The Central Idea
Suppose a differentiable function
\[ y=f(x) \]passes through the point
\[ (a,f(a)). \]The derivative at that point,
\[ f'(a), \]is the slope of the tangent line. If the tangent slope is finite and nonzero, the normal line has the negative reciprocal slope.
\[ \boxed{m_{\text{tan}}=f'(a)}, \qquad \boxed{m_{\text{normal}}=-\frac{1}{f'(a)}}. \]These two lines meet at the point on the curve and are perpendicular.
What Is a Tangent Line?
A tangent line is a line that matches the local direction of a curve at a specified point.
For a graph of a differentiable function, the tangent slope at \(x=a\) is
\[ m_{\text{tan}}=f'(a). \]This makes the tangent line the best linear description of the curve's direction at that point.
A Tangent Line Does Not Mean “Touches Only Once”
A common informal description says that a tangent line “touches” a curve. That language can be useful visually, but it is not a reliable definition.
A tangent line may cross the curve at the point of tangency, and it may intersect the same curve again elsewhere. What matters is that its slope matches the curve's local slope at the specified point.
Derivatives provide the precise slope relationship.
The Tangent-Line Equation
A line through \((a,f(a))\) with slope \(f'(a)\) can be written in point-slope form:
\[ \boxed{y-f(a)=f'(a)(x-a)}. \]The procedure is therefore:
- find the point on the curve;
- differentiate the function;
- evaluate the derivative at the point;
- use that slope in a line equation.
Example: Tangent Line to a Parabola
Find the tangent line to
\[ f(x)=x^2 \]at \(x=2\).
First find the point:
\[ f(2)=4. \]Differentiate:
\[ f'(x)=2x. \]At \(x=2\),
\[ m_{\text{tan}}=f'(2)=4. \]Therefore,
\[ y-4=4(x-2), \]or
\[ \boxed{y=4x-4}. \]What Is a Normal Line?
A normal line is a line perpendicular to the tangent line at the point of tangency.
When two nonvertical lines are perpendicular, their slopes are negative reciprocals. Thus, if
\[ m_{\text{tan}}\neq0, \]then
\[ \boxed{m_{\text{normal}}=-\frac{1}{m_{\text{tan}}}}. \]Equivalently,
\[ m_{\text{tan}}m_{\text{normal}}=-1. \]The Normal-Line Equation
If \(f'(a)\) is finite and nonzero, the normal line through \((a,f(a))\) is
\[ \boxed{ y-f(a)=-\frac{1}{f'(a)}(x-a) }. \]This formula is convenient, but it should not be used mechanically when \(f'(a)=0\) or when the tangent is vertical. Those special cases are best handled geometrically.
Example: Normal Line to a Parabola
Return to
\[ f(x)=x^2 \]at \((2,4)\). We already found
\[ m_{\text{tan}}=4. \]The normal slope is
\[ m_{\text{normal}}=-\frac14. \]Therefore,
\[ \boxed{y-4=-\frac14(x-2)}. \]Why the Negative Reciprocal Works
A line with finite slope \(m\) has a direction vector such as
\[ \langle 1,m\rangle. \]A line with slope \(-1/m\) has direction vector
\[ \left\langle 1,-\frac1m\right\rangle. \]Their dot product is
\[ 1+m\left(-\frac1m\right)=0. \]A zero dot product indicates perpendicular directions. The negative reciprocal is therefore a geometric consequence of perpendicularity.
Horizontal Tangents
A horizontal tangent occurs when
\[ \boxed{f'(a)=0}. \]The tangent line through \((a,f(a))\) is then
\[ \boxed{y=f(a)}. \]The line perpendicular to a horizontal line is vertical, so the normal line is
\[ \boxed{x=a}. \]We do not interpret \(-1/0\) as an ordinary numerical slope. We recognize the vertical normal geometrically.
