Title: Default for 12.1 Tangents and Normals
Course: HELM Workbooks 2023-2024

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Perpendicular lines

One form for the equation of a straight line is

\(\qquad y=mx+c\)

where \(m\) and \(c\) are constants. We remember that \(m\) is the gradient of the line and its value is the tangent of the angle \(\theta\) that the line makes with the positive \(x\)-axis. The constant \(c\) is the value obtained where the line intersects the \(y\)-axis. See figure below:

[Image]

If we have a second line, with equation

\(\qquad y=nx+d\)

then, unless \(m=n\), the two lines will intersect at one point. These are drawn together in the figure below. The second line makes an angle \(\psi\) with the positive \(x\)-axis.

[Image]

A simple question to ask is "what is the relation between \(m\) and \(n\) if the lines are perpendicular?" If the lines are perpendicular, as shown in the figure below, the angles \(\theta\) and \(\psi\) must satisfy the relation:

\(\qquad \psi -\theta=90^{\circ}\)

[Image]

This is true since the angles in a triangle add up to \(180^{\circ}\). According to the figure, the three angles are \(90^{\circ}\), \(\theta\) and \(180^{\circ}-\psi\). Therefore

\(\qquad 180^{\circ}= 90^{\circ}+\theta +(180^{\circ}-\psi)\qquad \text{implying}\qquad \psi -\theta =90^{\circ}\)

In this special case that the lines are perpendicular or normal to each other, the relation between the gradients \(m\) and \(n\) is easily obtained. In this deduction we use the following basic trigonometric relations and identities:

\(\qquad\sin(A-B)\equiv\sin A\cos B-\cos A\sin B\qquad \cos (A-B)\equiv\cos A\cos B +\sin A\sin B\)

\(\qquad\tan(A)\equiv\dfrac{\sin A}{\cos A}\qquad \sin 90^{\circ}=1 \qquad \cos 90^{\circ}=0\)

Now \(\begin{aligned} m&=\tan\theta\\ &=\tan(\psi-90^o)\qquad\text{(see Figure)}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &=\frac{\sin(\psi-90^o)}{\cos(\psi-90^o)}\\ &=-\frac{\cos\psi}{\sin\psi}=-\frac{1}{\tan\psi}=-\frac{1}{n}\\ \text{So}\qquad mn&=-1\end{aligned}\)

Key Point

Two straight lines \(y = mx + c,\ y = nx + d\) are perpendicular if

\(m = -\frac{1}{n} \) or equivalently \(mn = -1\)

This result assumes that neither of the lines are parallel to the \(x\)-axis or to the \(y\)-axis, as in such cases one gradient will be zero and the other infinite.

Exercise

Are the following pairs of lines perpendicular?

1. \(y = [value], \quad y = \displaystyle [value]\)

2. \(y = [value], \quad y = \displaystyle [value]\)

3. \(y = [value], \quad y = \displaystyle [value]\)
Feedback: 1. We are given the lines: \(y = [value], \quad y = \displaystyle [value]\) For the first line \( m = [value]\) and for the second line \(n= [value]\). Thus \(mn= [value]\) so the lines are perpendicular. 2. We are given the lines: \(y = [value], \quad y = \displaystyle [value]\) For the first line \( m = [value]\) and for the second line \(n= [value]\). Thus \(mn= [value]\) so the lines are not perpendicular. 3. We are given the lines: \(y = [value], \quad y = \displaystyle [value]\) For the first line \( m = [value]\) and for the second line \(n= [value]\). Thus \(mn= [value]\) so the lines are perpendicular.

Tangents and normals to a curve

As we know, the relationship between an independent variable \(x\) and a dependent variable \(y\) is denoted by

 \(\qquad y=f(x)\)

As we also know, the geometrical interpretation of this relation takes the form of a curve in an \(xy\) plane as illustrated in the Figure below.

[Image]

We know how to calculate a value of \(y\) given a value of \(x\). We can either do this graphically (which is inaccurate) or else use the function itself. So, at an \(x\) value of \(x_0\), the corresponding \(y\) value is \(y_0\) where

\(\qquad y_0=f(x_0)\)

Let us examine the curve in the neighbourhood of the point \((x_0,y_0)\). There are two important constructions of interest

- the tangent line at \((x_0,y_0)\)
- the normal line at \((x_0,y_0)\)

These are shown in the following Figure:

[Image]

We note the geometrically obvious fact: the tangent and normal lines at any given point on a curve are perpendicular to each other.

