Differential Equations: AP Calculus AB Math Test Answer Key – PDF Download

AP Calculus Differential Test Answer Key: questions 1–5

Question 1: Verify a solution (2 points)

1. \(\frac{dy}{dx}=-6e^{-3x}\). Also \(-3(y-1)=-3(2e^{-3x})=-6e^{-3x}\). The sides are equal.
2. \(y(0)=2e^{0}+1=3\).

(b) \(y(0)=3\)

Scoring: 1 point for differentiating and substituting to show equality; 1 point for \(y(0)=3\).

Question 2: Slope field reading (3 points)

1. At \((2,1)\): \(2-1=1\). At \((-1,2)\): \(-1-2=-3\).
2. \(f^{\prime}(0)=0-1=-1\). Tangent line: \(y=1-x\). \(f(0.5)\approx 0.5\).
3. \(\frac{d^{2}y}{dx^{2}}=1-\frac{dy}{dx}=1-(x-y)=1-x+y\). At \((0,1)\) this is \(2>0\), so the graph is concave up and the tangent line lies below it: the estimate is too small.

Scoring: 1 point for both slopes (\(1\) and \(-3\)); 1 point for the tangent line and \(0.5\); 1 point for \(\frac{d^{2}y}{dx^{2}}=1-x+y\) with the correct conclusion.

Question 3: Separable equation (3 points)

\(y^{2}\,dy=2x\,dx\), so \(\frac{y^{3}}{3}=x^{2}+C\).

\(y(0)=2\): \(\frac{8}{3}=C\). Then \(y^{3}=3x^{2}+8\).

\(y=(3x^{2}+8)^{1/3}\)

Scoring: 1 point for separating the variables; 1 point for both antiderivatives (with \(C\)); 1 point for \(C=\frac{8}{3}\) and the final function.

Question 4: Bacteria model (3 points)

1. \(P(t)=400e^{kt}\).
2. \(400e^{3k}=1600\), so \(e^{3k}=4\) and \(k=\frac{\ln 4}{3}\) (about \(0.4621\)).
3. \(P(5)=400\cdot 4^{5/3}\approx 400(10.0794)\approx 4032\).

(c) about 4,032 bacteria

Scoring: 1 point for \(P=400e^{kt}\); 1 point for \(k=\frac{\ln 4}{3}\); 1 point for \(4032\).

Question 5: Write a differential equation (2 points)

1. \(\frac{dT}{dt}=k(T-72)\).
2. The pie is hotter than the room, so \(T-72>0\), but its temperature falls, so \(\frac{dT}{dt}<0\). Therefore \(k<0\).

\(\frac{dT}{dt}=k(T-72)\) with \(k<0\)

Scoring: 1 point for the equation; 1 point for \(k<0\) with the reason.

AP Calculus Differential Test Answer Key: questions 6–8

Question 6: Solution and domain (3 points)

1. \(y^{-2}\,dy=x\,dx\), so \(-\frac{1}{y}=\frac{x^{2}}{2}+C\). With \(f(0)=1\), \(C=-1\). Then \(\frac{1}{y}=1-\frac{x^{2}}{2}\) and \(f(x)=\frac{2}{2-x^{2}}\).
2. The denominator is zero when \(x=\pm\sqrt{2}\), so the interval is \(-\sqrt{2}< x<\sqrt{2}\).

(a) \(f(x)=\frac{2}{2-x^{2}}\); (b) \((-\sqrt{2},\sqrt{2})\)

Scoring: part (a) 1 point for separation and antiderivatives, 1 point for \(C\) and \(f\); part (b) 1 point.

Question 7: Radioactive decay (2 points)

Half-life 30 years: 80 g becomes 40 g, 20 g, then 10 g after three half-lives. Equivalently \(80\cdot 2^{-t/30}=10\) gives \(2^{-t/30}=\frac{1}{8}\), so \(t=90\).

90 years

Scoring: 1 point for a correct model such as \(m=80\cdot 2^{-t/30}\) (or \(k=-\frac{\ln 2}{30}\)); 1 point for \(t=90\).

Question 8: Identify an equation (2 points)

Segments are parallel along each horizontal row, so the slope depends only on \(y\). The slope is positive above the \(x\)-axis and negative below it, so the equation is \(\frac{dy}{dx}=\frac{y}{2}\) (not \(-\frac{y}{2}\), which has the opposite signs).

The solution through \((0,1)\) is \(f(x)=e^{x/2}\), so \(\lim_{x\to\infty}f(x)=\infty\).

\(\frac{dy}{dx}=\frac{y}{2}\); the limit is \(\infty\).

Scoring: 1 point for the equation with a valid reason; 1 point for the limit.

Scoring summary

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