Lesson Explainer: Tangents and Normals to the Graph of a Function | Nagwa Lesson Explainer: Tangents and Normals to the Graph of a Function | Nagwa

Lesson Explainer: Tangents and Normals to the Graph of a Function Mathematics

In this explainer, we will learn how to find the equations of tangents and normals to trigonometric, parametric, and implicitly defined curves using derivatives.

We know that the derivative dd𝑦𝑥 at a point, if it exists, gives the slope of the tangent to the curve at the given point. Using the coordinates of the point with the slope, we can obtain the point–slope form of the equation of the tangent to the curve at that point. Let us begin by reviewing how to find the equation of the tangent to a curve.

How To: Finding the Equation of the Tangent to a Curve

Given a curve on the 𝑥𝑦-plane, the slope of the tangent to the curve at (𝑥,𝑦) is obtained by evaluating dd𝑦𝑥 at the point (𝑥,𝑦) if it exists at this point. This is denoted by dd𝑦𝑥|||().

The equation of the tangent is then 𝑦=𝑦𝑥|||(𝑥𝑥)+𝑦.dd()

When we are given an implicit curve on the 𝑥𝑦-plane, we can obtain an expression for dd𝑦𝑥 by applying implicit differentiation. Let us recall the chain rule, which is the main ingredient of implicit differentiation.

Rule: Chain Rule

If the variable of differentiation does not match variable of the function being differentiated, we need to change variables by applying the chain rule. In particular, if the derivatives 𝑓 and dd𝑦𝑥 exist, then we have dddddddd𝑥𝑓(𝑦)=𝑦𝑓(𝑦)×𝑦𝑥=𝑓(𝑦)𝑦𝑥.

In our first example, we will find the equation of the tangent to an implicitly defined curve and also identify where the tangent intersects the curve other than at the point of tangency.

Example 1: Implicit Function Theorem

The equation 𝑦24𝑥+24𝑥=0 describes a curve in the plane.

  1. Find the coordinates of two points on this curve, where 𝑥=12.
  2. Determine the equation of the tangent at the point where 𝑥=12 and the 𝑦-coordinate is positive.
  3. Find the coordinates of another point, if it exists, at which the tangent meets the curve.

Answer

Part 1

Since we are given the 𝑥-coordinates, we need to identify all possible 𝑦-coordinates when 𝑥=12. Substituting the given 𝑥-coordinate into the implicit equation, we have 𝑦2412+2412=0𝑦+312=0𝑦=9𝑦=±3.

Hence, the possible 𝑦-coordinates are 3 and 3. The coordinates of the two points on this curve with 𝑥=12 are 12,312,3.and

Part 2

To determine the equation of the tangent, we need to evaluate the expression dd𝑦𝑥 at the point of tangency. We begin by implicitly differentiating the equation: dddddddddd𝑥𝑦24𝑥+24𝑥=𝑥(0)𝑥𝑦𝑥24𝑥+𝑥(24𝑥)=0.

The second and the third terms in the equation above are regular derivatives since we are differentiating the function of 𝑥 with respect to 𝑥. Applying the power rule for differentiation, the terms are dddd𝑥24𝑥=24×3𝑥=72𝑥𝑥(24𝑥)=24.

For the first term, dd𝑥𝑦, we must apply the chain rule since we are differentiating the function of 𝑦 with respect to 𝑥: dddddddd𝑥𝑦=𝑦𝑦𝑦𝑥=2𝑦𝑦𝑥.

Substituting these expressions into our differentiated equation, we get 2𝑦𝑦𝑥72𝑥+24=0.dd

To find the slope of the tangent, we need to evaluate the expression dd𝑦𝑥 at the point of tangency. Since we want the 𝑦-coordinate to be positive, this line is tangent to the curve at 12,3. Hence, we can substitute 𝑥=12 and 𝑦=3 into the equation above: 2𝑦𝑦𝑥72𝑥+24=02(3)𝑦𝑥7212+24=06𝑦𝑥18+24=06𝑦𝑥=6𝑦𝑥=1.dddddddddd

This tells us that the slope of the tangent at this point is 1. We also know that this line passes through the point 12,3. Using the point–slope form of the equation of a line, the equation of the tangent is 𝑦=𝑥+12+3.