Example

1. The curve \(y=x^2\) is drawn below. On the Figure below, draw the tangent line and the normal line at the point \((x_0=1,y_0=1)\):
[Image]
2. From the graph, estimate the values of \(\theta\) and \(\psi\) in degrees. (You will need a protractor.)

Solution

1. [Image]
2. \(\theta ≈ 63.4^{\circ}, \qquad \psi ≈ 153.4^{\circ}\)

Returning to the curve \(y=f(x)\) we know, from the geometrical interpretation of the derivative, that

\(\qquad \dfrac{df}{dx}\bigg|_{x_0}=\tan\theta\)

(the notation \(\dfrac{df}{dx}\bigg|_{x_0}\) means evaluate \(\dfrac{df}{dx}\) at the value \(x=x_0\))

Here \(\theta\) is the angle the tangent line to the curve \(y=f(x)\) makes with the positive \(x\)-axis. This is highlighted in the following Figure:

[Image]

The tangent line to a curve

Let the equation of the tangent line to the curve \(y=f(x)\) at the point \((x_0,y_0)\) be: \[ y=mx+c,\] where \(m\) and \(c\) are constants to be found. The line just touches the curve \(y=f(x)\) at the point \((x_0,y_0)\). So, at this point, both must have the same value for the derivative. That is: \[m=\dfrac{df}{dx}\bigg|_{x_0}.\]

Since we know (in any particular case) \(f(x)\) and the value \(x_0\), we can readily calculate the value for \(m\). The value of \(c\) is found by using the fact that the tangent line and the curve pass through the same point \((x_0,y_0)\). \[y_0=mx_0+c,\qquad \textrm{where} \qquad y_0=f(x_0).\]

Thus \( mx_0+c=f(x_0)\) leading to \(c=f(x_0)-mx_0\).

Key Point

The equation of the tangent line to the curve \(y = f(x)\) at the point \((x_0, y_0)\) is \(y = mx + c\), where \[m = \dfrac{df}{dx}\bigg|_{ x_0} \qquad \textrm{and} \qquad c = f(x_0) - mx_0\] Alternatively, the equation is \(y - y_0 = m(x - x_0)\) where \(m = \dfrac{df}{dx}\bigg|_{ x_0}\) and \(y_0 = f(x_0)\)

Example

Find the equation of the tangent line to the curve \(y = x^2\) at the point \((1,1)\).

Solution

Method 1

Here \(f(x) = x^2\) and \(x_0 = 1\) thus \(\dfrac{df}{dx} = 2x\) therefore \(m =\dfrac{df}{dx}\bigg|_{x_0} = 2\).

Also \(c = f(x_0) - mx_0 = f(1) - m = 1 - 2 = -1\). The tangent line has equation \(y = 2x - 1\).

Method 2

\[ y_0 = f(x_0) = f(1) = 1^2 = 1.\]

The tangent line has equation \(y - 1 = 2(x - 1) \to y = 2x - 1\).

Exercise

In this exercise we will find the equation of the tangent line to the curve \(y = e^x\) at the point \(x = [value]\). The curve and the line are displayed in the following figure:

[Image]

1. We first specify \(x_0\) and \(f(x_0)\):
\(x_0=\)
\(f(x_0)=\)
2. We now obtain the values of the slope at \(x_0\), i.e. \(m=\dfrac{df}{ dx} \bigg|_{ x_0}\) and the \(y\)-intercept \(f(x_0) - mx_0\):
\(\dfrac{df}{ dx} \bigg|_{ x_0}=\)
\(f(x_0) - mx_0 =\)
3. Finally, we use these values to obtain the equation of the tangent line:
\(y =\)
Feedback: We are asked to find the equation of the tangent line to the curve \(y = e^x\) at the point \(x = [value]\). [Image] 1. First we specify that \(x_0 = [value]\) and \( f(x_0) = [value].\) 2. Then we obtain the values of \(\dfrac{df}{ dx} \bigg|_{ x_0}\) and \(f(x_0) - mx_0\): \(\dfrac{df}{dx} = e^x\) therefore \(\dfrac{df}{dx}\bigg|_[value] = e^{[value]}\) and \(f([value]) - [value]([value]) = [value] .\) 3. Hence we obtain the equation of the tangent line: \(y= [value] .\)

The normal line to a curve

We have already noted that, at any point \((x_0, y_0)\) on a curve \(y = f(x)\) (so that \(y_0=f(x_0)\)), the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are respectively:

\[\qquad y = mx + c \qquad \textrm{and} \qquad y = nx + d, \]

then recalling that two lines are perpendicular if the product of theirs slopes is \(-1\) we can see that \(m = -\dfrac{1}{ n}\) or, equivalently \(n = - \dfrac{1}{ m}\).

