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## Differential Calculus

### Course: Differential Calculus>Unit 5

Lesson 12: Analyzing implicit relations

# Horizontal tangent to implicit curve

AP.CALC:
FUN‑4 (EU)
,
FUN‑4.D (LO)
,
FUN‑4.D.1 (EK)
,
FUN‑4.E (LO)
,
FUN‑4.E.1 (EK)
,
FUN‑4.E.2 (EK)
Finding the equation of a horizontal tangent to a curve that is defined implicitly as an equation in x and y.

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• What happen if you had a variable another than x in the numerator, such as y? • 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.
• 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" ?   