Related rates

Answer: (B).

The volume of a sphere in terms of its radius is

\[V=\frac{4}{3}\pi r^3 .\]

Both \(V\) and \(r\) are functions of time, so differentiating with respect to \(t\) requires the chain rule:

\[\frac{dV}{dt}=4\pi r^2\,\frac{dr}{dt}.\]

The instant is described by \(r=4\) and \(dr/dt=0.5\), and both values are substituted only now, after the differentiation:

\[\frac{dV}{dt}=4\pi(4)^2(0.5)=32\pi \ \text{cm}^3\!/\text{s}.\]

Where the other options come from.

• (A) differentiates \(r^3\) as though the exponent were 2, producing \(4\pi r\) in place of \(4\pi r^2\).
• (C) differentiates correctly and then omits \(dr/dt\). That omission is the chain rule itself, and it is the single most common error in this archetype.
• (D) substitutes \(r=4\) before differentiating. The number that results is the volume at that instant, not a rate, and its units give it away.

Answer: (D).

Let \(x\) be the eastward distance, \(y\) the northward distance, and \(z\) the distance between the vehicles. The paths are perpendicular, so

\[z^2=x^2+y^2 .\]

Differentiating with respect to \(t\) gives

\[2z\,\frac{dz}{dt}=2x\,\frac{dx}{dt}+2y\,\frac{dy}{dt}.\]

At \(t=2\) the distances are \(x=60\) and \(y=80\), so \(z=100\). Substituting now,

\[\frac{dz}{dt}=\frac{60(30)+80(40)}{100}=\frac{5000}{100}=50 \ \text{mi/h}.\]

Note that 50 exceeds both speeds. It has to: each vehicle contributes to the separation, so the distance grows faster than either one travels. Adding the speeds to get 70 would be correct only for two vehicles moving in opposite directions along one line, where the geometry collapses to a subtraction of coordinates.

Where the other options come from.

• (A) averages the two speeds. Averaging would answer a question about a single vehicle, not about a distance built from two.
• (B) reports the speed of the faster vehicle, on the reasoning that the separation grows at least that fast. The reasoning is sound and the conclusion is a lower bound, not the rate.
• (C) pairs each distance with the other vehicle’s speed. The arithmetic is otherwise identical, which is exactly why the bookkeeping deserves the care.

Answer: (C).

The water forms a cone similar to the tank, so its radius and depth stay in a fixed ratio:

\[\frac{r}{h}=\frac{3}{6}=\frac{1}{2}\qquad\Longrightarrow\qquad r=\frac{h}{2}.\]

That relation holds at every instant, so it may be used before differentiating. Eliminating \(r\) first leaves one variable:

\[V=\frac{1}{3}\pi r^2 h=\frac{1}{3}\pi\left(\frac{h}{2}\right)^{2}h=\frac{\pi h^3}{12}.\]

Now differentiate and substitute the snapshot values \(dV/dt=2\) and \(h=4\):

\[\frac{dV}{dt}=\frac{\pi h^2}{4}\,\frac{dh}{dt}\qquad\Longrightarrow\qquad 2=4\pi\,\frac{dh}{dt}\qquad\Longrightarrow\qquad \frac{dh}{dt}=\frac{1}{2\pi}.\]

Where the other options come from.

• (A) uses \(r=h\), dropping the similar-triangle relation entirely.
• (B) omits the factor \(\tfrac13\) in the volume of a cone.
• (D) holds the radius fixed at 3. A vessel whose radius does not change with depth is a cylinder, and the whole difficulty of the item disappears with it.

Answer: (D).

Let \(x\) be the distance from the base of the lamp to the person and \(s\) the length of the shadow. The lamp and the person cut off similar right triangles, so

\[\frac{6}{s}=\frac{15}{x+s}.\]

Cross-multiplying and collecting,

\[6(x+s)=15s\qquad\Longrightarrow\qquad 6x=9s\qquad\Longrightarrow\qquad s=\frac{2}{3}x.\]

The tip of the shadow sits at distance \(x+s\) from the base, so

\[x+s=\frac{5}{3}x\qquad\Longrightarrow\qquad \frac{d}{dt}(x+s)=\frac{5}{3}\,\frac{dx}{dt}=\frac{5}{3}(4)=\frac{20}{3}\ \text{ft/s}.\]

Notice that \(x\) never appears in the answer. The tip moves at the same rate wherever the person is, which is worth knowing because a question that supplies a distance is often supplying a number the solution does not need.

