AP Calculus BC 4.5 Solving Related Rates Problems - FRQs - Exam Style Questions
No-Calc Question
| \(t\) (days) | 0 | 3 | 7 | 10 | 12 |
|---|---|---|---|---|---|
| \(r'(t)\) (centimeters per day) | \(-6.1\) | \(-5.0\) | \(-4.4\) | \(-3.8\) | \(-3.5\) |
An ice sculpture melts in such a way that it can be modeled as a cone that maintains a conical shape as it decreases in size. The radius of the base of the cone is given by a twice-differentiable function \(r\), where \(r(t)\) is measured in centimeters and \(t\) is measured in days. The table above gives selected values of \(r'(t)\), the rate of change of the radius, over the time interval \(0 \le t \le 12\).
Most-appropriate topic codes (CED):
• TOPIC 1.16: Working with the Intermediate Value Theorem (existence on an interval) — part (b)
• TOPIC 6.2: Approximating Areas with Riemann Sums (from tabular data) — part (c)
• TOPIC 4.5: Solving Related Rates Problems (product/chain rule in context) — part (d)
▶️ Answer/Explanation
(a)
\[ r”(8.5)\approx \frac{r'(10)-r'(7)}{10-7} = \frac{-3.8-(-4.4)}{3} = \frac{0.6}{3} = 0.2\ \text{cm/day}^2. \]
(b)
Since \(r\) is twice-differentiable, \(r’\) is continuous on \([0,3]\). We have \(r'(0)=-6.1\) and \(r'(3)=-5.0\), and \(-6\) lies between \(-6.1\) and \(-5.0\). By the Intermediate Value Theorem, there exists \(t\in(0,3)\) such that \(r'(t)=-6\).
(c)
Right-endpoint Riemann sum on \([0,3],[3,7],[7,10],[10,12]\): \[ \int_{0}^{12} r'(t)\,dt \approx 3\times(-5.0)+4\times(-4.4)+3\times(-3.8)+2\times(-3.5) = -15-17.6-11.4-7 = -51\ (\text{cm}). \]
(d)
\(V=\tfrac13\pi r^2 h\Rightarrow \displaystyle \frac{dV}{dt}=\frac13\pi\!\left(2r h\,\frac{dr}{dt}+r^2\frac{dh}{dt}\right).\)
At \(t=3\): \(r=100\ \text{cm},\ h=50\ \text{cm},\ \frac{dr}{dt}=r'(3)=-5.0\ \text{cm/day},\ \frac{dh}{dt}=-2\ \text{cm/day}\). \[ \frac{dV}{dt} =\frac13\pi\!\left(2(100)(50)(-5)+100^2(-2)\right) =\frac13\pi\,\big(-50{,}000-20{,}000\big) =-\frac{70{,}000\pi}{3}\ \text{cm}^3/\text{day}. \]