AP Calculus BC 4.4 Introduction to Related Rates - FRQs - Exam Style Questions

(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.