Where the other options come from.

• (A) inverts the ratio, multiplying by \(\tfrac{6}{15}\) rather than by the \(\tfrac53\) the similar triangles produce. An answer smaller than the person’s own speed should end this option before the arithmetic is checked.
• (B) gives \(ds/dt\), the rate at which the shadow lengthens. The tip carries the person’s own motion as well, so the two rates differ by exactly \(dx/dt\).
• (C) reports the person’s speed.

Answer: (B).

The relation is already given, so no geometry is needed. Differentiate it as a product, since both factors vary with time:

\[P\,\frac{dV}{dt}+V\,\frac{dP}{dt}=0.\]

The instant supplies \(V=20\) and \(dV/dt=4\). It does not supply \(P\), which has to be recovered from the relation itself:

\[P=\frac{200}{20}=10.\]

Substituting all three values,

\[10(4)+20\,\frac{dP}{dt}=0\qquad\Longrightarrow\qquad \frac{dP}{dt}=-2\ \text{atm/min}.\]

Where the other options come from.

• (A) uses 200 in place of \(P\). The constant on the right side of the relation is the product of the two quantities, not one of them, and a missing value has to be recovered before it can be substituted.
• (C) inverts the last division.
• (D) drops the sign. A relation of the form \(PV=c\) forces the two quantities to move in opposite directions, so a positive answer here is wrong before any arithmetic is checked.

Answer: (C).

Write the length as \(f(t)\) and the width as \(g(t)\), so the area is the product \(A=fg\) and

\[\frac{dA}{dt}=f'g+fg'.\]

The instant gives \(f=10\), \(f'=3\), \(g=6\), and \(g'=-2\). The word decreasing is what makes \(g'\) negative, and carrying that sign is the whole content of the item:

\[\frac{dA}{dt}=3(6)+10(-2)=18-20=-2\ \text{cm}^2\!/\text{s}.\]

The area is shrinking even though the length is growing faster than the width is shrinking, because each rate is weighted by the other dimension.

Where the other options come from.

• (A) keeps only the second term, differentiating the width and holding the length fixed.
• (B) multiplies the two rates. The product rule adds two terms, each of which holds one factor fixed, and never multiplies the derivatives.
• (D) ignores the sign on the width, so the two effects reinforce each other instead of opposing.

Answer: (A).

Let \(\theta\) be the angle of elevation and \(h\) the height of the balloon. The horizontal distance is fixed at 150, so

\[\tan\theta=\frac{h}{150}.\]

Differentiating with respect to \(t\),

\[\sec^2\theta\,\frac{d\theta}{dt}=\frac{1}{150}\,\frac{dh}{dt}.\]

At the instant in question \(h=200\), so \(\tan\theta=\tfrac43\) and

\[\sec^2\theta=1+\tan^2\theta=1+\frac{16}{9}=\frac{25}{9}.\]

Therefore

\[\frac{d\theta}{dt}=\frac{12}{150}\cdot\frac{9}{25}=0.0288\ \text{rad/s}.\]

Where the other options come from.

• (B) uses \(\sin^2\theta\) where \(\cos^2\theta\) belongs. The reciprocal of \(\sec^2\theta\) is \(\cos^2\theta\), and here \(\cos\theta=0.6\) rather than \(\sin\theta=0.8\).
• (C) omits \(\sec^2\theta\) entirely, differentiating \(\tan\theta\) as though it were \(\theta\).
• (D) is the correct rate reported in degrees per second. The question asks for radians, and a calculator left in degree mode produces this without any warning.

Answer: (C).

The surface of the water narrows as the level falls, so the width \(w\) at depth \(h\) is not constant. Similar triangles give

\[\frac{w}{h}=\frac{3}{2}\qquad\Longrightarrow\qquad w=\frac{3}{2}h.\]

The cross-section is a triangle of width \(w\) and height \(h\), and the trough is a prism of length 10, so

\[V=10\cdot\frac{1}{2}wh=10\cdot\frac{1}{2}\left(\frac{3}{2}h\right)h=\frac{15}{2}h^2 .\]

Differentiating and substituting \(dV/dt=5\) and \(h=1.2\),

\[\frac{dV}{dt}=15h\,\frac{dh}{dt}\qquad\Longrightarrow\qquad 5=15(1.2)\,\frac{dh}{dt}\qquad\Longrightarrow\qquad \frac{dh}{dt}=\frac{5}{18}=0.278\ \text{ft/min}.\]

Where the other options come from.

