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AP Calculus BC 6.3 Riemann Sums, Summation Notation, and Definite Integral Notation - FRQs - Exam Style Questions

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No-Calc Question

\(t\) (years)235710
\(H(t)\) (meters)1.5261115
The height of a tree at time \(t\) is given by a twice-differentiable function \(H\), where \(H(t)\) is measured in meters and \(t\) is measured in years. Selected values of \(H(t)\) are given in the table above.
(a) Use the data in the table to estimate \(H'(6)\). Using correct units, interpret the meaning of \(H'(6)\) in the context of the problem.
(b) Explain why there must be at least one time \(t\), for \(2<t<10\), such that \(H'(t)=2\).
(c) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the average height of the tree over the time interval \(2\le t\le 10\).
(d) The height of the tree, in meters, can also be modeled by the function \(G\), given by \(G(x)=\dfrac{100x}{1+x}\), where \(x\) is the diameter of the base of the tree, in meters. When the tree is \(50\) meters tall, the diameter of the base of the tree is increasing at a rate of \(0.03\) meter per year. According to this model, what is the rate of change of the height of the tree with respect to time, in meters per year, at the time when the tree is \(50\) meters tall?

Most-appropriate topic codes (CED):

• TOPIC 4.3: Rates of Change in Applied Contexts Other Than Motion — parts (a), (b)
• TOPIC 6.3: Riemann Sums & Trapezoidal Rule — part (c)
• TOPIC 4.4: Introduction to Related Rates — part (d) setup
• TOPIC 4.5: Solving Related Rates Problems (Chain Rule) — part (d) computation

▶️ Answer/Explanation

(a) Estimate \(H'(6)\) and interpret
Use a symmetric secant around \(t=6\): points \(t=5\) and \(t=7\).
\[ H'(6)\approx\frac{H(7)-H(5)}{7-5}=\frac{11-6}{2}=\boxed{\tfrac{5}{2}}. \]
Interpretation: At \(t=6\) years, the tree’s height is increasing at about \(\boxed{2.5\ \text{meters per year}}\).

(b) Show a time \(t\) with \(H'(t)=2\) for \(2<t<10\)
On \([3,5]\), the average rate of change is \(\dfrac{H(5)-H(3)}{5-3}=\dfrac{6-2}{2}=2\).
\(H\) is continuous on \([3,5]\) and differentiable on \((3,5)\).
By the Mean Value Theorem, there exists \(c\in(3,5)\) such that \(\boxed{H'(c)=2}\).

(c) Average height on \(2\le t\le 10\) via trapezoids
Subintervals: \([2,3], [3,5], [5,7], [7,10]\) with widths \(1,2,2,3\).
Trapezoidal estimate for the integral: \[ \int_{2}^{10} H(t)\,dt \approx 1\cdot\frac{1.5+2}{2} +2\cdot\frac{2+6}{2} +2\cdot\frac{6+11}{2} +3\cdot\frac{11+15}{2} = 65.75=\frac{263}{4}. \]
Average height \(=\dfrac{1}{10-2}\int_{2}^{10}H(t)\,dt =\dfrac{1}{8}\cdot\frac{263}{4} =\boxed{\tfrac{263}{32}\ \text{meters}}\;(\approx 8.219).\)

(d) Rate of change of height when the tree is 50 m tall
Model: \(G(x)=\dfrac{100x}{1+x}\), with \(x’=\dfrac{dx}{dt}=0.03\ \text{m/yr}\).
First find \(x\) when \(G(x)=50\): \[ \frac{100x}{1+x}=50 \;\Rightarrow\; 100x=50(1+x)\;\Rightarrow\; x=\boxed{1}. \]
Differentiate \(G\): \(G'(x)=\dfrac{100(1+x)-100x}{(1+x)^2}=\dfrac{100}{(1+x)^2}\).
Chain rule: \[ \frac{dG}{dt}=G'(x)\,x’ =\frac{100}{(1+1)^2}\cdot 0.03 =25\cdot 0.03 =\boxed{\tfrac{3}{4}\ \text{m/yr}}. \]
So when the tree is \(50\) m tall, its height is increasing at \(\tfrac{3}{4}\) meter per year.

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