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 dd๐‘ฆ๐‘ฅ 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 pointโ€“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 ๐‘ฅ๐‘ฆ-plane, the slope of the tangent to the curve at (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ is obtained by evaluating dd๐‘ฆ๐‘ฅ at the point (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ if it exists at this point. This is denoted by dd๐‘ฆ๐‘ฅ|||(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ.

The equation of the tangent is then ๐‘ฆ=๐‘ฆ๐‘ฅ|||(๐‘ฅโˆ’๐‘ฅ)+๐‘ฆ.dd(๏—๏Ž•๏˜)๏Šฆ๏Šฆ๏ŽŸ๏ŽŸ

When we are given an implicit curve on the ๐‘ฅ๐‘ฆ-plane, we can obtain an expression for dd๐‘ฆ๐‘ฅ 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 ๐‘“โ€ฒ and dd๐‘ฆ๐‘ฅ exist, then we have dddddddd๐‘ฅ๐‘“(๐‘ฆ)=๐‘ฆ๐‘“(๐‘ฆ)ร—๐‘ฆ๐‘ฅ=๐‘“โ€ฒ(๐‘ฆ)๐‘ฆ๐‘ฅ.

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 ๐‘ฆโˆ’24๐‘ฅ+24๐‘ฅ=0๏Šจ๏Šฉ describes a curve in the plane.

  1. Find the coordinates of two points on this curve, where ๐‘ฅ=โˆ’12.
  2. Determine the equation of the tangent at the point where ๐‘ฅ=โˆ’12 and the ๐‘ฆ-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 ๐‘ฅ-coordinates, we need to identify all possible ๐‘ฆ-coordinates when ๐‘ฅ=โˆ’12. Substituting the given ๐‘ฅ-coordinate into the implicit equation, we have ๐‘ฆโˆ’24๏€ผโˆ’12๏ˆ+24๏€ผโˆ’12๏ˆ=0๐‘ฆ+3โˆ’12=0๐‘ฆ=9๐‘ฆ=ยฑ3.๏Šจ๏Šฉ๏Šจ๏Šจ

Hence, the possible ๐‘ฆ-coordinates are 3 and โˆ’3. The coordinates of the two points on this curve with ๐‘ฅ=โˆ’12 are ๏€ผโˆ’12,3๏ˆ๏€ผโˆ’12,โˆ’3๏ˆ.and

Part 2

To determine the equation of the tangent, we need to evaluate the expression dd๐‘ฆ๐‘ฅ at the point of tangency. We begin by implicitly differentiating the equation: dddddddddd๐‘ฅ๏€น๐‘ฆโˆ’24๐‘ฅ+24๐‘ฅ๏…=๐‘ฅ(0)๐‘ฅ๏€น๐‘ฆ๏…โˆ’๐‘ฅ๏€น24๐‘ฅ๏…+๐‘ฅ(24๐‘ฅ)=0.๏Šจ๏Šฉ๏Šจ๏Šฉ

The second and the third terms in the equation above are regular derivatives since we are differentiating the function of ๐‘ฅ with respect to ๐‘ฅ. Applying the power rule for differentiation, the terms are dddd๐‘ฅ๏€น24๐‘ฅ๏…=24ร—3๐‘ฅ=72๐‘ฅ๐‘ฅ(24๐‘ฅ)=24.๏Šฉ๏Šจ๏Šจ

For the first term, dd๐‘ฅ๏€น๐‘ฆ๏…๏Šจ, we must apply the chain rule since we are differentiating the function of ๐‘ฆ with respect to ๐‘ฅ: dddddddd๐‘ฅ๏€น๐‘ฆ๏…=๐‘ฆ๏€น๐‘ฆ๏…๐‘ฆ๐‘ฅ=2๐‘ฆ๐‘ฆ๐‘ฅ.๏Šจ๏Šจ

Substituting these expressions into our differentiated equation, we get 2๐‘ฆ๐‘ฆ๐‘ฅโˆ’72๐‘ฅ+24=0.dd๏Šจ

