AP Calculus BC 4.5 Solving Related Rates Problems Exam Style Questions - FRQ
Question
(a) Topic-3.6 Calculating Higher-Order Derivatives
(b) Topic-2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
(c) Topic-6.3 Riemann Sums, Summation Notation, and Definite Integral Notation
(d) Topic-4.5 Solving Related Rates Problems
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 $0 \leq t \leq 12$.
\(\text{(a)}\) Approximate $r”(8.5)$ using the average rate of change of $r’$ over the interval $7 \leq t \leq 10$. Show the computations that lead to your answer, and indicate units of measure.
\(\text{(b)}\) Is there a time $t$, $0 \leq t \leq 3$, for which $r'(t) = -6$? Justify your answer.
\(\text{(c)}\) Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of
\(
\int_{0}^{12} r'(t) \, dt.
\)
\(\text{(d)}\) The height of the cone decreases at a rate of $2$ centimeters per day. At time $t = 3$ days, the radius is $100$ centimeters and the height is $50$centimeters. Find the rate of change of the volume of the cone with respect to time, in cubic centimeters per day, at time $t = 3$ days. (The volume $V$ of a cone with radius $r$ and height $h$ is $V = \frac{1}{3}\pi r^2 h$.)
▶️Answer/Explanation
\(\textbf{4(a)}\)
\(
r^{”}(8.5) \approx \frac{r'(10) – r'(7)}{10 – 7} = \frac{-3.8 – (-4.4)}{10 – 7} = \frac{0.6}{3} \, \frac{\text{cm}}{\text{day}} = 0.2 \, \frac{\text{cm}}{\text{day}^2}
\)
\(\textbf{4(b)}\)
Since $r$ is twice differentiable, then $r’$ is differentiable and, hence, $r’$ is also continuous.
So, by the Intermediate Value Theorem,
$r’$ takes on all values between $r'(0) = -6.1$ and $r'(3) = -5.0$.
Hence, since $-6.1 = r'(0) < -6 < -5 = r'(3)$, there is a time on $[0, 3]$ for which $r'(t) = -6$.
\(\textbf{4(c)}\)
Right Riemann sum with 4 subintervals:
\(
\int_{0}^{12} r'(t) \, dt \approx (3 – 0)r'(3) + (7 – 3)r'(7) + (10 – 7)r'(10) + (12 – 10)r'(12)
\)
\(
= 3(-5) + 4(-4.4) + 3(-3.8) + 2(-3.5)
\)
\(
= -51 \, \text{cm}
\)
\(\textbf{4(d)}\)
\(
\frac{dh}{dt} = -2, \quad \text{when } t = 3, \, r = 100, \, h = 50. \quad \text{Find } \frac{dV}{dt} \text{ when } t = 3.
\)
\(
V = \frac{1}{3}\pi r^2 h
\)
\(
\frac{dV}{dt} = \frac{1}{3}\pi \left( r^2 \frac{dh}{dt} + h 2r \frac{dr}{dt} \right) \quad \text{(using the product rule).}
\)
\(
\text{Note: } \frac{dr}{dt} \big|_{t=3} = -5 \text{ (from the table).}
\)
\(
\frac{dV}{dt} = \frac{1}{3}\pi \left( (100)^2(-2) + (50)(2)(100)(-5) \right)
\)
\(
= \frac{1}{3}\pi \left( -20000 – 50000 \right) = \frac{-70000\pi}{3}
\)