Simplifying gives 𝑦=𝑥12+3𝑦=52𝑥.

So, the equation of the tangent is 𝑦=52𝑥.

Part 3

We want to find the intersection point of the tangent other than the point of tangency. Since we know the equation of the tangent, 𝑦=52𝑥, we can substitute this expression into the given implicit equation of the curve: 𝑦24𝑥+24𝑥=052𝑥24𝑥+24𝑥=02545𝑥+𝑥24𝑥+24𝑥=0254+19𝑥+𝑥24𝑥=0.

We know that one of the possible solutions of this equation is 𝑥=12 since this is the point of tangency. Therefore, by the remainder theorem, we have that 𝑥+12 is a factor of the cubic polynomial on the left-hand side of the equation. Let us perform a long division:

This leads to the factored expression 𝑥+1224𝑥+13𝑥+252.

Factoring out 12 from the second factor, we obtain 12𝑥+1248𝑥+26𝑥+25.

Factoring the quadratic term into (24𝑥+25)(2𝑥+1) leads to the equation 12𝑥+12(24𝑥+25)(2𝑥+1)=0.

The first and the third factors give the root 𝑥=12, which is the point of tangency. The middle factor gives 𝑥=2524, which is a new point of intersection between the tangent and the curve. We can find the 𝑦-coordinate of this point by using the equation of the tangent: 𝑦=52𝑥=522524=602524=3524.

Hence, the other point of intersection between the tangent and the curve is 2524,3524.

So far, we have computed the equations of tangents to different curves described by implicit equations. In these cases, slopes of the tangents can usually be obtained by evaluating dd𝑦𝑥. Let us now consider examples where we find the equations of tangents to curves described by parametric equations. We recall parametric differentiation.

Theorem: Parametric Differentiation

Consider a curve on the 𝑥𝑦-plane described by the parametric equations 𝑥=𝑓(𝑡),𝑦=𝑔(𝑡).

If the derivatives 𝑓 and 𝑔 exist and are continuous, dd𝑦𝑥 is obtained by dd𝑦𝑥==𝑔(𝑡)𝑓(𝑡),dddd where the denominator is not equal to zero.

We can find the slope of the tangent to a parametric curve by evaluating dd𝑦𝑥 at the parameter value corresponding to the point of tangency. Let us consider an example where we will find the equation of the tangent to a parametric curve.

Example 2: Finding the Equation of the Tangent to a Given Curve

Find the equation of the tangent to the curve 𝑥=𝑡+1, 𝑦=𝑡+𝑡 at the point corresponding to the value 𝑡=1.

Answer

Let us begin by finding the coordinates of the point of tangency. We can find these coordinates by substituting 𝑡=1 into the parametric equations 𝑥=(1)+1=0,𝑦=(1)+(1)=0.

This tells us that the tangent line passes through the origin (0,0).

Next, let us find the slope of the tangent. We know that the slope of the tangent is obtained by evaluating the expression dd𝑦𝑥 at the point of tangency. Since we are given parametric equations for the curve, we need to apply parametric differentiation to obtain this expression. We recall that dd𝑦𝑥=.dddd

We can compute dddddddd𝑥𝑡=𝑡𝑡+1=3𝑡,𝑦𝑡=𝑡𝑡+𝑡=4𝑡+1.

Hence, dd𝑦𝑥==4𝑡+13𝑡.dddd

We are given that the point of tangency occurs at the parameter value 𝑡=1. Evaluating dd𝑦𝑥 at this point gives dd𝑦𝑥|||=4(1)+13(1)=33=1.

This tells us that the slope of the tangent to the curve when 𝑡=1 is 1.

Using the point–slope form with the point (0,0) and the slope 1, we obtain 𝑦=(𝑥0)+0.

Hence, the equation of the tangent is 𝑦=𝑥.

In previous examples, we computed the slope of the tangent of a curve by evaluating dd𝑦𝑥 at the point of tangency. Let us consider how to find the equation of the line that is orthogonal to the curve, which is called the normal to the curve.

