AP Calculus BC 5.6 Determining Concavity of Functions over Their Domains - FRQs - Exam Style Questions
No-Calc Question
(a) A slope field for \(\displaystyle \frac{dM}{dt}=\frac14(40-M)\) is shown. Sketch the solution curve through the point \((0,5)\).
(b) Use the line tangent to the graph of \(M\) at \(t=0\) to approximate \(M(2)\), the temperature at \(t=2\) minutes.
(c) Write an expression for \(\displaystyle \frac{d^{2}M}{dt^{2}}\) in terms of \(M\). Use \(\displaystyle \frac{d^{2}M}{dt^{2}}\) to decide whether the approximation from part (b) is an underestimate or an overestimate of \(M(2)\). Give a reason.
(d) Use separation of variables to find the particular solution \(M(t)\) to \(\displaystyle \frac{dM}{dt}=\frac14(40-M)\) with \(M(0)=5\).
Most-appropriate topic codes:
• TOPIC 4.6: Approximating Values Using Local Linearity & Linearization — part (b)
• TOPIC 5.6: Determining Concavity of Functions over Their Domains — part (c)
• TOPIC 7.7: Finding Particular Solutions Using Initial Conditions & Separation of Variables — part (d)
▶️ Answer/Explanation
Since \(\dfrac{dM}{dt}=\frac14(40-M)>0\) for \(M<40\), the solution is increasing. Also \(\dfrac{d^{2}M}{dt^{2}}<0\) (shown in part (c)), so the curve is concave down. The solution passes through \((0,5)\) and rises toward the horizontal asymptote \(M=40\) as \(t\) increases.
(b) Tangent-line approximation at \(t=0\)
Slope at \(t=0\): \[ M'(0)=\frac14\big(40-M(0)\big)=\frac14(40-5)=\frac{35}{4}=8.75. \] Tangent line at \(t=0\): \[ L(t)=M(0)+M'(0)\,(t-0)=5+\frac{35}{4}t. \] Approximate \(M(2)\): \[ M(2)\approx L(2)=5+\frac{35}{4}\cdot 2=5+\frac{35}{2}=5+17.5=\boxed{22.5^\circ\text{C}}. \]
Differentiate the DE: \[ \frac{d^{2}M}{dt^{2}}=\frac{d}{dt}\!\left[\tfrac14(40-M)\right] =-\tfrac14\,\frac{dM}{dt}. \] Using the original DE again, \[ \frac{d^{2}M}{dt^{2}}=-\tfrac14\Big(\tfrac14(40-M)\Big) =-\frac{1}{16}(40-M). \] Because \(M(t)<40\) for all \(t\), we have \(\dfrac{d^{2}M}{dt^{2}}<0\) (concave down). For an increasing, concave-down function, the tangent line at \(t=0\) lies above the curve for \(t>0\). Therefore the estimate in (b) is an overestimate of the actual \(M(2)\).
Start with \[ \frac{dM}{dt}=\frac14(40-M). \] Separate: \[ \frac{dM}{40-M}=\frac14\,dt. \] Integrate: \[ \int \frac{dM}{40-M}=\int \frac14\,dt \quad\Rightarrow\quad -\ln|40-M|=\frac{t}{4}+C. \] Write as \(\ln|40-M|=-\tfrac{t}{4}+C\). Apply \(M(0)=5\): \(\ln(40-5)=\ln 35=C\). Hence \(\ln(40-M)=-\tfrac{t}{4}+\ln 35\). Exponentiate: \[ 40-M=35\,e^{-t/4} \quad\Rightarrow\quad \boxed{\,M(t)=40-35\,e^{-t/4}\,}. \]