We have also noted that slope of the tangent line can be calculated from the derivative, i.e.,

\[ m =  \dfrac{df}{dx}\bigg|_{x_0}.\]

Thus we can also obtain \(n\) from this information. To find \(d\), we use the fact that the point \((x_0, y_0)\) lies on the normal line \(y = nx + d\)  so \(n x_0 + d = f(x_0)\) leading to \(d = f(x_0) - nx_0\). Though it is better to work this out in each case rather than trying to recall this a formula. The takeaway should be that we can do this and how.

Exercise

In this exercise we will find the equation of the normal line to curve \(y = \sin([value]x)\) at the point \(x =\displaystyle [value]\).

1. First find \(m\), the value of the slope of the tangent at \(\displaystyle x=[value]\):
 \(m =\)
2. Hence find \(n\), the value of the slope of the normal at \(\displaystyle x=[value]\):
\(n=\)
3. Recall that the equation of the normal line is \(y = nx + d\) and that the normal line must pass through the curve when \(\displaystyle x =[value]\).
Find the value of \(d\):
\(d = \)
4. Now using the values found above, the equation of the normal is
\(y=\)
Feedback: We are asked to find the equation of the normal line to curve \(y = \sin([value]x)\) at the point \(x =\displaystyle [value]\). This will be a line \(y=nx+d\) where \(n\) is slope of the normal and \(d\) is the \(y\)-intercept. Recall that the normal is perpendicular to the tangent. 1. To find the slope of the tangent at any point we find the derivative \( \dfrac{df}{dx}=y'= [value]\cos([value]x) .\) So when \(x =[value]\) we evaluate the derivative to find the slope of the tangent, so that\(m= \dfrac{df}{dx}\bigg|_[value]=\displaystyle [value].\) 2. Hence we find the value of \(n\) can be found: \(mn = -1 \Rightarrow n=-\dfrac{1}{m}\) so \(n = \displaystyle [value].\) 3. The equation of the normal line is then \(y = \displaystyle{[value]}x + d.\) We know the normal intersects the curve when \(x = [value]\), i.e. the point \(\left([value],\sin([value][value])\right) =\left([value],[value]\right).\) Substituting this point into the line gives: \([value] = [value][value] + d \quad \therefore \quad d=[value]\,\, (\approx [value]).\) 4. We thus find the line: \(y = \displaystyle{[value]}.\)

Exercise

Find the equations of the tangent and normal lines to the following curves at the points indicated.

1. \(y = [value], \text{ at } ([value], [value])\)
Give all numerical values as precise values.
Tangent line: \(y = \)
Normal line: \(y = \)
2. \(y = \sqrt{1 - x^2}, \text{ at }\left(\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)\)
Give all numerical values as precise values.
Tangent line: \(y = \)
Normal line: \(y = \)
3. \(y = [value], \text{ at } ([value], [value])\)
Round all numerical values to 2 decimal places.
Tangent line: \(y = \)
Normal line: \(y = \)
Feedback: 1. \(f(x) = [value] \quad \frac{df}{dx} = [value] \quad \frac{df}{dx}\bigg|_{[value]} = [value].\) Tangent line \(y = [value]x + c\). This passes through \(([value], [value])\) so \(y = [value].\) Normal line \(y = [value]x + d\). This passes through \(([value], [value])\) so \(y = [value].\) 2. \(f(x) = \sqrt{1 - x^2} \quad \frac{df}{dx} = \frac{-x}{\sqrt{1 - x^2}} \quad \frac{df}{dx} \bigg|_{x_0} = -1.\) Tangent line \(y = -x + c\). This passes through \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) so \(y = -x + \sqrt{2}.\) Normal line \(y = x + d\). This passes through \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) so \(y=x.\) 3. \( f(x) = [value] \quad \frac{df}{dx} = [value] \quad \frac{df}{dx} \bigg|_{[value]} = [value]\) Tangent line: \(y = [value]x + c\) passes through \(([value], [value])\) so \(y = [value]x + [value].\) Normal line: \(y = [value]x + d\). This passes through \(([value], [value])\) so \(y = [value]x + [value].\)

