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2022 AP Calculus AB FRQ Question 4: Melting Ice Sculpture — Table Data

Full worked solution for every part, with AP scoring notes. See all 6 questions from 2022.

Question 4

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Melting Ice Sculpture — Table Data

Rate & Data from Tables

Hard
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 rr, where r(t)r(t) is measured in centimeters and tt is measured in days. The table above gives selected values of r′(t)r'(t), the rate of change of the radius, over the time interval 0≤t≤120\leq t\leq12.

Values of r′(t)

t (days)0371012
r′(t) (centimeters per day)−6.1−5.0−4.4−3.8−3.5
Part AEasy2 points
Approximate r′′(8.5)r''(8.5) using the average rate of change of r′r' over the interval 7≤t≤107\leq t\leq10. Show the computations that lead to your answer, and indicate units of measure.

Answer

r′′(8.5)≈0.2r''(8.5)\approx0.2 cm/day²
Full Solution & Work

Set up the difference quotient

r′′(8.5)≈r′(10)−r′(7)10−7=−3.8−(−4.4)3=0.63=0.2r''(8.5) \approx \frac{r'(10)-r'(7)}{10-7} = \frac{-3.8-(-4.4)}{3} = \frac{0.6}{3}=0.2

State the units

The units are centimeters per day per day (cm/day²).

AP Scoring — 2 Points

**P1**: The difference quotient, correctly set up and evaluated.
**P2**: Correct units.
Part BMedium2 points
Is there a time tt, 0≤t≤30\leq t\leq3, for which r′(t)=−6r'(t)=-6? Justify your answer.

Answer

Yes, by the Intermediate Value Theorem.
Full Solution & Work

Set up the IVT argument

rr is twice-differentiable, so r′r' is differentiable, hence continuous, on [0,3][0,3].

Compare -6 to the endpoint values

r′(0)=−6.1<−6<−5.0=r′(3)r'(0)=-6.1<-6<-5.0=r'(3)

Conclude

By the **Intermediate Value Theorem**, since r′r' is continuous on [0,3][0,3] and −6-6 lies between r′(0)r'(0) and r′(3)r'(3), there must exist a t∈(0,3)t\in(0,3) with r′(t)=−6r'(t)=-6.

AP Scoring — 2 Points

**P1**: Establishes continuity of r′r' on [0,3][0,3].
**P2**: Correct comparison and conclusion citing the IVT.
Part CMedium2 points
Use a right Riemann sum with the four subintervals indicated in the table to approximate the value of ∫012r′(t) dt\displaystyle\int_0^{12} r'(t)\,dt.

Answer

−51-51
Full Solution & Work

Set up the right Riemann sum

∫012r′(t) dt≈3⋅r′(3)+4⋅r′(7)+3⋅r′(10)+2⋅r′(12)\int_0^{12} r'(t)\,dt \approx 3\cdot r'(3)+4\cdot r'(7)+3\cdot r'(10)+2\cdot r'(12)

Substitute and compute

=3(−5.0)+4(−4.4)+3(−3.8)+2(−3.5)=−15−17.6−11.4−7=−51= 3(-5.0)+4(-4.4)+3(-3.8)+2(-3.5) = -15-17.6-11.4-7=-51

AP Scoring — 2 Points

**P1**: Correct form of the right Riemann sum with correct widths.
**P2**: Correct answer, −51-51.
Part DHard3 points
The height of the cone decreases at a rate of 2 centimeters per day. At time t=3t=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=3t=3 days. (The volume VV of a cone with radius rr and height hh is V=13πr2hV=\dfrac13\pi r^2 h.)

Answer

−70000π3≈−73303.83-\dfrac{70000\pi}{3}\approx -73303.83 cm³/day
Full Solution & Work

Differentiate V with respect to t

dVdt=13π(2rdrdth+r2dhdt)\frac{dV}{dt} = \frac{1}{3}\pi\left(2r\frac{dr}{dt}h+r^2\frac{dh}{dt}\right)

Substitute values at t = 3

r=100r=100, drdt=r′(3)=−5\frac{dr}{dt}=r'(3)=-5 (from the table), h=50h=50, dhdt=−2\frac{dh}{dt}=-2:
dVdt=π3[2(100)(−5)(50)+(100)2(−2)]=π3[−50000−20000]\frac{dV}{dt} = \frac{\pi}{3}\Big[2(100)(-5)(50)+(100)^2(-2)\Big] = \frac{\pi}{3}\big[-50000-20000\big]

Simplify

=π3(−70000)=−70000π3≈−73303.83 cm3/day= \frac{\pi}{3}(-70000) = -\frac{70000\pi}{3} \approx -73303.83 \text{ cm}^3/\text{day}

AP Scoring — 3 Points

**P1**: Correct related-rates differentiation of V=13πr2hV=\frac13\pi r^2 h using the Product Rule.
**P2**: Correct substitution, including using the table value r′(3)=−5r'(3)=-5 for drdt\frac{dr}{dt}.
**P3**: Correct final answer, −70000π3-\frac{70000\pi}{3} (or its decimal equivalent).

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