## Sunday, February 5, 2017

### High School Math Solutions – Derivative Applications Calculator, Tangent Line

We learned in previous posts how to take the derivative of a function. Now, it’s time to see the applications of derivatives.

We use derivatives when we find the equation of a tangent line. A tangent line is a straight line that just touches the curve at a point (on the curve). Tangent lines can help us find the length of the curve and their slopes tell us what the curve looks like and where we can find maximum and minimums. When we take the derivative of the function of the curve at a particular point, we get the slope of the tangent line.

Let’s see how we can use taking the derivative to find the equation of a tangent line.

Steps to find the equation of a tangent line at a point:

1.  Find the tangent point

• Plug in the value for x into the function to find the y coordinate

2.  Compute the slope of the function

• Take the derivative of the function

3.  Compute the slope of the function at the given x coordinate

• Plug in the value for x into the derivative

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

• y-y_1=m(x-x_1)
• Get (x_1, y_1) from Step 1 and get m from Step 3

We’ll now go over some examples.

Find the tangent line of \f(x)=\sqrt{x^2+1}  at  x=-1

1. Find the tangent point

\f(-1)=\sqrt{(-1)^2+1}=\sqrt{2}

(-1,\sqrt{2})

2. Compute the slope of the function

\f(x)=(x^2+1)^(\frac{1}{2})

\f^' (x)=\frac{1}{2} (x^2+1)^(\frac{-1}{2})∙2x

\f^' (x)=\frac{x}{\sqrt{x^2+1}

3. Compute the slope of the function at the given x coordinate

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

m=\frac{-1}{\sqrt{2}}

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

(-1,\sqrt{2})            m=\frac{-1}{\sqrt{2}}

y-y_1=m(x-x_1)

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

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

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

Find the tangent line of \f(x)=\frac{1}{x^2}   at (-1,1)
1. Find the tangent point

We can skip this step because the tangent point is given

2. Compute the slope of the function

\f(x)=x^(-2)

\f^' (x)=-2x^(-3)

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

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

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

m=2

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

(-1,1)          m=2

y-1=2(x-(-1))

y=2x+3

Find the tangent line of \f(x)=x^2+2x+3 at  x=2
1. Find the tangent point

\f(2)=2^2+2(2)+3=11

2. Compute the slope of the function

\f^' (x)=2x+2

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

\f^' (2)=2(2)+2=6

m=6

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

(2,11)        m=6

y-11=6(x-2)

y=6x-1

As you can see, finding the equation of a tangent line of a point on a curve is not too hard. As long as you’ve mastered computing derivatives and the steps to finding the equation of the tangent line, you will be able to solve these problems quick and easily. For more help or practice on this topic, check out Symbolab’s Practice.

Until next time,

Leah

### 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