Drill 1 · Math · Related Rates
AP Calculus AB: Related Rates (Drill 1) is a Math practice drill covering Related Rates. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice setting up and solving related rates problems involving expanding circles, expanding spheres, a sliding ladder, and a filling conical tank. These AP Calculus AB problems require applying the chain rule to equations relating two or more changing quantities.
Question 1. The radius of a circle is increasing at a constant rate of 3 cm/sec. At the instant when the radius is 5 cm, what is the rate at which the area of the circle is increasing?
Explanation: Choice B is correct. The area of a circle is \( A = \pi r^2 \). Differentiating with respect to time: \( \dfrac{dA}{dt} = 2\pi r \cdot \dfrac{dr}{dt} \). Substituting \( r = 5 \) and \( \dfrac{dr}{dt} = 3 \): \( \dfrac{dA}{dt} = 2\pi(5)(3) = 30\pi \) cm²/sec. Choice A is off because it uses \( \pi r \cdot \dfrac{dr}{dt} \) instead of \( 2\pi r \cdot \dfrac{dr}{dt} \), omitting the factor of 2 from the power rule. Choice C doesn't fit: it multiplies the area formula itself by the rate, \( \pi r^2 \cdot \dfrac{dr}{dt} = \pi(25)(3) = 75\pi \), rather than differentiating first. Choice D is wrong because \( \pi\left(\dfrac{dr}{dt}\right)^2 = 9\pi \) results from squaring the rate rather than applying the chain rule.
Question 2. The volume of a sphere is increasing at a rate of \( 16\pi \) in³/sec. At the instant when the radius is 2 inches, what is the rate at which the radius is increasing?
Explanation: Choice A is correct. Volume of a sphere: \( V = \dfrac{4}{3}\pi r^3 \). Differentiating: \( \dfrac{dV}{dt} = 4\pi r^2 \cdot \dfrac{dr}{dt} \). Substituting \( \dfrac{dV}{dt} = 16\pi \) and \( r = 2 \): \( 16\pi = 4\pi(4)\dfrac{dr}{dt} = 16\pi\dfrac{dr}{dt} \), so \( \dfrac{dr}{dt} = 1 \) in/sec. Choice B falls short because it results from using \( r \) instead of \( r^2 \) in the derivative, \( \dfrac{dV}{dt} = 4\pi r \cdot \dfrac{dr}{dt} \) gives \( 16\pi = 8\pi\dfrac{dr}{dt} \Rightarrow \dfrac{dr}{dt} = 2 \). Choice C misses the mark: it results from dropping the \( r^2 \) factor entirely and computing \( \dfrac{dr}{dt} = \dfrac{16\pi}{4\pi} = 4 \). Choice D is off because it results from treating \( \dfrac{dr}{dt} = \dfrac{dV/dt}{V} \), dividing the rate of volume change by the volume formula rather than its derivative.
Question 3. A 10-foot ladder leans against a vertical wall. The base of the ladder slides away from the wall at 2 ft/sec. At the instant when the base is 6 feet from the wall, at what rate is the top of the ladder sliding down the wall?
Explanation: Choice A is correct. Let \( x \) = horizontal distance from wall to base, \( y \) = height of the top. The constraint is \( x^2 + y^2 = 100 \). When \( x = 6 \): \( y = \sqrt{100-36} = 8 \). Differentiating with respect to time: \( 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0 \). Substituting \( x = 6 \), \( y = 8 \), \( \dfrac{dx}{dt} = 2 \): \( 12(2) + 16\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{24}{16} = -\dfrac{3}{2} \). The top slides down at \( \dfrac{3}{2} \) ft/sec. Choice B doesn't fit: it results from inverting the ratio, writing \( \dfrac{dy}{dt} = -\dfrac{y}{x}\cdot\dfrac{dx}{dt} = -\dfrac{8}{6}(2) = -\dfrac{8}{3} \) instead of the correct \( -\dfrac{x}{y}\cdot\dfrac{dx}{dt} \). Choice C is wrong because it results from correctly writing \( \dfrac{dy}{dt} = -\dfrac{x}{y}\cdot\dfrac{dx}{dt} = -\dfrac{6}{8}(2) = -\dfrac{3}{2} \) but then erroneously dividing by 2 a second time, perhaps from confusing the factor of 2 that cancels in the differentiation step with an additional division, yielding \( \dfrac{3}{4} \). Choice D doesn't work because it results from swapping \( x \) and \( y \) in the substitution, using \( x = 8 \) and \( y = 6 \), which gives \( 16(2) + 12\dfrac{dy}{dt} = 0 \Rightarrow \dfrac{dy}{dt} = -\dfrac{32}{12} = -\dfrac{8}{3} \), and then halving by a further arithmetic error to get \( \dfrac{4}{3} \).