During our first example, we recalled that two lines are orthogonal to each other if their slopes multiply to 1. Exceptions to this rule are when one of the lines is vertical or horizontal since, in these cases, the slope of the line is either zero or undefined. However, these cases are simpler since the line orthogonal to a horizontal line must be vertical, and vice versa. Hence, if we know that the tangent is horizontal or vertical, then we can immediately find the equation of the normal.

If the tangent is neither vertical nor horizontal, that is, dd𝑦𝑥 evaluated at this point is defined and not equal to zero, then the slope of the normal to the line is the negative reciprocal of the slope of the tangent to the curve. We summarize this as follows.

How To: Finding the Equation of the Normal to a Curve

Given a curve on the 𝑥𝑦-plane, the normal to the curve at a point (𝑥,𝑦) is the line passing through the point that is orthogonal to the tangent at that point. The equation of the normal is given in the following cases:

  • If dd𝑦𝑥|||=0(), then the normal is vertical. In this case, the equation of the normal is 𝑥=𝑥.
  • If dd𝑦𝑥|||() is not defined, then either the tangent is vertical or the tangent is not defined. If the tangent is not defined, then the normal is not defined either. If the tangent is vertical, then the normal is horizontal. In this case, the equation of the normal is 𝑦=𝑦.
  • If dd𝑦𝑥|||0() and is defined, then the slope of the normal is given by the negative reciprocal of this expression, 1||dd(). The equation of the normal to the curve in this case is 𝑦=1||(𝑥𝑥)+𝑦.dd()

In the next example, we will find the normal to a curve using this method.

Example 3: Finding the Equation of the Normal to the Curve of a Function Involving Trigonometric Functions at a Given 𝑥-Coordinate

Find the equation of the normal to the curve 𝑦=8𝑥3𝑥cossec at 𝑥=𝜋3.

Answer

Let us begin by finding the coordinates of the point where the line is normal to the curve. We know that the normal passes through the point at 𝑥=𝜋3. We can find the 𝑦-coordinate of this point by substituting this value into the function: 𝑦=8𝜋33𝜋3=8×123×2=46=2.cossec

Hence, the coordinates of this point are 𝜋3,2.

Next, let us find the slope of the normal. We know that the normal to a curve at a point is the line that passes through the given point and is orthogonal to the tangent at the point. Let us first compute the slope of the tangent to the curve at this point, which is given by the derivative at this point.

We recall the following derivatives of the trigonometric and reciprocal trigonometric functions: ddcossinddsecsectan𝑥𝑥=𝑥,𝑥𝑥=𝑥𝑥.

Using these rules, we have ddddcossecddcosddsecsinsectan𝑦𝑥=𝑥(8𝑥3𝑥)=8𝑥𝑥3𝑥𝑥=8𝑥3𝑥𝑥.

Evaluating the derivative at 𝑥=𝜋3 gives ddsinsectan𝑦𝑥|||=8𝜋33𝜋3𝜋3=8×323×2×3=4363=103.

Hence, the slope of the tangent at this point is 103. We recall that if the slopes of two lines multiply to give 1, then the lines are orthogonal. Since the slope of the tangent is not zero, the slope of the normal must be its negative reciprocal. This leads to the slope of the normal: 1||=1103=330.dd

Using the point 𝜋3,2 and the slope 330, the point–slope form of the equation of the normal is 𝑦=330𝑥𝜋32.

Let us write the equation in general form. Simplifying gives 𝑦=330𝑥𝜋32𝑦=330𝑥𝜋3902𝑦330𝑥+𝜋390+2=0.

The equation of the normal is 𝑦330𝑥+𝜋390+2=0.

In the previous example, we found the slope of the normal to an implicit curve by first finding dd𝑦𝑥 through implicit differentiation and then taking the negative reciprocal of this expression evaluated at the given point. We can find the normal to a parametric curve using an analogous method, where the only difference is that we need to use parametric differentiation to obtain the expression dd𝑦𝑥.

In our next example, we will find the equation of the normal to a parametric curve.