To find the slope of the tangent, we need to evaluate the expression dd๐‘ฆ๐‘ฅ at the point of tangency. Since we want the ๐‘ฆ-coordinate to be positive, this line is tangent to the curve at ๏€ผโˆ’12,3๏ˆ. Hence, we can substitute ๐‘ฅ=โˆ’12 and ๐‘ฆ=3 into the equation above: 2๐‘ฆ๐‘ฆ๐‘ฅโˆ’72๐‘ฅ+24=02(3)๐‘ฆ๐‘ฅโˆ’72๏€ผโˆ’12๏ˆ+24=06๐‘ฆ๐‘ฅโˆ’18+24=06๐‘ฆ๐‘ฅ=โˆ’6๐‘ฆ๐‘ฅ=โˆ’1.dddddddddd๏Šจ๏Šจ

This tells us that the slope of the tangent at this point is โˆ’1. We also know that this line passes through the point ๏€ผโˆ’12,3๏ˆ. Using the pointโ€“slope form of the equation of a line, the equation of the tangent is ๐‘ฆ=โˆ’๏€ผ๐‘ฅ+12๏ˆ+3.

Simplifying gives ๐‘ฆ=โˆ’๐‘ฅโˆ’12+3๐‘ฆ=52โˆ’๐‘ฅ.

So, the equation of the tangent is ๐‘ฆ=52โˆ’๐‘ฅ.

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, ๐‘ฆ=52โˆ’๐‘ฅ, we can substitute this expression into the given implicit equation of the curve: ๐‘ฆโˆ’24๐‘ฅ+24๐‘ฅ=0๏€ผ52โˆ’๐‘ฅ๏ˆโˆ’24๐‘ฅ+24๐‘ฅ=0254โˆ’5๐‘ฅ+๐‘ฅโˆ’24๐‘ฅ+24๐‘ฅ=0254+19๐‘ฅ+๐‘ฅโˆ’24๐‘ฅ=0.๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ๏Šจ๏Šฉ

We know that one of the possible solutions of this equation is ๐‘ฅ=โˆ’12 since this is the point of tangency. Therefore, by the remainder theorem, we have that ๏€ผ๐‘ฅ+12๏ˆ 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 ๏€ผ๐‘ฅ+12๏ˆ๏€ผโˆ’24๐‘ฅ+13๐‘ฅ+252๏ˆ.๏Šจ

Factoring out 12 from the second factor, we obtain 12๏€ผ๐‘ฅ+12๏ˆ๏€นโˆ’48๐‘ฅ+26๐‘ฅ+25๏….๏Šจ

Factoring the quadratic term into (โˆ’24๐‘ฅ+25)(2๐‘ฅ+1) leads to the equation 12๏€ผ๐‘ฅ+12๏ˆ(โˆ’24๐‘ฅ+25)(2๐‘ฅ+1)=0.

The first and the third factors give the root ๐‘ฅ=โˆ’12, which is the point of tangency. The middle factor gives ๐‘ฅ=2524, which is a new point of intersection between the tangent and the curve. We can find the ๐‘ฆ-coordinate of this point by using the equation of the tangent: ๐‘ฆ=52โˆ’๐‘ฅ=52โˆ’2524=60โˆ’2524=3524.

Hence, the other point of intersection between the tangent and the curve is ๏€ผ2524,3524๏ˆ.

So far, we have computed the equations of tangents and normals to different curves described by implicit equations. In these cases, slopes of the tangents can usually be obtained by evaluating dd๐‘ฆ๐‘ฅ. Let us now consider examples where we find the equations of tangents and normals to curves described by parametric equations. We recall parametric differentiation.

Theorem: Parametric Differentiation

Consider a curve on the ๐‘ฅ๐‘ฆ-plane described by the parametric equations ๐‘ฅ=๐‘“(๐‘ก),๐‘ฆ=๐‘”(๐‘ก).

If the derivatives ๐‘“โ€ฒ and ๐‘”โ€ฒ exist and are continuous, dd๐‘ฆ๐‘ฅ is obtained by dd๐‘ฆ๐‘ฅ==๐‘”โ€ฒ(๐‘ก)๐‘“โ€ฒ(๐‘ก),dddd๏˜๏๏—๏ where the denominator is not equal to zero.