Example: Horizontal Tangent
Consider
\[ f(x)=x^2-4x+7. \]Differentiate:
\[ f'(x)=2x-4. \]The tangent is horizontal where
\[ 2x-4=0, \]so \(x=2\). The corresponding point is
\[ f(2)=3. \]Therefore,
\[ \boxed{\text{tangent: }y=3}, \qquad \boxed{\text{normal: }x=2}. \]Vertical Tangents
Some curves have a vertical tangent at a point. A vertical line does not have a finite slope, so it is better written directly as
\[ \boxed{x=a}. \]The normal line is perpendicular to that vertical tangent, so the normal is horizontal:
\[ \boxed{y=b}, \]where \((a,b)\) is the point of tangency.
Tangent and Normal Lines from an Implicit Equation
A curve does not need to be given explicitly as \(y=f(x)\). If \(x\) and \(y\) satisfy an implicit equation, differentiate implicitly to find \(dy/dx\).
For example, consider the circle
\[ x^2+y^2=25. \]Differentiate with respect to \(x\):
\[ 2x+2y\frac{dy}{dx}=0. \]Solve for the slope:
\[ \boxed{\frac{dy}{dx}=-\frac{x}{y}} \]whenever \(y\neq0\).
Example: Tangent and Normal to a Circle
Find the tangent and normal lines to
\[ x^2+y^2=25 \]at \((3,4)\).
From
\[ \frac{dy}{dx}=-\frac{x}{y}, \]the tangent slope is
\[ m_{\text{tan}}=-\frac34. \]Therefore,
\[ \boxed{y-4=-\frac34(x-3)}. \]The normal slope is
\[ m_{\text{normal}}=\frac43, \]so
\[ \boxed{y-4=\frac43(x-3)}. \]Normal Lines and Radius Lines of a Circle
For a circle centered at the origin, the radius from the center to a point on the circle is perpendicular to the tangent at that point.
At \((3,4)\), the radius line has slope
\[ \frac43, \]which is exactly the normal slope found by differentiation.
This is a special geometric feature of circles. A normal line to a general curve does not necessarily point toward a single fixed center.
The General Implicit Form \(F(x,y)=0\)
Suppose a curve is described by
\[ F(x,y)=0. \]Implicit differentiation gives
\[ F_x+F_y\frac{dy}{dx}=0. \]When \(F_y\neq0\),
\[ \boxed{\frac{dy}{dx}=-\frac{F_x}{F_y}}. \]Thus the tangent slope is \(-F_x/F_y\), while the corresponding finite normal slope is
\[ \boxed{\frac{F_y}{F_x}} \]when the required denominators are nonzero.
The Gradient Points in a Normal Direction
For a level curve
\[ F(x,y)=0, \]the vector
\[ \nabla F=\langle F_x,F_y\rangle \]points in a direction normal to the curve whenever the gradient is nonzero.
This gives a vector-based version of the same geometry. It is especially useful when slope formulas become awkward or when the tangent is vertical.
A Tangent Direction from the Gradient
If
\[ \nabla F=\langle F_x,F_y\rangle \]is normal to a level curve, then a perpendicular tangent direction can be chosen as
\[ \boxed{\langle F_y,-F_x\rangle}. \]Their dot product is zero:
\[ F_xF_y+F_y(-F_x)=0. \]This formulation handles horizontal and vertical directions without first converting everything into slopes.
Tangent Lines for Parametric Curves
Suppose a curve is given parametrically by
\[ x=x(t), \qquad y=y(t). \]When \(dx/dt\neq0\), the tangent slope is
\[ \boxed{ \frac{dy}{dx}=\frac{dy/dt}{dx/dt} }. \]Evaluate this ratio at the parameter value corresponding to the point, then use the resulting slope in the tangent-line equation.
Example: Parametric Tangent and Normal
Let
\[ x=t^2+1, \qquad y=t^3-1. \]At \(t=1\), the point is
\[ (x,y)=(2,0). \]Differentiate both coordinates:
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2. \]Therefore,
\[ \frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2} \]for \(t\neq0\). At \(t=1\),
\[ m_{\text{tan}}=\frac32. \]Hence
\[ \boxed{\text{tangent: }y=\frac32(x-2)} \]and
\[ \boxed{\text{normal: }y=-\frac23(x-2)}. \]Parametric Curves Also Have Tangent Vectors
For a parametric curve, the vector
\[ \boxed{\left\langle\frac{dx}{dt},\frac{dy}{dt}\right\rangle} \]gives a tangent direction whenever it is nonzero.
