## 17Calculus Derivatives - Normal Lines

Normal Lines

### Resources

tangent lines

The normal line is perpendicular to the curve and, therefore, also perpendicular to the tangent line.   To find the equation of the normal line at a point, follow the same procedure as you did to find the tangent line, except after finding the slope of the tangent line, take the negative reciprocal of the slope to get the slope of the normal line.   Then use the point to find the equation of normal line.

In the example on the tangent line page, the slope of the normal line is $$m=-1/15$$.   Using this and the point $$(x_1, y_1) = (3,23)$$, the equation of the normal line is

$$\begin{array}{rcl} y-y_1 & = & m(x-x_1) \\ y-23 & = & (-1/15)(x-3) \\ y & = & -x/15 + 1/5 + 23 \\ y & = & -x/15 + 116/5 \end{array}$$

The graph of this normal line is shown in the plot on the right.   It is shown as the green almost horizontal line.   Notice that it doesn't look exactly perpendicular to the tangent line.   This is because the scales on the axes are not equivalent.   However, you know from the equations that they are perpendicular.   The slope of the tangent line is $$m_t = 15$$ and the slope of the normal line is $$m_n = -1/15$$ and since $$m_n = -1/m_t$$, they are perpendicular.

Practice

Unless otherwise instructed, calculate the equation of the normal line to the graph that goes through the given point.  Make sure to check to make sure that the point is on the graph.  Give your answers in slope-intercept form.

$$y=x^2$$, $$(-2,4)$$

Problem Statement

Calculate the equation of the normal line to the graph given by the equation $$y=x^2$$ that goes through the point $$(-2,4)$$. Check to make sure that the point is on the graph and give your answer in slope-intercept form.

Solution

### Krista King Math - 964 video solution

video by Krista King Math

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$$y=2x^2+3x-5$$, $$(2,9)$$

Problem Statement

Calculate the equation of the normal line to the graph given by the equation $$y=2x^2+3x-5$$ that goes through the point $$(2,9)$$. Check to make sure that the point is on the graph and give your answer in slope-intercept form.

Solution

### Krista King Math - 965 video solution

video by Krista King Math

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$$y=5-x-2x^2$$, $$(-1,4)$$

Problem Statement

Calculate the equation of the normal line to the graph given by the equation $$y=5-x-2x^2$$ that goes through the point $$(-1,4)$$. Check to make sure that the point is on the graph and give your answer in slope-intercept form.

Solution

### Krista King Math - 966 video solution

video by Krista King Math

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$$y=x^4+2e^x$$, $$(0,2)$$

Problem Statement

Calculate the equation of the normal line to the graph given by the equation $$y=x^4+2e^x$$ that goes through the point $$(0,2)$$. Check to make sure that the point is on the graph and give your answer in slope-intercept form.

Solution

### Krista King Math - 967 video solution

video by Krista King Math

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$$y=(2+x)e^{-x}$$, $$(0,2)$$

Problem Statement

Calculate the equation of the normal line to the graph given by the equation $$y=(2+x)e^{-x}$$ that goes through the point $$(0,2)$$. Check to make sure that the point is on the graph and give your answer in slope-intercept form.

Solution

### Krista King Math - 968 video solution

video by Krista King Math

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Find the equations of the tangent line and the normal line to $$f(x) = x^3-2x+4$$ at $$x=1$$.

Problem Statement

Find the equations of the tangent line and the normal line to $$f(x) = x^3-2x+4$$ at $$x=1$$.

Solution

### The Organic Chemistry Tutor - 3697 video solution

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Find the equations of the tangent line and the normal line to $$f(x) = x^4-2x+1$$ at $$x=2$$.

Problem Statement

Find the equations of the tangent line and the normal line to $$f(x) = x^4-2x+1$$ at $$x=2$$.

Solution

### The Organic Chemistry Tutor - 3698 video solution

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Find the equations of the tangent line and the normal line to $$f(x) = x^4+4x-1$$ at $$x=1$$.

Problem Statement

Find the equations of the tangent line and the normal line to $$f(x) = x^4+4x-1$$ at $$x=1$$.

Solution

### The Organic Chemistry Tutor - 3702 video solution

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