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)
No-Calc Question
Shown above is a slope field for which of the following differential equations?
(A) \(\dfrac{dy}{dx}=|x+y|\)
(B) \(\dfrac{dy}{dx}=x^{3}\)
(C) \(\dfrac{dy}{dx}=y^{3}\)
(D) \(\dfrac{dy}{dx}=y^{2}\)
(B) \(\dfrac{dy}{dx}=x^{3}\)
(C) \(\dfrac{dy}{dx}=y^{3}\)
(D) \(\dfrac{dy}{dx}=y^{2}\)
▶️ Answer/Explanation
Slopes are constant along horizontal lines ⇒ depend only on \(y\).
Slopes are \(0\) on the \(x\)-axis and positive above/below (same sign).
That behavior matches \(\dfrac{dy}{dx}=y^{2}\) (nonnegative; zero at \(y=0\)).
✅ Answer: (D)
Slopes are \(0\) on the \(x\)-axis and positive above/below (same sign).
That behavior matches \(\dfrac{dy}{dx}=y^{2}\) (nonnegative; zero at \(y=0\)).
✅ Answer: (D)