AP Calculus AB 7.3 Sketching Slope Fields - MCQs - Exam Style Questions
No-Calc Question
The figure shown is a slope field for which of the following differential equations?
(A) \(\displaystyle \frac{dy}{dx} = -\frac{y^{2}}{(x-2)^{2}}\)
(B) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x-2)^{3}}\)
(C) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x+2)^{2}}\)
(D) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x-2)^{2}}\)
(B) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x-2)^{3}}\)
(C) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x+2)^{2}}\)
(D) \(\displaystyle \frac{dy}{dx} = \frac{1}{(x-2)^{2}}\)
▶️ Answer/Explanation
In the slope field, slopes are the same along vertical lines (depend only on \(x\), not on \(y\)).
Slopes are positive everywhere and become extremely steep near \(x=2\) (vertical column of nearly vertical segments).
A model that is positive for all \(x\neq 2\) and blows up as \(x\to 2\) is \[ \frac{dy}{dx}=\frac{1}{(x-2)^{2}}. \] Therefore the slope field corresponds to choice (D).
✅ Answer: (D)
Slopes are positive everywhere and become extremely steep near \(x=2\) (vertical column of nearly vertical segments).
A model that is positive for all \(x\neq 2\) and blows up as \(x\to 2\) is \[ \frac{dy}{dx}=\frac{1}{(x-2)^{2}}. \] Therefore the slope field corresponds to choice (D).
✅ Answer: (D)