Exercise

Find the value of \(a\) if the two curves \(y = e^{-x}\) and \(y = e^{ax}\) are to intersect at right-angles

\(a = \)
Feedback: The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular. We begin by finding the point of intersection: \[e^{-x} = e^{ax} \implies -x = ax,\] which means that \( x=0 \), since having \(a = -1\) is not sensible (why?). The tangent line to \(y = e^{ax}\) is \(y = mx + c\), where \(m = ae^{ax} |_{x=0} = a.\) The tangent line to \(y = e^{-x}\) is \(y = kx + g\), where \( k = -e^{-x}|_{x=0} = -1.\) These two lines are perpendicular if \(mk = a \cdot (-1) = -1\), which gives \(a = 1\).

Exercise

In this exercise we will approximate the equation of the tangent line to the curve \(y = [value]\) at the point \(x = \displaystyle [value]\) and find where this line intersects the \(x\)-axis. Note that the following figure is for illustration only and the gradient can be either positive or negative:

[Image: graph of sin(x)]

1. We first specify \(x_0\) and \(f(x_0)\) from the information above:
\(x_0 =\)
\(f(x_0) =\)
2. We now obtain the values of the slope of the tangent at \(x_0\), i.e., \(m=\dfrac{df}{dx}\bigg|_{x_0}\) and the \(y\)-intercept \(f(x_0) - mx_0\). Give your answers correct to \(2\) decimal places
\(\dfrac{df}{dx}\bigg|_{x_0}=\)
\(f(x_0) - mx_0=\)
3. Use these approximate values to obtain an equation approximating the tangent line:
\(y=\)
4. Where does this line intersect the \(x\)-axis?
\(x=\)
Feedback: We are asked to find the equation of the tangent line to the curve \(y = [value]\) at the point \(x = \displaystyle [value].\) 1. First we specify that \(\displaystyle x_0 = [value]\) and \(\displaystyle f(x_0) = [value].\) 2. Then we obtain the values of \(\dfrac{df}{ dx} \bigg|_{ x_0}\) and \(f(x_0) - mx_0\). We have \(\dfrac{df}{dx} = [value]\cos([value]x)\), and therefore \[\frac{df}{dx} \bigg|_[value] = [value]\cos\left([value]\right) = [value]=[value] ,\] and \(\displaystyle f\left([value]\right) - \left([value]\right) \cdot [value] = [value] \) to 2 d.p. 3. Hence we obtain the equation of the tangent line: \(\displaystyle y = [value]x + [value]\) to 2 d.p. 4. When \(y = 0\), we have \([value]x + [value] = 0\) and then \(x = [value]\) to 2 d.p.

Exercise

In this exercise we will find the equation of the normal line to the curve \(y = [value]\) at \(x = [value]\).

1. First identify \(f(x)\) and \( x_0\):
\(f(x) =\)
\(x_0 =\)
2. Now find the derivative of \(f(x)\) and the slopes \( m,\, n\), the slopes of the tangent and normal when \(x=x_0\) respectively:
\(\dfrac{df}{dx} =\)
\(m =\)
\(n =\)
3. Now use the property that the normal line \(y = nx + d\) and the curve \(y = [value]\) pass through the point \((x_0, f(x_0))\) to find \(d\) and so obtain the equation of the normal line:
\(d =\)
\(y =\)
Feedback: We find the equation of the normal line to the curve \(y = [value]\) at \(x = [value]\). 1. First we find \(\displaystyle f(x) = [value]\) and \(x_0 = [value]\). 2. We then have that \[\frac{df}{dx} = [value].\] Then \(\displaystyle m = \frac{df}{dx}\bigg|_{[value]} = [value]\) and \(\displaystyle n = -\frac{1}{m} = [value]\). 3. Now we use the property that the normal line \(y = nx + d\) and the curve \(y = [value]\) pass through the point \(([value], [value])\) to find \(d\), and so obtain the equation of the normal line: \[[value]=[value]+d \implies d= [value] \implies d = [value]. \] Thus, the equation of the normal line is \[ y = [value].\]