We can find the slope of the tangent to a parametric curve by evaluating dd๐‘ฆ๐‘ฅ 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 ๐‘ฅ=๐‘ก+1๏Šฉ, ๐‘ฆ=๐‘ก+๐‘ก๏Šช at the point corresponding to the value ๐‘ก=โˆ’1.

Answer

Let us begin by finding the coordinates of the point of tangency. We can find these coordinates by substituting ๐‘ก=โˆ’1 into the parametric equations ๐‘ฅ=(โˆ’1)+1=0,๐‘ฆ=(โˆ’1)+(โˆ’1)=0.๏Šฉ๏Šช

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 dd๐‘ฆ๐‘ฅ 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 dd๐‘ฆ๐‘ฅ=.dddd๏˜๏๏—๏

We can compute dddddddd๐‘ฅ๐‘ก=๐‘ก๏€น๐‘ก+1๏…=3๐‘ก,๐‘ฆ๐‘ก=๐‘ก๏€น๐‘ก+๐‘ก๏…=4๐‘ก+1.๏Šฉ๏Šจ๏Šช๏Šฉ

Hence, dd๐‘ฆ๐‘ฅ==4๐‘ก+13๐‘ก.dddd๏˜๏๏—๏๏Šฉ๏Šจ

We are given that the point of tangency occurs at the parameter value ๐‘ก=โˆ’1. Evaluating dd๐‘ฆ๐‘ฅ at this point gives dd๐‘ฆ๐‘ฅ|||=4(โˆ’1)+13(โˆ’1)=โˆ’33=โˆ’1.๏๏Šฒ๏Šฑ๏Šง๏Šฉ๏Šจ

This tells us that the slope of the tangent to the curve when ๐‘ก=โˆ’1 is โˆ’1.

Using the pointโ€“slope form with the point (0,0) and the slope โˆ’1, we obtain ๐‘ฆ=โˆ’(๐‘ฅโˆ’0)+0.

Hence, the equation of the tangent is ๐‘ฆ=โˆ’๐‘ฅ.

In previous examples, we computed the slope of the tangent of a curve by evaluating dd๐‘ฆ๐‘ฅ 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 โˆ’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, dd๐‘ฆ๐‘ฅ 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 ๐‘ฅ๐‘ฆ-plane, the normal to the curve at a point (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ 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 dd๐‘ฆ๐‘ฅ|||=0(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ, then the normal is vertical. In this case, the equation of the normal is ๐‘ฅ=๐‘ฅ.๏Šฆ
  • If dd๐‘ฆ๐‘ฅ|||(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ 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 ๐‘ฆ=๐‘ฆ.๏Šฆ
  • If dd๐‘ฆ๐‘ฅ|||โ‰ 0(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ and is defined, then the slope of the normal is given by the negative reciprocal of this expression, โˆ’1||dd๏˜๏—(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ. The equation of the normal to the curve in this case is ๐‘ฆ=โˆ’1||(๐‘ฅโˆ’๐‘ฅ)+๐‘ฆ.dd๏˜๏—(๏—๏Ž•๏˜)๏Šฆ๏Šฆ๏ŽŸ๏ŽŸ

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 ๐‘ฅ-Coordinate

Find the equation of the normal to the curve ๐‘ฆ=8๐‘ฅโˆ’3๐‘ฅcossec at ๐‘ฅ=๐œ‹3.

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 ๐‘ฅ=๐œ‹3. We can find the ๐‘ฆ-coordinate of this point by substituting this value into the function: ๐‘ฆ=8๐œ‹3โˆ’3๐œ‹3=8ร—12โˆ’3ร—2=4โˆ’6=โˆ’2.cossec

Hence, the coordinates of this point are ๏€ป๐œ‹3,โˆ’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: ddcossinddsecsectan๐‘ฅ๐‘ฅ=โˆ’๐‘ฅ,๐‘ฅ๐‘ฅ=๐‘ฅ๐‘ฅ.

Using these rules, we have ddddcossecddcosddsecsinsectan๐‘ฆ๐‘ฅ=๐‘ฅ(8๐‘ฅโˆ’3๐‘ฅ)=8๐‘ฅ๐‘ฅโˆ’3๐‘ฅ๐‘ฅ=โˆ’8๐‘ฅโˆ’3๐‘ฅ๐‘ฅ.