A perpendicular normal direction can be chosen as
\[ \boxed{\left\langle-\frac{dy}{dt},\frac{dx}{dt}\right\rangle}. \]Vector descriptions can remain useful even when a slope is undefined.
Horizontal Tangents on Parametric Curves
A parametric curve often has a horizontal tangent when
\[ \frac{dy}{dt}=0 \]while
\[ \frac{dx}{dt}\neq0. \]Then
\[ \frac{dy}{dx}=0. \]The tangent is horizontal and the normal is vertical.
Vertical Tangents on Parametric Curves
A parametric curve often has a vertical tangent when
\[ \frac{dx}{dt}=0 \]while
\[ \frac{dy}{dt}\neq0. \]The slope ratio is then not finite, but the tangent direction is vertical. The normal line is horizontal.
If both derivatives are zero, more analysis is required; the simple ratio alone does not determine the tangent direction.
Tangent Direction versus Motion Direction
For a parametrized path, the tangent line describes the geometric direction of the curve, while the tangent vector also carries orientation information.
If increasing \(t\) reverses direction along the same line, the tangent line itself is unchanged even though the tangent vector changes sign.
A line has no preferred travel direction unless additional information is supplied.
Tangent Lines as Local Linear Models
Near \(x=a\), a differentiable function behaves approximately like its tangent line:
\[ f(x)\approx f(a)+f'(a)(x-a). \]The tangent line therefore does more than describe orientation. It also gives a local linear approximation to the function.
The later lesson on Linearization and Differentials will develop this approximation role systematically.
Normal Lines as Perpendicular Local Directions
The normal line is useful whenever the direction perpendicular to a curve matters. Examples include geometric construction, surface orientation, reflection models, force directions, and optimization constraints.
At this stage, the essential relationship is geometric:
\[ \boxed{\text{tangent direction}\perp\text{normal direction}}. \]Finding a Point of Tangency from a Required Slope
Sometimes the slope is given and the point must be found.
Suppose
\[ f(x)=x^2-3x \]and we want a tangent with slope \(5\). Since
\[ f'(x)=2x-3, \]solve
\[ 2x-3=5. \]Thus \(x=4\). The point is
\[ (4,f(4))=(4,4). \]The tangent line is
\[ \boxed{y-4=5(x-4)}. \]Finding a Point Where the Normal Has a Required Slope
Suppose a desired normal slope is \(m_n\neq0\). Then the corresponding tangent slope must be
\[ m_t=-\frac1{m_n}. \]For a graph \(y=f(x)\), solve
\[ f'(x)=m_t \]to find candidate points. Then evaluate \(f(x)\) and construct the normal line through each resulting point.
Parallel Tangents
Two tangent lines are parallel when they have the same slope.
For a function \(f\), points \(x=a\) and \(x=b\) have parallel tangents when
\[ \boxed{f'(a)=f'(b)}. \]Different points on the same curve can therefore share the same local direction.
Perpendicular Tangents
If two nonvertical tangent lines have slopes \(m_1\) and \(m_2\), they are perpendicular when
\[ \boxed{m_1m_2=-1}. \]Since tangent slopes come from derivatives, this can become a condition on derivative values at two different points.
Angle of a Tangent Line
If a nonvertical tangent line makes an angle \(\theta\) with the positive \(x\)-axis, then
\[ \boxed{\tan\theta=m_{\text{tan}}=f'(a)}. \]This connects derivative slope with geometric angle. Because tangent is periodic, the line direction is understood up to an angle of \(\pi\).
Common Error: Using the Function Value as the Slope
At \(x=a\), the value
\[ f(a) \]gives the vertical coordinate of the point. It does not usually give the tangent slope.
The tangent slope is
\[ \boxed{f'(a)}. \]Point information and slope information play different roles in the line equation.