• (A) omits the factor \(\tfrac12\) in the area of a triangle.
• (B) treats the cross-section as a rectangle of fixed width 3.
• (D) keeps the triangle but fixes the width at 3, so the surface never narrows toward the vertex.

Answer: (A).

Let \(D\) be the distance from the particle to the origin. Squaring first avoids differentiating a square root:

\[D^2=x^2+y^2=x^2+x,\]

using \(y^2=x\), which holds along the whole curve and so may be substituted before differentiating.

Differentiate with respect to \(t\):

\[2D\,\frac{dD}{dt}=(2x+1)\,\frac{dx}{dt}.\]

At the given instant \(x=4\), so \(D=\sqrt{20}\), and \(dx/dt=3\):

\[\frac{dD}{dt}=\frac{(2(4)+1)(3)}{2\sqrt{20}}=\frac{27}{2\sqrt{20}}=3.019\ \text{units/s}.\]

Where the other options come from.

• (B) adds \(dx/dt\) and \(dy/dt\). A distance is not the sum of the coordinates, and its rate is not the sum of their rates.
• (C) divides by \(D\) rather than by \(2D\), dropping the factor the power rule puts in front of \(D^2\).
• (D) reports \(d(D^2)/dt=27\), the rate at which the square of the distance changes. Squaring to avoid the radical is the right move, and undoing it at the end is the step it costs.

Answer: (D).

Let \(\theta\) be the angle between the beam and the perpendicular from the lighthouse to the shore, and let \(x\) be the distance along the shore from that perpendicular foot to the beam. Then

\[x=3\tan\theta .\]

The rotation rate is given in revolutions, and every derivative of a trigonometric function assumes radians, so convert first:

\[\frac{d\theta}{dt}=4\ \text{rev/min}=8\pi\ \text{rad/min}.\]

Differentiating and using \(\tan\theta=\tfrac43\) at \(x=4\), so that \(\sec^2\theta=\tfrac{25}{9}\),

\[\frac{dx}{dt}=3\sec^2\theta\,\frac{d\theta}{dt}=3\cdot\frac{25}{9}\cdot 8\pi=\frac{200\pi}{3}=209.439\ \text{km/min}.\]

Where the other options come from.

• (A) uses 4 as a rate in radians per minute. One revolution is \(2\pi\) radians, and the factor of \(2\pi\) is the most common oversight on any rotating-beam item.
• (B) omits \(\sec^2\theta\).
• (C) uses \(\sec\theta\) where \(\sec^2\theta\) belongs.

Part (a).

The water in the tank forms a cone similar to the tank itself, so its radius and depth hold a fixed ratio:

\[\frac{r}{h}=\frac{4}{12}=\frac{1}{3}\qquad\Longrightarrow\qquad r=\frac{h}{3}.\]

Substituting into the volume of a cone,

\[V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\left(\frac{h}{3}\right)^{2}h=\frac{\pi h^3}{27}.\]

Part (b).

The volume changes at the rate water enters minus the rate it leaves:

\[\frac{dV}{dt}=6-0.4\sqrt{t}.\]

At \(t=9\),

\[\frac{dV}{dt}=6-0.4(3)=4.8\ \text{ft}^3\!/\text{min}.\]

Differentiating the result of part (a) connects that to the depth:

\[\frac{dV}{dt}=\frac{\pi h^2}{9}\,\frac{dh}{dt}.\]

At the given instant \(h=5\), so

\[4.8=\frac{25\pi}{9}\,\frac{dh}{dt}\qquad\Longrightarrow\qquad \frac{dh}{dt}=\frac{43.2}{25\pi}=0.550\ \text{ft/min}.\]

Part (c).

The depth is increasing. At \(t=9\) water enters at 6 cubic feet per minute and leaves at 1.2, so \(dV/dt=4.8>0\). Since \(dh/dt\) is a positive multiple of \(dV/dt\), the depth is increasing as well.

Part (d).

The level stops rising at the instant the two rates balance:

\[6=0.4\sqrt{t}\qquad\Longrightarrow\qquad \sqrt{t}=15\qquad\Longrightarrow\qquad t=225\ \text{minutes}.\]

Rubric pattern.