Evaluating the derivative at ๐‘ฅ=๐œ‹3 gives ddsinsectan๐‘ฆ๐‘ฅ|||=โˆ’8๐œ‹3โˆ’3๐œ‹3๐œ‹3=โˆ’8ร—โˆš32โˆ’3ร—2ร—โˆš3=โˆ’4โˆš3โˆ’6โˆš3=โˆ’10โˆš3.๏—๏Šฒ๏‘ฝ๏Žข

Hence, the slope of the tangent at this point is โˆ’10โˆš3. We recall that if the slopes of two lines multiply to give โˆ’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: โˆ’1||=โˆ’1โˆ’10โˆš3=โˆš330.dd๏˜๏—๏—๏Šฒ๏‘ฝ๏Žข

Using the point ๏€ป๐œ‹3,โˆ’2๏‡ and the slope โˆš330, the pointโ€“slope form of the equation of the normal is ๐‘ฆ=โˆš330๏€ป๐‘ฅโˆ’๐œ‹3๏‡โˆ’2.

Let us write the equation in general form. Simplifying gives ๐‘ฆ=โˆš330๏€ป๐‘ฅโˆ’๐œ‹3๏‡โˆ’2๐‘ฆ=โˆš330๐‘ฅโˆ’๐œ‹โˆš390โˆ’2๐‘ฆโˆ’โˆš330๐‘ฅ+๐œ‹โˆš390+2=0.

The equation of the normal is ๐‘ฆโˆ’โˆš330๐‘ฅ+๐œ‹โˆš390+2=0.

In the previous example, we found the slope of the normal to an implicit curve by first finding dd๐‘ฆ๐‘ฅ 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 dd๐‘ฆ๐‘ฅ.

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 ๐‘ฅ=โˆ’4๐œƒ+3cot, ๐‘ฆ=3๐œƒ+โˆš2๐œƒsinsec๏Šจ at ๐œƒ=๐œ‹4.

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 ๐œƒ=๐œ‹4. We can find the coordinates of this point by substituting this value into the parametric equations ๐‘ฅ=โˆ’4๐œ‹4+3=โˆ’4ร—1+3=โˆ’1,๐‘ฆ=3๐œ‹4+โˆš2๐œ‹4=3๏€ฟโˆš22๏‹+โˆš2ร—2โˆš2=32+2=72.cotsinsec๏Šจ๏Šจ

Hence, the coordinates of this point are ๏€ผโˆ’1,72๏ˆ.

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 dd๐‘ฆ๐‘ฅ at this point, if it exists. To find the expression dd๐‘ฆ๐‘ฅ, we apply parametric differentiation: dd๐‘ฆ๐‘ฅ=.dddd๏˜๏ผ๏—๏ผ

We recall the following derivatives of the trigonometric and reciprocal trigonometric functions: ddcotcscddsincosddsecsectan๐œƒ๐œƒ=โˆ’๐œƒ,๐œƒ๐œƒ=๐œƒ,๐œƒ๐œƒ=๐œƒ๐œƒ.๏Šจ

Differentiating the parametric equation for the ๐‘ฅ-variable with respect to ๐œƒ, we obtain ddddcotddcotddcsccsc๐‘ฅ๐œƒ=๐œƒ(โˆ’4๐œƒ+3)=โˆ’4๐œƒ๐œƒ+๐œƒ(3)=โˆ’4๏€นโˆ’๐œƒ๏…+0=4๐œƒ.๏Šจ๏Šจ

Differentiating the ๐‘ฆ-equation, we obtain ddddsinsecddsinddsec๐‘ฆ๐œƒ=๐œƒ๏€ป3๐œƒ+โˆš2๐œƒ๏‡=3๐œƒ๐œƒ+โˆš2๐œƒ๐œƒ.๏Šจ๏Šจ

We need to apply the chain rule for the derivative of sin๏Šจ๐œƒ. We can write this expression as a composition ๐‘“โˆ˜๐‘”, where ๐‘“(๐œƒ)=๐œƒ๏Šจ and ๐‘”(๐œƒ)=๐œƒsin. Then, ๐‘“โ€ฒ(๐œƒ)=2๐œƒ and ๐‘”โ€ฒ(๐œƒ)=๐œƒcos, which leads to ddsinsincos๐œƒ๐œƒ=๐‘“โ€ฒ(๐‘”(๐œƒ))๐‘”โ€ฒ(๐œƒ)=2๐œƒ๐œƒ.๏Šจ

Substituting this expression and also the derivative of sec๐œƒ into our expression for dd๐‘ฆ๐œƒ, we obtain ddsincossectan๐‘ฆ๐œƒ=6๐œƒ๐œƒ+โˆš2๐œƒ๐œƒ.

