AP Calculus AB 4.5 Solving Related Rates Problems - MCQs - Exam Style Questions
Calc-Ok Question
The right circular cone shown has a radius of \(r\) meters and a constant height of \(6\) meters. The lateral surface area of the cone, in square meters, is \(S(r)=\pi\sqrt{36r^{2}+r^{4}}\). At the instant when \(r=0.75\) meter, the radius is increasing at \(dr/dt=1.15\) meters per minute. At what rate is the lateral surface area changing, in square meters per minute, at that instant?
(A) \(14.247\)
(B) \(16.384\)
(C) \(19.288\)
(D) \(22.182\)
▶️ Answer/Explanation
Chain rule: \(\dfrac{dS}{dt}=\dfrac{dS}{dr}\dfrac{dr}{dt}\).
\(S'(r)=\pi\cdot\dfrac{1}{2}(36r^{2}+r^{4})^{-1/2}\cdot(72r+4r^{3})\).
Simplify: \(S'(r)=\dfrac{\pi}{2}(36r^{2}+r^{4})^{-1/2}(72r+4r^{3})\).
Evaluate at \(r=0.75\) and multiply by \(dr/dt=1.15\).
Compute → \(\dfrac{dS}{dt}\approx 22.182\ \text{m}^{2}/\text{min}\).
✅ Answer: (D)
Calc-Ok Question
A truck is traveling south toward an intersection while a car travels east away from it. At an instant the truck is \(50\) m from the intersection and moving toward it at \(15\) m/s; the car is \(120\) m from the intersection and moving away at \(25\) m/s. At that moment, at what rate is the distance \(h\) between them changing?
(A) \(1.308\)
(B) \(10.000\)
(C) \(17.308\)
(D) \(28.846\)
▶️ Answer/Explanation
With \(x=120\) (car), \(y=50\) (truck), \(h^{2}=x^{2}+y^{2}\).
\(dx/dt=+25\), \(dy/dt=-15\).
\(2h\,dh/dt=2x\,dx/dt+2y\,dy/dt\Rightarrow dh/dt=\dfrac{x\,dx/dt+y\,dy/dt}{h}\).
\(h=\sqrt{120^{2}+50^{2}}=130\).
\(dh/dt=(120\cdot25+50\cdot(-15))/130=2250/130\approx 17.308\ \text{m/s}.\)
✅ Answer: (C)
\(dx/dt=+25\), \(dy/dt=-15\).
\(2h\,dh/dt=2x\,dx/dt+2y\,dy/dt\Rightarrow dh/dt=\dfrac{x\,dx/dt+y\,dy/dt}{h}\).
\(h=\sqrt{120^{2}+50^{2}}=130\).
\(dh/dt=(120\cdot25+50\cdot(-15))/130=2250/130\approx 17.308\ \text{m/s}.\)
✅ Answer: (C)