the function at the x coordinate

\f^' (x)=4x^3+2e^x

\f^' (0)=4(0)^3+2e^0=2

m_1=2

3. Compute the slope of the perpendicular line at the x coordinate

m=\frac{-1}{2}

4. Use the point-slope formula to find the equation of the normal line

(0,2)      m=\frac{-1}{2}

y-2=\frac{-1}{2}(x-0)

y=\frac{-1}{2} x+2

Find the normal line of \f(x)=\frac{1}{x^2}   at x=-1

1. Find the normal point

\f(-1)=\frac{1}{(-1)^2 }=1

(-1,1)

2. Compute the slope of the function at the x coordinate

\f^' (x)=\frac{-2}{x^3}

\f^' (-1)=\frac{-2}{(-1)^3} =2

m_1=2

3. Compute the slope of the perpendicular line at the x coordinate

m=\frac{-1}{2}

4. Use point-slope formula to find the equation of the normal line

(-1,1)     m=\frac{-1}{2}

y-1=\frac{-1}{2}(x-(-1))

y=\frac{-1}{2} x+\frac{1}{2}

Finding the equation of a normal line is very similar to finding the equation of a tangent line. Since the steps are similar, make sure you don’t confuse and mix up the definitions of a tangent line and a normal line. For more practice and help with normal lines, check out Symbolab’s Practice.

Until next time,

Leah

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