Home / AP calculus AB and BC concise summary notes / AP Calculus BC 5.12 Exploring Behaviors of Implicit Relations Study Notes

AP Calculus BC 5.12 Exploring Behaviors of Implicit Relations Study Notes - New Syllabus

AP Calculus BC 5.12 Exploring Behaviors of Implicit Relations Study Notes- New syllabus

AP Calculus BC 5.12 Exploring Behaviors of Implicit Relations Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • A function’s derivative can be used to understand some behaviors of the function.

Key Concepts: 

  • Exploring Behaviors of Implicit Relations

AP Calculus BC-Concise Summary Notes- All Topics

AP Calculus BC Exam Questions MCQs and FRQs

Exploring Behaviors of Implicit Relations

Exploring Behaviors of Implicit Relations

In calculus, some functions or curves are not expressed in the explicit form \( y = f(x) \), but instead given implicitly, such as:

This means that the variable \( y \) is not isolated, and we must use implicit differentiation to analyze the curve’s behavior, including slope, concavity, and special features like vertical/horizontal tangents.

Steps for Exploring Implicit Relations:

  • Differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \).
  • Apply the chain rule: whenever you differentiate \( y \), multiply by \( \dfrac{dy}{dx} \).
  • Solve for \( \dfrac{dy}{dx} \) to find the slope of the tangent.
  • Use the second derivative \( \dfrac{d^2y}{dx^2} \) to analyze concavity or inflection points.

Example:

Analyze the behavior of the curve \( x^2 + xy + y^2 = 7 \) Find:

  • \( \dfrac{dy}{dx} \) the slope of the tangent line
  • The equation of the tangent at point (1, 2)
▶️Answer/Explanation

Differentiate both sides implicitly:

\( \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(y^2) = \dfrac{d}{dx}(7) \)
\( 2x + \left( x \dfrac{dy}{dx} + y \right) + 2y \dfrac{dy}{dx} = 0 \)

Combine like terms:
\( 2x + y + x \dfrac{dy}{dx} + 2y \dfrac{dy}{dx} = 0 \)
\( (x + 2y) \dfrac{dy}{dx} = – (2x + y) \)

So, \( \dfrac{dy}{dx} = \dfrac{-(2x + y)}{x + 2y} \)

At point (1, 2):
\( \dfrac{dy}{dx} = \dfrac{-(2(1) + 2)}{1 + 2(2)} = \dfrac{-4}{5} \)
So the slope is \( -\dfrac{4}{5} \).

Tangent line at (1, 2):
Use point-slope form:
\( y – 2 = -\dfrac{4}{5}(x – 1) \)

Example:

Consider the curve defined by \( x^2 – y^2 – 5xy = 25 \).

  1. Show that \( \dfrac{dy}{dx} = \dfrac{2x – 5y}{5x + 2y} \).
  2. Find the slope of the line tangent to the curve at each point on the curve when \( x = 2 \).
  3. Find the positive value of \( x \) at which the curve has a vertical tangent line. Show the work that leads to your answer.
  4. Let \( x \) and \( y \) be functions of time \( t \) that are related by the equation \( x^2 – y^2 – 5xy = 25 \). At time \( t = 3 \), the value of \( x \) is 5, the value of \( y \) is 0, and the value of \( \dfrac{dy}{dt} = -2 \). Find the value of \( \dfrac{dx}{dt} \) at time \( t = 3 \).
▶️Answer/Explanation

Part (a):
Differentiate both sides implicitly with respect to \( x \):

\( \dfrac{d}{dx}(x^2 – y^2 – 5xy) = \dfrac{d}{dx}(25) \)
\( 2x – 2y \dfrac{dy}{dx} – 5\left( x \dfrac{dy}{dx} + y \right) = 0 \)

Expand and rearrange:
\( 2x – 2y \dfrac{dy}{dx} – 5x \dfrac{dy}{dx} – 5y = 0 \)

Group \( \dfrac{dy}{dx} \) terms:
\( 2x – 5y – (2y + 5x)\dfrac{dy}{dx} = 0 \)

Solve for \( \dfrac{dy}{dx} \):
\( \dfrac{dy}{dx} = \dfrac{2x – 5y}{5x + 2y} \)

Part (b):
Use the formula \( \dfrac{dy}{dx} = \dfrac{2x – 5y}{5x + 2y} \) and plug in \( x = 2 \).
We must find all points on the curve where \( x = 2 \).