Hence, by parametric differentiation, we have ddsincossectancsc๐‘ฆ๐‘ฅ==6๐œƒ๐œƒ+โˆš2๐œƒ๐œƒ4๐œƒ.dddd๏˜๏ผ๏—๏ผ๏Šจ

Since the line is normal at the parameter value ๐œƒ=๐œ‹4, we need to evaluate the expression above at this point: ddsincossectancsc๐‘ฆ๐‘ฅ|||=6+โˆš24=6ร—ร—+โˆš2ร—ร—14ร—๏€ผ๏ˆ=3+28=58.๏ผ๏Šฒ๏Ž„๏Šช๏Ž„๏Šช๏Ž„๏Šช๏Ž„๏Šช๏Šจ๏Ž„๏Šชโˆš๏Šจ๏Šจโˆš๏Šจ๏Šจ๏Šจโˆš๏Šจ๏Šจโˆš๏Šจ๏Šจ๏‘ฝ๏Žฃ

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: โˆ’1||=โˆ’1=โˆ’85.dd๏˜๏—๏ผ๏Šฒ๏Šซ๏Šฎ๏‘ฝ๏Žฃ

Using the point ๏€ผโˆ’1,72๏ˆ and the slope โˆ’85, the pointโ€“slope form of the equation of the normal is ๐‘ฆ=โˆ’85(๐‘ฅ+1)+72.

Let us write this equation in general form. Simplifying gives ๐‘ฆ=โˆ’85(๐‘ฅ+1)+72๐‘ฆ=โˆ’85๐‘ฅโˆ’85+72๐‘ฆ=โˆ’85๐‘ฅโˆ’1610+3510๐‘ฆ=โˆ’85๐‘ฅ+191085๐‘ฅ+๐‘ฆโˆ’1910=0.

This is one form of the equation we are looking for. Alternatively, we can simplify this equation so that the coefficient of ๐‘ฅ is equal to 1. Multiplying the equation through by 58 to simplify the coefficient of ๐‘ฅ gives 58ร—๏€ผ85๐‘ฅ+๐‘ฆโˆ’1910๏ˆ=58ร—0๐‘ฅ+58๐‘ฆโˆ’58ร—1910=0๐‘ฅ+58๐‘ฆโˆ’1916=0.

Hence, the equation of the normal to the given parametric curve at ๐œƒ=๐œ‹4 is ๐‘ฅ+58๐‘ฆโˆ’1916=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 (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ. 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 9๐‘ฆโˆ’8๐‘ฆ=6๐‘ฅ๏Šช and โˆ’5๐‘ฅโˆ’3๐‘ฆ=โˆ’4๐‘ฅ๏Šจ 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 ๐‘ฅ=0 and ๐‘ฆ=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 dddd๐‘ฅ๏€น9๐‘ฆโˆ’8๐‘ฆ๏…=๐‘ฅ(6๐‘ฅ).๏Šช

The expression on the left-hand side of the equation is in terms of the variable ๐‘ฆ, so we need to apply the chain rule: dddddddd๐‘ฅ๏€น9๐‘ฆโˆ’8๐‘ฆ๏…=๐‘ฆ๏€น9๐‘ฆโˆ’8๐‘ฆ๏…๐‘ฆ๐‘ฅ=๏€น36๐‘ฆโˆ’8๏…๐‘ฆ๐‘ฅ.๏Šช๏Šช๏Šฉ

On the other hand, the right-hand side is dd๐‘ฅ(6๐‘ฅ)=6. Hence, we have ๏€น36๐‘ฆโˆ’8๏…๐‘ฆ๐‘ฅ=6.๏Šฉdd