Common Error: Forgetting to Evaluate the Derivative
If
\[ f'(x)=3x^2-2, \]then \(3x^2-2\) is a slope function, not the final numerical slope at a specified point.
At \(x=2\), the tangent slope is
\[ f'(2)=10. \]A particular tangent line requires a particular slope value.
Common Error: Negating without Taking the Reciprocal
If the tangent slope is
\[ m_{\text{tan}}=3, \]the normal slope is not merely \(-3\). It is
\[ \boxed{-\frac13}. \]Perpendicular slopes require both a sign change and a reciprocal, except in the special horizontal/vertical cases handled geometrically.
Common Error: Applying the Negative Reciprocal to Zero
When the tangent slope is zero, writing
\[ -\frac10 \]does not produce an ordinary real-number slope.
Instead, recognize the geometry: a horizontal tangent has a vertical normal.
Common Error: Assuming Every Point Has a Tangent Line
A curve can fail to have a unique tangent direction at a point. Corners, cusps, crossings, or other singular behavior may prevent the ordinary derivative from defining one tangent line.
Before constructing a tangent from \(f'(a)\), make sure the relevant derivative or tangent direction actually exists.
Common Error: Treating a Vertical Tangent as “No Tangent”
A vertical tangent does not have a finite slope, but it can still be a perfectly meaningful tangent line.
The equation is written directly as
\[ x=a. \]Undefined slope and nonexistent tangent are not the same statement.
A Reliable Procedure for Explicit Functions
- identify the input value \(x=a\);
- compute \(f(a)\) to locate the point;
- find \(f'(x)\);
- evaluate \(f'(a)\);
- write the tangent line through \((a,f(a))\);
- find the perpendicular direction for the normal;
- handle horizontal or vertical special cases geometrically;
- check that both lines pass through the required point.
A Reliable Procedure for Implicit Curves
- verify that the given point lies on the curve;
- differentiate the equation implicitly;
- solve for \(dy/dx\) when appropriate;
- substitute the point into the slope formula;
- construct the tangent line;
- construct the perpendicular normal line;
- use gradient or vector reasoning if slope notation becomes singular.
A Reliable Procedure for Parametric Curves
- identify the relevant parameter value \(t=t_0\);
- compute the point \((x(t_0),y(t_0))\);
- find \(dx/dt\) and \(dy/dt\);
- form \((dy/dt)/(dx/dt)\) when \(dx/dt\neq0\);
- evaluate the tangent direction at \(t_0\);
- construct the tangent and normal lines;
- check separately for horizontal, vertical, or indeterminate cases.
Practice 1: Tangent Line
Find the tangent line to
\[ f(x)=x^3-2x \]at \(x=1\).
Practice 2: Normal Line
Find the normal line to
\[ f(x)=x^2+1 \]at \(x=1\).
Practice 3: Horizontal Tangent
Find the tangent and normal lines to
\[ f(x)=x^2-6x+5 \]at its horizontal tangent point.
Practice 4: Implicit Circle
Find the tangent and normal lines to
\[ x^2+y^2=13 \]at \((2,3)\).
Practice 5: General Implicit Curve
For
\[ x^2+xy+y^2=7, \]find the tangent slope at \((1,2)\).
Practice 6: Parametric Curve
Let
\[ x=t^2, \qquad y=t^3+t. \]Find the tangent and normal lines at \(t=1\).
Practice 7: Required Tangent Slope
Find the point on
\[ f(x)=x^2+2x \]where the tangent slope is \(6\), then find the tangent line there.
Practice 8: Required Normal Slope
Find the point on
\[ f(x)=x^2 \]where the normal slope is \(-1/4\), then find the normal line.
Practice 9: Vertical Parametric Tangent
Let
\[ x=t^2, \qquad y=t+1. \]Determine the tangent and normal lines at \(t=0\).
Practice 10: Identify the Error
A student finds that a curve has tangent slope \(m=2\) at a point and says the normal slope is \(-2\). Explain the error and give the correct normal slope.