Question 4. Water is poured into an inverted conical tank at 2 ft³/min. The tank has a height of 10 ft and a base radius of 5 ft. At the instant when the water is 4 ft deep, at what rate is the water level rising?
Explanation: Choice A is correct. By similar triangles, \( \dfrac{r}{h} = \dfrac{5}{10} = \dfrac{1}{2} \), so \( r = \dfrac{h}{2} \). Substituting into the cone volume formula: \( V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi\left(\dfrac{h}{2}\right)^2 h = \dfrac{\pi h^3}{12} \). Differentiating: \( \dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\cdot\dfrac{dh}{dt} \). Substituting \( \dfrac{dV}{dt} = 2 \) and \( h = 4 \): \( 2 = \dfrac{16\pi}{4}\cdot\dfrac{dh}{dt} = 4\pi\cdot\dfrac{dh}{dt} \), so \( \dfrac{dh}{dt} = \dfrac{1}{2\pi} \) ft/min. Choice B doesn't fit: it results from correctly applying similar triangles and arriving at \( V = \dfrac{\pi h^3}{12} \), but then differentiating as \( \dfrac{dV}{dt} = \dfrac{\pi h^2}{2}\cdot\dfrac{dh}{dt} \), a factor-of-2 error in the power rule on \( h^3 \), giving \( 2 = 8\pi\cdot\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{1}{4\pi} \). Choice C is wrong because it results from failing to apply similar triangles and instead treating \( r = 5 \) as a fixed constant, then using the cylinder-like formula \( V = \pi(25)h \) and differentiating to get \( \dfrac{dV}{dt} = 25\pi\cdot\dfrac{dh}{dt} \), giving \( \dfrac{dh}{dt} = \dfrac{2}{25\pi} \). Choice D doesn't work because it results from correctly applying similar triangles to get \( V = \dfrac{\pi h^3}{12} \) but substituting \( h = 8 \) (the diameter of the water surface) instead of \( h = 4 \) (the depth), giving \( 2 = \dfrac{\pi(64)}{4}\cdot\dfrac{dh}{dt} = 16\pi\cdot\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt} = \dfrac{1}{8\pi} \).
Question 5. A streetlight is mounted at the top of a 15-foot pole. A person 6 feet tall walks away from the base of the pole at 4 ft/sec along a straight path. At what rate is the tip of the person's shadow moving away from the base of the pole?
Explanation: Choice A is correct. Let \( x \) = distance from the pole to the person, and \( s \) = length of the shadow. By similar triangles: \( \dfrac{15}{x+s} = \dfrac{6}{s} \Rightarrow 15s = 6x + 6s \Rightarrow 9s = 6x \Rightarrow s = \dfrac{2x}{3} \). The tip of the shadow is at distance \( x + s = x + \dfrac{2x}{3} = \dfrac{5x}{3} \) from the pole. Differentiating: \( \dfrac{d}{dt}(x+s) = \dfrac{5}{3}\cdot\dfrac{dx}{dt} = \dfrac{5}{3}(4) = \dfrac{20}{3} \) ft/sec. Choice B is wrong because \( \dfrac{ds}{dt} = \dfrac{2}{3}(4) = \dfrac{8}{3} \) is the rate at which the shadow length grows, not the rate at which the tip moves from the pole, the most common error on this problem type. Choice C doesn't work because it results from setting up the similar triangles as \( \dfrac{15}{x} = \dfrac{6}{s} \), placing only the person's distance in the denominator on the left rather than the full tip distance \( x + s \), giving \( s = \dfrac{6x}{15} = \dfrac{2x}{5} \), so the tip is at \( x + \dfrac{2x}{5} = \dfrac{7x}{5} \) from the pole, and \( \dfrac{d}{dt}\!\left(\dfrac{7x}{5}\right) = \dfrac{7}{5}(4) = \dfrac{28}{5} \). Choice D falls short because it simply restates the person's walking speed, reflecting a failure to recognize that the shadow tip moves faster than the person due to the geometry of the projection.