We know that dd๐‘ฆ๐‘ฅ 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 ๏€น36ร—0โˆ’8๏…๐‘ฆ๐‘ฅ|||=6โˆ’8๐‘ฆ๐‘ฅ|||=6๐‘ฆ๐‘ฅ|||=6โˆ’8=โˆ’34.๏Šฉ(๏Šฆ๏Ž•๏Šฆ)(๏Šฆ๏Ž•๏Šฆ)(๏Šฆ๏Ž•๏Šฆ)dddddd

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

Next, let us find the slope of the tangent to the second curve. Implicitly differentiating the second equation gives us dddddd๐‘ฅ๏€นโˆ’5๐‘ฅ๏…โˆ’๐‘ฅ(3๐‘ฆ)=๐‘ฅ(โˆ’4๐‘ฅ).๏Šจ

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, ๐‘ฅ. For the second term on the left-hand side, we need to use the chain rule. We have dddddddddddd๐‘ฅ๏€นโˆ’5๐‘ฅ๏…=โˆ’10๐‘ฅ๐‘ฅ(3๐‘ฆ)=๐‘ฆ(3๐‘ฆ)๐‘ฆ๐‘ฅ=3๐‘ฆ๐‘ฅ๐‘ฅ(โˆ’4๐‘ฅ)=โˆ’4.๏Šจ

Substituting these into the equation above, we obtain โˆ’10๐‘ฅโˆ’3๐‘ฆ๐‘ฅ=โˆ’4.dd

Evaluating this equation at the origin (0,0) and simplifying, we have โˆ’10ร—0โˆ’3๐‘ฆ๐‘ฅ|||=โˆ’4โˆ’3๐‘ฆ๐‘ฅ|||=โˆ’4๐‘ฆ๐‘ฅ|||=โˆ’4โˆ’3=43.dddddd(๏Šฆ๏Ž•๏Šฆ)(๏Šฆ๏Ž•๏Šฆ)(๏Šฆ๏Ž•๏Šฆ)

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 โˆ’1, then they are orthogonal. We obtained the slopes of the tangents for these two curves, โˆ’34 and 43, and we see that โˆ’34ร—43=โˆ’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 ๐‘ฅ-Axis and the Tangent and the Normal to the Curve of an Ellipse Using Differentiation

Find the area of the triangle bounded by the ๐‘ฅ-axis, the tangent, and the normal to the curve ๐‘ฅ+5๐‘ฆ=101๏Šจ๏Šจ 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 ๐‘ฅ=9 and ๐‘ฆ=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 ๐‘ฅ-axis. The height of the triangle is given by the ๐‘ฆ-coordinate of the point (9,2), which is 2. Recall that the area of the triangle is given by areaofatrianglebaseheight=12ร—.

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 ๐‘ฅ-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 and the tangent, which will lead us to the ๐‘ฅ-intercept of the tangent. Implicitly differentiating the given equation with respect to ๐‘ฅ, we have dddddddd๐‘ฅ๏€น๐‘ฅ+5๐‘ฆ๏…=๐‘ฅ(101)๐‘ฅ๏€น๐‘ฅ๏…+5๐‘ฅ๏€น๐‘ฆ๏…=0.๏Šจ๏Šจ๏Šจ๏Šจ

The first term in the equation above is a regular derivative since we are differentiating a function of ๐‘ฅ with respect to ๐‘ฅ. Applying the power rule for differentiation, we obtain dd๐‘ฅ๏€น๐‘ฅ๏…=2๐‘ฅ.๏Šจ

For the second term, dd๐‘ฅ๏€น๐‘ฆ๏…๏Šจ, we must apply the chain rule since we are differentiating the function of ๐‘ฆ with respect to ๐‘ฅ: dddddddd๐‘ฅ๏€น๐‘ฆ๏…=๐‘ฆ๏€น๐‘ฆ๏…๐‘ฆ๐‘ฅ=2๐‘ฆ๐‘ฆ๐‘ฅ.๏Šจ๏Šจ

Substituting these expressions into our differentiated equation 2๐‘ฅ+10๐‘ฆ๐‘ฆ๐‘ฅ=0dd leads to the expression dd๐‘ฆ๐‘ฅ=โˆ’2๐‘ฅ10๐‘ฆ=โˆ’๐‘ฅ5๐‘ฆ.