Example 4: Finding the Equation of the Normal to the Curve of a Function Defined by Parametric Equations Containing Reciprocal Trigonometric Functions

Determine the equation of the normal to the curve 𝑥=4𝜃+3cot, 𝑦=3𝜃+2𝜃sinsec at 𝜃=𝜋4.

Answer

Let us begin by finding the coordinates of the point where the line is normal to the curve. We are given that the normal passes through the point at the parameter value 𝜃=𝜋4. We can find the coordinates of this point by substituting this value into the parametric equations 𝑥=4𝜋4+3=4×1+3=1,𝑦=3𝜋4+2𝜋4=322+2×22=32+2=72.cotsinsec

Hence, the coordinates of this point are 1,72.

Next, let us find the slope of the normal. We know that the normal to a curve at a point is the line that passes through the given point and is orthogonal to the tangent at the point. Let us first compute the slope of the tangent to the curve at this point, which we can obtain by evaluating dd𝑦𝑥 at this point, if it exists. To find the expression dd𝑦𝑥, we apply parametric differentiation: dd𝑦𝑥=.dddd

We recall the following derivatives of the trigonometric and reciprocal trigonometric functions: ddcotcscddsincosddsecsectan𝜃𝜃=𝜃,𝜃𝜃=𝜃,𝜃𝜃=𝜃𝜃.

Differentiating the parametric equation for the 𝑥-variable with respect to 𝜃, we obtain ddddcotddcotddcsccsc𝑥𝜃=𝜃(4𝜃+3)=4𝜃𝜃+𝜃(3)=4𝜃+0=4𝜃.

Differentiating the 𝑦-equation, we obtain ddddsinsecddsinddsec𝑦𝜃=𝜃3𝜃+2𝜃=3𝜃𝜃+2𝜃𝜃.

We need to apply the chain rule for the derivative of sin𝜃. We can write this expression as a composition 𝑓𝑔, where 𝑓(𝜃)=𝜃 and 𝑔(𝜃)=𝜃sin. Then, 𝑓(𝜃)=2𝜃 and 𝑔(𝜃)=𝜃cos, which leads to ddsinsincos𝜃𝜃=𝑓(𝑔(𝜃))𝑔(𝜃)=2𝜃𝜃.

Substituting this expression and also the derivative of sec𝜃 into our expression for dd𝑦𝜃, we obtain ddsincossectan𝑦𝜃=6𝜃𝜃+2𝜃𝜃.

Hence, by parametric differentiation, we have ddsincossectancsc𝑦𝑥==6𝜃𝜃+2𝜃𝜃4𝜃.dddd

Since the line is normal at the parameter value 𝜃=𝜋4, we need to evaluate the expression above at this point: ddsincossectancsc𝑦𝑥|||=6+24=6××+2××14×=3+28=58.

Hence, the slope of the tangent at this point is 58. We recall that we can find the slope of the normal by taking the negative reciprocal of this value: 1||=1=85.dd

Using the point 1,72 and the slope 85, the point–slope form of the equation of the normal is 𝑦=85(𝑥+1)+72.

Let us write this equation in general form. Simplifying gives 𝑦=85(𝑥+1)+72𝑦=85𝑥85+72𝑦=85𝑥1610+3510𝑦=85𝑥+191085𝑥+𝑦1910=0.

This is one form of the equation we are looking for. Alternatively, we can simplify this equation so that the coefficient of 𝑥 is equal to 1. Multiplying the equation through by 58 to simplify the coefficient of 𝑥 gives 58×85𝑥+𝑦1910=58×0𝑥+58𝑦58×1910=0𝑥+58𝑦1916=0.

Hence, the equation of the normal to the given parametric curve at 𝜃=𝜋4 is 𝑥+58𝑦1916=0.

In previous equations, we computed the equations of tangents and normals to a curve. Using tangents, we can determine when two curves intersect orthogonally.

Definition: Curves Intersecting Orthogonally

Say that two curves intersect at a point (𝑥,𝑦). The curves intersect orthogonally if the tangents to both curves are well defined at this point and the tangents are orthogonal to each other.

In our next example, we will find the slopes of the tangents to two implicit curves at the point of intersection to determine whether they intersect orthogonally.