Part	Points	Earned for
(a)	1	similar triangles giving \(r=h/3\)
(a)	1	substitution and simplification to \(\pi h^3/27\)
(b)	1	\(dV/dt=6-0.4\sqrt{t}\), or the value 4.8
(b)	1	chain rule: \(dV/dt=\tfrac{\pi h^2}{9}\,dh/dt\)
(b)	1	answer 0.550
(b)	1	units, feet per minute
(c)	1	increasing, with a reason resting on the sign of \(dV/dt\)
(d)	1	the equation \(6=0.4\sqrt{t}\)
(d)	1	\(t=225\) minutes

Where the points go.

• Reporting 4.8 as the answer to part (b) is the most common loss. It is a rate of volume, and the question asks for a rate of depth.
• The units point in part (b) is free and is lost more often than any computational point on the paper.
• In part (c), because water is being pumped in does not earn the point. The tank is also leaking, so the reason has to compare the two rates or cite the sign of their difference.
• Part (d) does not require solving a differential equation. The level stops rising when the net rate is zero, which is an algebraic condition on \(t\) alone.

Part (a).

The pole and the person are both vertical, and the light travels in a straight line from the top of the pole past the person’s head to the tip of the shadow. The two right triangles that result are similar, so corresponding sides are proportional:

\[\frac{6}{s}=\frac{18}{x+s}.\]

Cross-multiplying and collecting terms,

\[6(x+s)=18s\qquad\Longrightarrow\qquad 6x=12s\qquad\Longrightarrow\qquad s=\frac{x}{2}.\]

Part (b).

Differentiating the relation from part (a),

\[\frac{ds}{dt}=\frac{1}{2}\,\frac{dx}{dt}=\frac{1}{2}(5)=\frac{5}{2}\ \text{ft/s}.\]

The shadow lengthens at a constant rate. The relation between \(s\) and \(x\) is linear, so \(ds/dt\) is a fixed multiple of \(dx/dt\), and \(dx/dt\) is constant by hypothesis.

Part (c).

The tip of the shadow sits at distance \(x+s\) from the base of the pole, and part (a) gives

\[x+s=x+\frac{x}{2}=\frac{3}{2}x.\]

Therefore

\[\frac{d}{dt}(x+s)=\frac{3}{2}\,\frac{dx}{dt}=\frac{3}{2}(5)=\frac{15}{2}\ \text{ft/s}.\]

Part (d).

Let \(D\) be the distance from the person to the top of the pole. The person stands on the ground at horizontal distance \(x\), and the top of the pole is 18 feet up, so

\[D^2=x^2+18^2=x^2+324.\]

Differentiating with respect to \(t\),

\[2D\,\frac{dD}{dt}=2x\,\frac{dx}{dt}.\]

At the instant \(x=24\),

\[D=\sqrt{576+324}=\sqrt{900}=30,\]

and the speed at that instant is \(dx/dt=3\), so

\[\frac{dD}{dt}=\frac{x}{D}\,\frac{dx}{dt}=\frac{24(3)}{30}=\frac{12}{5}\ \text{ft/s}.\]

Rubric pattern.

Part	Points	Earned for
(a)	1	a correct proportion from similar triangles
(a)	1	algebra reaching \(s=x/2\)
(b)	1	\(ds/dt=5/2\) feet per second
(b)	1	constant, with a reason naming the linear relation
(c)	1	identifying the tip as \(x+s\)
(c)	1	answer \(15/2\) feet per second
(d)	1	\(D^2=x^2+324\)
(d)	1	implicit differentiation with respect to \(t\)
(d)	1	answer \(12/5\) feet per second

Where the points go.

• Part (c) answered with \(ds/dt\) earns nothing. The shadow lengthens at \(5/2\) while its tip moves at \(15/2\), and the difference is exactly the person’s own speed. Reading the question for which quantity is moving is the whole task.
• In part (d), 18 is a constant, not a variable. Differentiating it as though the pole were growing is a fast way to lose the second point.
• Part (b) asks two things, a value and a judgment with a reason. A number alone earns one point of the two.
• The change of speed in part (d) applies only to that instant. Nothing computed in parts (b) and (c) needs revisiting, and a solution that recomputes them has misread the question rather than made an error.