Practice Answers
Practice 1
Differentiate:
\[ f'(x)=3x^2-2. \]At \(x=1\),
\[ f(1)=-1, \qquad f'(1)=1. \]Therefore,
\[ \boxed{y+1=x-1}, \]or \(\boxed{y=x-2}\).
Practice 2
At \(x=1\), the point is \((1,2)\). Since
\[ f'(x)=2x, \]the tangent slope is \(2\), so the normal slope is \(-1/2\). Thus
\[ \boxed{y-2=-\frac12(x-1)}. \]Practice 3
Since
\[ f'(x)=2x-6, \]the horizontal tangent occurs at \(x=3\). The point is
\[ (3,-4). \]Therefore,
\[ \boxed{\text{tangent: }y=-4}, \qquad \boxed{\text{normal: }x=3}. \]Practice 4
Implicit differentiation gives
\[ \frac{dy}{dx}=-\frac{x}{y}. \]At \((2,3)\),
\[ m_{\text{tan}}=-\frac23, \qquad m_{\text{normal}}=\frac32. \]Hence
\[ \boxed{\text{tangent: }y-3=-\frac23(x-2)} \]and
\[ \boxed{\text{normal: }y-3=\frac32(x-2)}. \]Practice 5
Differentiate
\[ x^2+xy+y^2=7: \] \[ 2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0. \]Therefore,
\[ \frac{dy}{dx}=-\frac{2x+y}{x+2y}. \]At \((1,2)\),
\[ \boxed{m_{\text{tan}}=-\frac45}. \]Practice 6
At \(t=1\), the point is \((1,2)\). Also,
\[ \frac{dx}{dt}=2t, \qquad \frac{dy}{dt}=3t^2+1. \]Thus
\[ m_{\text{tan}}=\frac{4}{2}=2, \qquad m_{\text{normal}}=-\frac12. \]Therefore,
\[ \boxed{\text{tangent: }y-2=2(x-1)} \]and
\[ \boxed{\text{normal: }y-2=-\frac12(x-1)}. \]Practice 7
Since
\[ f'(x)=2x+2, \]solve
\[ 2x+2=6. \]Thus \(x=2\), and
\[ f(2)=8. \]The point is \((2,8)\), and the tangent line is
\[ \boxed{y-8=6(x-2)}. \]Practice 8
A normal slope of \(-1/4\) corresponds to a tangent slope of \(4\). Since
\[ f'(x)=2x, \]solve \(2x=4\), giving \(x=2\). The point is \((2,4)\). Therefore,
\[ \boxed{y-4=-\frac14(x-2)}. \]Practice 9
At \(t=0\), the point is \((0,1)\). Also,
\[ \frac{dx}{dt}=2t=0, \qquad \frac{dy}{dt}=1. \]The tangent direction is vertical, so
\[ \boxed{\text{tangent: }x=0}. \]The normal is horizontal:
\[ \boxed{\text{normal: }y=1}. \]Practice 10
The student changed the sign but did not take the reciprocal. The normal slope is
\[ \boxed{m_{\text{normal}}=-\frac12}. \]What to Remember
For a graph \(y=f(x)\) at \(x=a\),
\[ \boxed{m_{\text{tan}}=f'(a)}. \]The tangent line is
\[ \boxed{y-f(a)=f'(a)(x-a)}. \]If the tangent slope is finite and nonzero, the normal slope is
\[ \boxed{m_{\text{normal}}=-\frac1{f'(a)}}. \]- the tangent follows the curve's local direction;
- the normal is perpendicular to the tangent;
- a horizontal tangent has a vertical normal;
- a vertical tangent has a horizontal normal;
- implicit differentiation supplies slopes for implicit curves;
- the gradient of a level curve gives a normal direction;
- parametric derivatives and tangent vectors describe parametric curves;
- a tangent line may cross the curve or meet it again elsewhere.
Where We Go Next
Tangent slopes are derivative information attached to a specific point. Many applied problems reverse the perspective: several changing quantities are connected by an equation, and a known rate for one quantity is used to determine an unknown rate for another.
The next lesson, Related Rates, develops that method by differentiating relationships with respect to time and connecting the rates at which linked quantities change.