Remember, the slope of the tangent to a curve at a point (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ is given by dd๐‘ฆ๐‘ฅ|||(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ. Hence, we can find the slope of our tangent by substituting the point (9,2) into the expression dd๐‘ฆ๐‘ฅ: dd๐‘ฆ๐‘ฅ|||=โˆ’95ร—2=โˆ’910.(๏Šฏ๏Ž•๏Šจ)

Thus, the equation of the tangent going through the point (9,2) has the slope โˆ’910. The pointโ€“slope form of the equation of the line gives ๐‘ฆ=โˆ’910(๐‘ฅโˆ’9)+2๐‘ฆ=โˆ’910๐‘ฅ+8110+2๐‘ฆ=โˆ’910๐‘ฅ+8110+2010๐‘ฆ=โˆ’910๐‘ฅ+10110.

We can find the ๐‘ฅ-intercept of the tangent by substituting ๐‘ฆ=0 into this equation and solving for the ๐‘ฅ-variable: 0=โˆ’910๐‘ฅ+10110โˆ’10110=โˆ’910๐‘ฅ๐‘ฅ=โˆ’109ร—๏€ผโˆ’10110๏ˆ=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 โˆ’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 โˆ’910, we obtain the slope of the normal to be 109. This line also passes through the point (9,2). The pointโ€“slope form of the equation of the normal is ๐‘ฆ=109(๐‘ฅโˆ’9)+2๐‘ฆ=109๐‘ฅโˆ’10+2๐‘ฆ=109๐‘ฅโˆ’8.

Setting ๐‘ฆ=0, we can obtain the ๐‘ฅ-intercept: 0=109๐‘ฅโˆ’88=109๐‘ฅ๐‘ฅ=910ร—8=7210=365.

We can compute the base of the triangle by finding the positive difference between these two ๐‘ฅ-intercepts: 1019โˆ’365=50545โˆ’32445=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 12ร—18145ร—2=18145=4.0222โ€ฆ.

The area of the triangle bounded by the ๐‘ฅ-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 ๐‘ฅ๐‘ฆ-plane, we can find dd๐‘ฆ๐‘ฅ by applying implicit differentiation. Given a parametric curve on the ๐‘ฅ๐‘ฆ-plane, we can find dd๐‘ฆ๐‘ฅ by applying parametric differentiation.
  • Given a curve on the ๐‘ฅ๐‘ฆ-plane, the slope of the tangent to the curve at (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ is obtained by evaluating dd๐‘ฆ๐‘ฅ at the point (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ, if it exists. This is denoted by dd๐‘ฆ๐‘ฅ|||(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ.
    In this case, the equation of the tangent at the curve at (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ is ๐‘ฆ=๐‘ฆ๐‘ฅ|||(๐‘ฅโˆ’๐‘ฅ)+๐‘ฆ.dd(๏—๏Ž•๏˜)๏Šฆ๏Šฆ๏ŽŸ๏ŽŸ
  • Given a curve on the ๐‘ฅ๐‘ฆ-plane, the normal to the curve at a point (๐‘ฅ,๐‘ฆ)๏Šฆ๏Šฆ 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 dd๐‘ฆ๐‘ฅ|||=0(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ, then the normal is vertical. In this case, the equation of the normal is ๐‘ฅ=๐‘ฅ.๏Šฆ
    • If dd๐‘ฆ๐‘ฅ|||(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ 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 ๐‘ฆ=๐‘ฆ.๏Šฆ
    • If dd๐‘ฆ๐‘ฅ|||โ‰ 0(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ and is defined, then the slope of the normal is given by the negative reciprocal of this expression, โˆ’1||dd๏˜๏—(๏—๏Ž•๏˜)๏ŽŸ๏ŽŸ. The equation of the normal to the curve in this case is ๐‘ฆ=โˆ’1||(๐‘ฅโˆ’๐‘ฅ)+๐‘ฆ.dd๏˜๏—(๏—๏Ž•๏˜)๏Šฆ๏Šฆ๏ŽŸ๏ŽŸ

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