Plug \( x = 2 \) into the original equation:
\( 4 – y^2 – 10y = 25 \)
\( -y^2 – 10y – 21 = 0 \Rightarrow y^2 + 10y + 21 = 0 \)
Factor: \( (y + 3)(y + 7) = 0 \Rightarrow y = -3, -7 \)

So, points are \( (2, -3) \) and \( (2, -7) \).

At \( (2, -3) \):
\( \dfrac{dy}{dx} = \dfrac{2(2) – 5(-3)}{5(2) + 2(-3)} = \dfrac{4 + 15}{10 – 6} = \dfrac{19}{4} \)
At \( (2, -7) \):
\( \dfrac{dy}{dx} = \dfrac{4 + 35}{10 – 14} = \dfrac{39}{-4} = -\dfrac{39}{4} \)

Part (c):
A vertical tangent occurs when the denominator of \( \dfrac{dy}{dx} \) is 0 (since slope is undefined).
Solve \( 5x + 2y = 0 \Rightarrow y = -\dfrac{5x}{2} \)

Substitute into the original equation:
\( x^2 – y^2 – 5xy = 25 \)
Replace \( y \):
\( x^2 – \left( \dfrac{25x^2}{4} \right) – 5x \left( -\dfrac{5x}{2} \right) = 25 \) \( x^2 – \dfrac{25x^2}{4} + \dfrac{25x^2}{2} = 25 \) Get common denominators:
\( \left( \dfrac{4x^2 – 25x^2 + 50x^2}{4} \right) = 25 \Rightarrow \dfrac{29x^2}{4} = 25 \) Multiply both sides by 4:
\( 29x^2 = 100 \Rightarrow x^2 = \dfrac{100}{29} \)
Take positive root:
\( x = \sqrt{\dfrac{100}{29}} = \dfrac{10}{\sqrt{29}} \)

Part (d):
Use implicit differentiation with respect to \( t \):

Differentiate: \( \dfrac{d}{dt}(x^2 – y^2 – 5xy) = \dfrac{d}{dt}(25) \) \( 2x \dfrac{dx}{dt} – 2y \dfrac{dy}{dt} – 5\left( \dfrac{dx}{dt}y + x \dfrac{dy}{dt} \right) = 0 \)

Plug in \( x = 5 \), \( y = 0 \), \( \dfrac{dy}{dt} = -2 \):
\( 2(5) \dfrac{dx}{dt} – 0 – 5( \dfrac{dx}{dt}(0) + 5(-2) ) = 0 \Rightarrow 10 \dfrac{dx}{dt} + 50 = 0 \Rightarrow \dfrac{dx}{dt} = -5 \)

Example:

Let the curve be defined by \( x^2 + xy + y^2 = 7 \). What is the slope of the tangent line to the curve at the point \( (1, 2) \)?

  1. \( \dfrac{1}{5} \)
  2. \( \dfrac{2}{5} \)
  3. \( \dfrac{3}{5} \)
  4. \( \dfrac{-4}{5} \)
  5. \( \dfrac{5}{6} \)
▶️Answer/Explanation

Given: \( x^2 + xy + y^2 = 7 \)
Differentiate both sides with respect to \( x \):

\( 2x + x \dfrac{dy}{dx} + y + 2y \dfrac{dy}{dx} = 0 \)

Grouping terms:

\( (x + 2y)\dfrac{dy}{dx} = -2x – y \Rightarrow \dfrac{dy}{dx} = \dfrac{-2x – y}{x + 2y} \)

Plug in \( x = 1 \), \( y = 2 \):

\( \dfrac{dy}{dx} = \dfrac{-2(1) – 2}{1 + 2(2)} = \dfrac{-4}{5} \)

Scroll to Top