AP Calculus BC 7.8 Exponential Models with Differential Equations - FRQs - Exam Style Questions
No-Calc Question
At time \(t=0\), a boiled potato is taken from a pot on a stove and left to cool in a kitchen. The internal temperature of the potato is \(91^\circ\)C at time \(t=0\), and the internal temperature of the potato is greater than \(27^\circ\)C for all times \(t>0\). The internal temperature of the potato at time \(t\) minutes can be modeled by the function \(H\) that satisfies the differential equation \[ \frac{dH}{dt}=-\frac14\,(H-27), \] where \(H(t)\) is measured in degrees Celsius and \(H(0)=91\).
(a) Write an equation for the line tangent to the graph of \(H\) at \(t=0\). Use this equation to approximate the internal temperature of the potato at time \(t=3\).
(b) Use \(\dfrac{d^2H}{dt^2}\) to determine whether your answer in part (a) is an underestimate or an overestimate of the internal temperature of the potato at time \(t=3\).
(c) For \(t<10\), an alternate model for the internal temperature of the potato at time \(t\) minutes is the function \(G\) that satisfies the differential equation \[ \frac{dG}{dt}=-(G-27)^{2/3}, \] where \(G(t)\) is measured in degrees Celsius and \(G(0)=91\). Find an expression for \(G(t)\). Based on this model, what is the internal temperature of the potato at time \(t=3\)?
Most-appropriate topic codes (CED):
• TOPIC 4.6: Local Linearity & Linearization — part (a)
• TOPIC 5.4: Using the Second Derivative to Analyze Behavior — part (b)
• TOPIC 7.7: Solving Differential Equations by Separation of Variables — part (c)
• TOPIC 7.8: Particular Solutions to Differential Equations — part (c)
• TOPIC 5.4: Using the Second Derivative to Analyze Behavior — part (b)
• TOPIC 7.7: Solving Differential Equations by Separation of Variables — part (c)
• TOPIC 7.8: Particular Solutions to Differential Equations — part (c)
▶️ Answer/Explanation
(a) Tangent line at \(t=0\) and linear estimate at \(t=3\)
Slope at \(t=0\):\quad \(H'(0)=-\tfrac14\big(H(0)-27\big)=-\tfrac14(91-27)=-16.\)
Point: \((0,\,H(0))=(0,91)\). Tangent line: \[ L(t)=91-16t. \] Approximation at \(t=3\): \[ H(3)\approx L(3)=91-16(3)=\boxed{43^\circ\text{C}}. \]
(b) Underestimate or overestimate?
Differentiate the DE: \[ \frac{d^2H}{dt^2}=-\frac14\,\frac{dH}{dt} =-\frac14\!\left(-\frac14(H-27)\right)=\frac{1}{16}(H-27). \] Given \(H(t)>27\) for \(t>0\), we have \(H”(t)>0\) (concave up). For a concave-up curve, the tangent line lies below the curve near the point of tangency. Therefore the linear estimate in (a) is an \(\boxed{\text{underestimate}}\).
(c) Solve \(G'(t)=-(G-27)^{2/3}\) with \(G(0)=91\) and evaluate \(G(3)\)
Separate variables: \[ \frac{-dG}{(G-27)^{2/3}}=dt. \] Integrate: \[ \int\frac{-dG}{(G-27)^{2/3}}=\int 1\,dt \;\Rightarrow\; -3(G-27)^{1/3}=t+C. \] Apply \(G(0)=91\Rightarrow (G-27)^{1/3}=64^{1/3}=4\): \(-3\cdot 4=0+C\Rightarrow C=-12.\)
Thus, \[ -3(G-27)^{1/3}=t-12 \;\Rightarrow\; (G-27)^{1/3}=\frac{12-t}{3} \;\Rightarrow\; \boxed{\,G(t)=27+\Big(\tfrac{12-t}{3}\Big)^3\,},\quad (t<10). \] At \(t=3\): \[ G(3)=27+\Big(\tfrac{12-3}{3}\Big)^3 =27+3^3=27+27=\boxed{54^\circ\text{C}}. \]
Slope at \(t=0\):\quad \(H'(0)=-\tfrac14\big(H(0)-27\big)=-\tfrac14(91-27)=-16.\)
Point: \((0,\,H(0))=(0,91)\). Tangent line: \[ L(t)=91-16t. \] Approximation at \(t=3\): \[ H(3)\approx L(3)=91-16(3)=\boxed{43^\circ\text{C}}. \]
(b) Underestimate or overestimate?
Differentiate the DE: \[ \frac{d^2H}{dt^2}=-\frac14\,\frac{dH}{dt} =-\frac14\!\left(-\frac14(H-27)\right)=\frac{1}{16}(H-27). \] Given \(H(t)>27\) for \(t>0\), we have \(H”(t)>0\) (concave up). For a concave-up curve, the tangent line lies below the curve near the point of tangency. Therefore the linear estimate in (a) is an \(\boxed{\text{underestimate}}\).
(c) Solve \(G'(t)=-(G-27)^{2/3}\) with \(G(0)=91\) and evaluate \(G(3)\)
Separate variables: \[ \frac{-dG}{(G-27)^{2/3}}=dt. \] Integrate: \[ \int\frac{-dG}{(G-27)^{2/3}}=\int 1\,dt \;\Rightarrow\; -3(G-27)^{1/3}=t+C. \] Apply \(G(0)=91\Rightarrow (G-27)^{1/3}=64^{1/3}=4\): \(-3\cdot 4=0+C\Rightarrow C=-12.\)
Thus, \[ -3(G-27)^{1/3}=t-12 \;\Rightarrow\; (G-27)^{1/3}=\frac{12-t}{3} \;\Rightarrow\; \boxed{\,G(t)=27+\Big(\tfrac{12-t}{3}\Big)^3\,},\quad (t<10). \] At \(t=3\): \[ G(3)=27+\Big(\tfrac{12-3}{3}\Big)^3 =27+3^3=27+27=\boxed{54^\circ\text{C}}. \]