If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: AP®︎/College Calculus BC>Unit 5

Lesson 12: Exploring behaviors of implicit relations

# Horizontal tangent to implicit curve

Finding the equation of a horizontal tangent to a curve that is defined implicitly as an equation in x and y.

## Want to join the conversation?

• What happen if you had a variable another than x in the numerator, such as y?
• Solve the numerator for y to find an equation for when the derivative is equal to zero. Substitute this equation for y into the original equation to find where the equation has a derivative equal to zero (horizontal tangent).
• Why do we often need to follow these two rules when solving the exercises?
1) The denominator of the derivative must equal 0.
2) The numerator of the derivative must not equal 0.
I don't get it, especially letting the denominator equal zero.
• If a curve has a vertical asymptote at 𝑥 = 𝑐,
then the slope of the tangent line (i.e. the derivative) there is ±∞,
which means that the denominator of the derivative approaches zero as 𝑥 approaches 𝑐, while the numerator approaches a non-zero number.

– – –

In the video we are given the curve 𝑥² + 𝑦⁴ + 6𝑥 = 7
and its derivative 𝑑𝑦∕𝑑𝑥 = −2(𝑥 + 3)∕(4𝑦³)

So, to find where the curve has a vertical asymptote we need to find the 𝑥-value/s for which the denominator of the derivative is zero, while the numerator isn't – that is 𝑦 = 0 and 𝑥 ≠ −3

We do this by solving the equation 𝑥² + 0 + 6𝑥 = 7,
which gives us 𝑥 = −7 or 𝑥 = 1

– – – EDIT (12/3/2022) – – –
The numerator can approach zero, as long as the denominator approaches zero faster.
For example 𝑓(𝑥) = 𝑥²∕𝑥³ has a vertical asymptote at 𝑥 = 0 even though 𝑓′(0) = 0∕0.
• Why is it that sometimes the numerator can not be zero and then other times it has to be zero? It shouldn't matter which one is Zero because zero divided by a number or a number divided by zero will still give you a dy/dx of zero. Also, what are you supposed to do when the tangent line is the y axis? Then, the derivative would be undefined since it would have a vertical slope.
• "because zero divided by a number or a number divided by zero will still give you a dy/dx of zero."

This is incorrect, a number divided by zero is undefined, which would give a vertical slope. Thus, when we are trying to find when the slope is vertical, (parallel to the y axis), we set the denominator equal to 0, which means that the derivative must be undefined (vertical slope). When we want to find the horizontal line, we set the numerator equal to zero, which means that the derivative must equal zero (horizontal slope).
• How do you know that the initial curve is a circle?
• It isn't a circle, but it is a closed loop. You can see this if you parameterize the curve as

x(t)=4cos(t)-3
y(t)=√|4sin(t)| ·sin(t)/|sin(t)|

The sin(t)/|sin(t)| factor makes y(t) negative when sin(t) is negative, so that you get both halves of the loop.

Because we've expressed the curve in terms of sine and cosine, periodic functions, a particle tracing the path of the curve will repeat every 2π. Because the path repeats, it must form a loop.
• Is there also a video for vertical tangents anywhere?
• If the question is: "Write the equation of the horizontal line that is tangent to the curve and is above the x-axis", shouldn't the answer be: "y = x + 2", no the "y = 2" ?
• Could we get a video on how to do the vertical line too? I know it might seem self explanatory considering the video, but honestly I'm struggling to put it together.
• I don't know I am baffled. I got everything, but when I took the quiz, one of them made different assumptions, such as the denominator being equal to zero. Why is that?

it is written like this
Let's start by finding the point on the curve where the line tangent to the curve is vertical. Then we can find the line's equation based on that point's

xx-coordinate.
We have three conditions that the point must meet.
The point must be on the curve.
The denominator of the derivative must equal
0
00.
The numerator of the derivative must not equal
0
0
(1 vote)
• Well, if you need points where the tangent is vertical, the slope must be undefined. For that to happen, the numerator must be non-negative but the denominator must be negative. For horizontal slopes, these conditions are the opposite.
• Sal chooses to solve for 0 using the value of the dy/dx, rather than finding the derivative of the whole curve and solving from there.

Is this because if the answer is meant to be a horizontal line we only need to care about how the y-value of the derivative changes?
(1 vote)
• The derivative of the whole curve is dy/dx, its equation gives us the slope of the curve when it is given values of x and y. That means making it equal to 0 shows us the values we need to input to find the points on the curve where the slope is 0, which is a horizontal line.
(1 vote)
• Is this sketch proof without using the derivative ok?
g(x) = x^2 + y^2 + 6x = 7
g(x) = (x+3)^2 + y^2 = 16 (circle of center (-3, 0))
Horizontal tangent of this circle is y=4 or y=-4, above x-axis is y=4, going to the question: y = sqrt(4) = 2
If the question is under the x-axis it can be done by symmetry of the positive solution.
(1 vote)

## Video transcript

- [Instructor] We're told to consider the curve given by the equation. They give this equation. It can be shown that the derivative of y with respect to x is equal to this expression, and you could figure that out with just some implicit differentiation and then solving for the derivative of y with respect to x. We've done that in other videos. Write the equation of the horizontal line that is tangent to the curve and is above the x-axis. Pause this video, and see if you can have a go at it. So let's just make sure we're visualizing this right. So let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there. And then another one might be maybe right over here. And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know? What is true if this tangent line is horizontal? Well, that tells us that, at this point, dy/dx is equal to zero. In fact, that would be true at both of these points. And we know what dy/dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y. And so when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero. So what is going to be the corresponding y value when x is equal to negative three? And, if we know that, well, this equation is just going to be y is equal to something. It's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y. So let's do that. So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven. This is nine. This is negative 18. And so we're going to get y to the fourth minus nine is equal to seven, or, adding nine to both sides, we get y to the fourth power is equal to 16. And this would tell us that y is going to be equal to plus or minus two. Well, there would be then two horizontal lines. One would be y is equal to two. The other is y is equal to negative two. But they want us, the equation of the horizontal line that is tangent to the curve and is above the x-axis, so only this one is going to be above the x-axis. And we're done. It's going to be y is equal to two.