Example 5: Determining Whether Two Curves Defined Implicitly Are Orthogonal or Not

Do the curves 9𝑦8𝑦=6𝑥 and 5𝑥3𝑦=4𝑥 intersect orthogonally at the origin?

Answer

Before we start, we can verify that the curves intersect at the origin by checking both equations evaluated at (0,0). When 𝑥=0 and 𝑦=0, we can see that both equations become 0=0, which tells us that both curves pass through the origin.

Remember, two curves intersect orthogonally at a point if their tangents are well defined at that point and are orthogonal to each other. Let us find the slopes of the tangents at the origin for each curve by implicitly differentiating the given equations. For the first curve, we have dddd𝑥9𝑦8𝑦=𝑥(6𝑥).

The expression on the left-hand side of the equation is in terms of the variable 𝑦, so we need to apply the chain rule: dddddddd𝑥9𝑦8𝑦=𝑦9𝑦8𝑦𝑦𝑥=36𝑦8𝑦𝑥.

On the other hand, the right-hand side is dd𝑥(6𝑥)=6. Hence, we have 36𝑦8𝑦𝑥=6.dd

We know that dd𝑦𝑥 evaluated at the origin gives the slope of the tangent. Substituting the origin (0,0) into the equation above to find this value gives us 36×08𝑦𝑥|||=68𝑦𝑥|||=6𝑦𝑥|||=68=34.()()()dddddd

Thus, the slope of the tangent to the first curve at the origin is 34.

Next, let us find the slope of the tangent to the second curve. Implicitly differentiating the second equation gives us dddddd𝑥5𝑥𝑥(3𝑦)=𝑥(4𝑥).

The first term on the left-hand side and the term on the right-hand side of the equation are regular derivatives since the variable of these expressions matches the variable of differentiation, 𝑥. For the second term on the left-hand side, we need to use the chain rule. We have dddddddddddd𝑥5𝑥=10𝑥𝑥(3𝑦)=𝑦(3𝑦)𝑦𝑥=3𝑦𝑥𝑥(4𝑥)=4.

Substituting these into the equation above, we obtain 10𝑥3𝑦𝑥=4.dd

Evaluating this equation at the origin (0,0) and simplifying, we have 10×03𝑦𝑥|||=43𝑦𝑥|||=4𝑦𝑥|||=43=43.dddddd()()()

Thus, the slope of the tangent to the second curve at the origin is 43.

Remember, if the slopes of two lines multiply to give 1, then they are orthogonal. We obtained the slopes of the tangents for these two curves, 34 and 43, and we see that 34×43=1.

We conclude that the two curves intersect orthogonally at the origin.

In our final example, we will apply the concepts of tangents and normals to a curve to find the area of a triangle.

Example 6: Finding the Area of the Triangle Formed from the 𝑥-Axis and the Tangent and the Normal to the Curve of an Ellipse Using Differentiation

Find the area of the triangle bounded by the 𝑥-axis, the tangent, and the normal to the curve 𝑥+5𝑦=101 at the point (9,2) to the nearest thousandth.

Answer

Let us begin by visualizing this problem. We know that the given equation describes an ellipse centered at the origin. By substituting 𝑥=9 and 𝑦=2 into the equation, we can verify that the point (9,2) lies on this ellipse. We sketch the tangent (green) and the normal (red) of the ellipse at this point in the diagram below.

We want to find the area of the triangle formed by the tangent, the normal, and the 𝑥-axis. The height of the triangle is given by the 𝑦-coordinate of the point (9,2), which is 2. Recall that the area of the triangle is given by areaofatrianglebaseheight=12×.

Since we know that height is 2 length unit, we need to find the base of the triangle. The base of this triangle is given by the positive difference between the 𝑥-intercepts of the tangent and the normal, for which we need the equations of the tangent and the normal to the ellipse at (9,2).

Let us begin by finding the equation of the tangent, which will lead us to the 𝑥-intercept of the tangent. Implicitly differentiating the given equation with respect to 𝑥, we have dddddddd𝑥𝑥+5𝑦=𝑥(101)𝑥𝑥+5𝑥𝑦=0.

