(DC7) Equation of a Tangent Line#

By the end of the lesson you will be able to:

  • find the equation of the tangent line to a curve at a specified point \(P\).

Lecture Videos#

No video for this lesson (other than the prep video).

Equation of a Tangent Lines#

Now that we know how to find the derivative of a function, we can reliably use this new function to help us find the equation of a tangent line.

Ingredients

In order to use the point-slope formula to find the equation of a line we need three things:

  1. \(x_c\) the \(x\)-coordinate of the point of tangency.

  2. \(y_c= f(x_c)\) the \(y\)-coordinate of the point of tangency. Note that we can use the original function to find this value.

  3. \(m = f'(x_c)\) the slope of our tangent line. This is where we use the derivative.

Point-Slope Formula

The equation of a line with slope \(m\) and point \((x_c, y_c)\) is given by:

\[ y-y_c = m(x-x_c) \]

Example 1#

Find the equation of the tangent line to the function \(f(x)=\sqrt{x}\) at \(x=4\) using the derivative \(f'(x)=\dfrac{1}{2\sqrt{x}}\).

Click through the tabs to see the steps of our solution.

Since we want to find the equation of the tangent line, let’s start with the point-slope formula for the equation of a line:

\[ y-y_c=m(x-x_c) \]

This tells us we need three things, \(x\)-coordinate: \(x_c\), \(y\)-coordinate: \(y_c\), and slope \(m\). For a tangent line:

  • \(x_c = a\)

  • \(y_c = f(a)\)

  • \(m = f'(a) \)

We are only given the \(x\) coordinate for our point of tangency.

\[ x_c=4 \]

This means we need to find the \(y\) coordinate by plugging our \(x\)-coordinate into the original function:

\[ y_c = f(x_c) = f(4) = \sqrt{4}=2 \]

Next, we our going to use the first derivative to find the slope:

\[ m = f' (x_c)=f'(4)=\dfrac{1}{2\sqrt{4}} = \dfrac{1}{4} \]

Now we just need to plug all of the pieces into the point-slope equation for our line:

\[ y-y_c=m(x-x_c) \]

We have found:

  • \(x_c = 4\)

  • \(y_c = 2\)

  • \(m = \tfrac{1}{4} \)

Giving us the tangent line equation:

\[ y-2=\dfrac{1}{4}(x-4) \]

Or solving for \(y\):

\[ y= \dfrac{1}{4}x+1 \]