A separable differential-equation problem set in a Newton-style warming context: a quantity M (a temperature in degrees Celsius) increases over time t and satisfies a first-order autonomous differential equation of the form dM/dt = k(L - M), with a stated initial condition and the fact that M stays below the limiting value L. Students sketch a particular solution curve on a given slope field, use the tangent line at t = 0 to make a linear approximation of M at a later time, determine via the second derivative whether that tangent-line approximation is an under- or overestimate (concavity argument), and finally solve the differential equation explicitly by separation of variables to obtain the particular solution.
A separable differential-equation problem set in a Newton-style warming context: a quantity M (a temperature in degrees Celsius) increases over time t and satisfies a first-order autonomous differential equation of the form dM/dt = k(L - M), with a stated initial condition and the fact that M stays below the limiting value L. Students sketch a particular solution curve on a given slope field, use the tangent line at t = 0 to make a linear approximation of M at a later time, determine via the second derivative whether that tangent-line approximation is an under- or overestimate (concavity argument), and finally solve the differential equation explicitly by separation of variables to obtain the particular solution.
Read the complete question in the official College Board FRQ PDF ›
Original Tian2 solution. Each part identifies which rubric points are earned and why.
On the given slope field for $\frac{dM}{dt}=\frac{1}{4}(40-M)$, sketch the particular solution curve passing through $(0,5)$. Because $M(0)=5<40$, the slope $\frac{1}{4}(40-M)$ is positive, so the curve rises from $(0,5)$ and stays **below** the equilibrium line $M=40$ (where every slope mark is horizontal). The curve approaches $M=40$ asymptotically from below but never reaches or crosses it. Draw it so it threads consistently along the slope marks and extends to both the left and right edges of the field.
1 point (P1): the sketched solution curve must pass through (0,5), follow the slope marks without obvious conflicts, extend reasonably to the left and right edges, and lie entirely below the line M=40. Common loss: drawing a curve that touches or crosses M=40, or that does not pass through the given point.
Use the tangent line to $M$ at $t=0$ to approximate $M(2)$. The slope at $t=0$ is $$\frac{dM}{dt}\Big|_{t=0}=\frac{1}{4}(40-5)=\frac{35}{4}.$$ The tangent line through $(0,5)$ is $y=5+\frac{35}{4}(t-0)$, so $$M(2)\approx 5+\frac{35}{4}\cdot 2 = 5+17.5 = 22.5.$$ The milk's temperature at $t=2$ minutes is approximately $22.5^{\circ}\mathrm{C}$.
1 point for computing the slope dM/dt|_{t=0} = 35/4; 1 point for the tangent-line approximation evaluated at t=2, i.e. 5 + (35/4)(2) = 22.5. The minimal sufficient response for both points is 5 + (35/4)*2. An unsupported value of 22.5 with no slope shown does not earn the approximation point. The approximation need not be simplified, but an incorrect simplification forfeits the second point.
Differentiate $\frac{dM}{dt}=\frac{1}{4}(40-M)$ again with respect to $t$, using the chain rule: $$\frac{d^2M}{dt^2}=-\frac{1}{4}\frac{dM}{dt}=-\frac{1}{4}\cdot\frac{1}{4}(40-M)=-\frac{1}{16}(40-M).$$ Since $M(t)<40$ for all $t$, we have $40-M>0$, so $\frac{d^2M}{dt^2}<0$. Thus the graph of $M$ is **concave down**, which means the tangent line lies **above** the curve. Therefore the part (b) tangent-line value $22.5$ is an **overestimate** of the true $M(2)$.
1 point for the correct expression d^2M/dt^2 = -(1/16)(40-M) in terms of M (or the equivalent -(1/4)(1/4)(40-M)). 1 point for the overestimate conclusion supported by a concavity reason: because M<40, d^2M/dt^2<0, so M is concave down and the tangent line overestimates. An argument based on a single point, or concluding 'underestimate', does not earn the second point.
Solve $\frac{dM}{dt}=\frac{1}{4}(40-M)$ with $M(0)=5$ by separation of variables. Separate and integrate: $$\frac{dM}{40-M}=\frac{1}{4}\,dt \;\Rightarrow\; -\ln|40-M| = \frac{1}{4}t + C.$$ Apply the initial condition $M(0)=5$: $-\ln|40-5| = 0 + C$, so $C = -\ln 35$. Since $M<40$, $40-M>0$ and $|40-M|=40-M$, giving $$-\ln(40-M) = \frac{1}{4}t - \ln 35 \;\Rightarrow\; \ln(40-M) = -\frac{1}{4}t + \ln 35.$$ Exponentiate: $$40-M = 35e^{-t/4} \;\Rightarrow\; \boxed{M(t) = 40 - 35e^{-t/4}}.$$ As a check, $M(0)=40-35=5$ matches the initial condition, and $M(t)\to 40$ as $t\to\infty$, consistent with the limiting temperature.
4 points, awarded sequentially: P1 separates the variables (dM/(40-M) = (1/4)dt); P2 finds correct antiderivatives (-ln|40-M| = t/4 + C); P3 (eligible only after P1, P2) includes the constant of integration and uses M(0)=5 to get C = -ln 35; P4 (eligible only after P1-P3) solves explicitly for M to reach M(t) = 40 - 35e^{-t/4}. No separation of variables earns 0/4. No constant of integration caps the score at 2/4.
Our worked solutions and practice questions are original instructional content created by Tian2 AP. They are aligned to the concepts and skills described in College Board’s Course and Exam Description and are not reproductions of, or affiliated with, College Board’s official materials.