The first term in the equation above is a regular derivative since we are differentiating a function of 𝑥 with respect to 𝑥. Applying the power rule for differentiation, we obtain dd𝑥𝑥=2𝑥.

For the second term, dd𝑥𝑦, we must apply the chain rule since we are differentiating the function of 𝑦 with respect to 𝑥: dddddddd𝑥𝑦=𝑦𝑦𝑦𝑥=2𝑦𝑦𝑥.

Substituting these expressions into our differentiated equation 2𝑥+10𝑦𝑦𝑥=0dd leads to the expression dd𝑦𝑥=2𝑥10𝑦=𝑥5𝑦.

Remember, the slope of the tangent to a curve at a point (𝑥,𝑦) is given by dd𝑦𝑥|||(). Hence, we can find the slope of our tangent by substituting the point (9,2) into the expression dd𝑦𝑥: dd𝑦𝑥|||=95×2=910.()

Thus, the equation of the tangent going through the point (9,2) has the slope 910. The point–slope form of the equation of the line gives 𝑦=910(𝑥9)+2𝑦=910𝑥+8110+2𝑦=910𝑥+8110+2010𝑦=910𝑥+10110.

We can find the 𝑥-intercept of the tangent by substituting 𝑦=0 into this equation and solving for the 𝑥-variable: 0=910𝑥+1011010110=910𝑥𝑥=109×10110=1019.

Next, we will find the equation of the normal to the ellipse at the same point. Recall that the normal of a curve is the line that orthogonally intersects with the tangent at the point of tangency. We know that as long as two lines are not horizontal or vertical and their slopes multiply to 1, then they intersect orthogonally. In other words, the slope of the normal is the negative reciprocal of the slope of the tangent. Since we know that the slope of the tangent is 910, we obtain the slope of the normal to be 109. This line also passes through the point (9,2). The point–slope form of the equation of the normal is 𝑦=109(𝑥9)+2𝑦=109𝑥10+2𝑦=109𝑥8.

Setting 𝑦=0, we can obtain the 𝑥-intercept: 0=109𝑥88=109𝑥𝑥=910×8=7210=365.

We can compute the base of the triangle by finding the positive difference between these two 𝑥-intercepts: 1019365=5054532445=18145.

Hence, the base of the triangle is 18145 length units. We also know that the height of the triangle is 2 length units. The area of the triangle is 12×18145×2=18145=4.0222.

The area of the triangle bounded by the 𝑥-axis, the tangent, and the normal to the given ellipse at the point (9,2) is 4.022 square units, rounded to the nearest thousandth.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • Given an implicit curve on the 𝑥𝑦-plane, we can find dd𝑦𝑥 by applying implicit differentiation. Given a parametric curve on the 𝑥𝑦-plane, we can find dd𝑦𝑥 by applying parametric differentiation.
  • Given a curve on the 𝑥𝑦-plane, the slope of the tangent to the curve at (𝑥,𝑦) is obtained by evaluating dd𝑦𝑥 at the point (𝑥,𝑦), if it exists. This is denoted by dd𝑦𝑥|||().
    In this case, the equation of the tangent at the curve at (𝑥,𝑦) is 𝑦=𝑦𝑥|||(𝑥𝑥)+𝑦.dd()
  • Given a curve on the 𝑥𝑦-plane, the normal to the curve at a point (𝑥,𝑦) is the line passing through the point that is orthogonal to the tangent at that point. The equation of the normal is given in the following cases:
    • If dd𝑦𝑥|||=0(), then the normal is vertical. In this case, the equation of the normal is 𝑥=𝑥.
    • If dd𝑦𝑥|||() is not defined, then either the tangent is vertical or the tangent is not defined. If the tangent is not defined, then the normal is not defined either. If the tangent is vertical, the normal is horizontal. In this case, the equation of the normal is 𝑦=𝑦.
    • If dd𝑦𝑥|||0() and is defined, then the slope of the normal is given by the negative reciprocal of this expression, 1||dd(). The equation of the normal to the curve in this case is 𝑦=1||(𝑥𝑥)+𝑦.dd()

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