AP Calculus AB Unit 4: Contextual Applications of Differentiation — Free Review Games.

dV/dt = 4pi*r^2 * dr/dt = 4pi(25)(2) = 200pi cm^3/s.

The linearization (tangent line approximation) is L(x) = f(a) + f'(a)(x-a), which approximates f near x = a.

L'Hopital's rule applies to limits of the form 0/0 or infinity/infinity, allowing you to differentiate numerator and denominator separately.

This is \(\frac{0}{0}\) form. L'Hopital: \(\lim_{x \to 0} \frac{\cos(x)}{1} = \cos(0) = 1\).

The derivative \(\frac{dy}{dx}\) represents the rate of change of \(y\) with respect to \(x\), so its units are meters/second (velocity).

\(x^2 + y^2 = 169\). When \(x=5\), \(y=12\). Differentiating: \(2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\). \(2(5)(2) + 2(12)\frac{dy}{dt} = 0\), \(\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}\) ft/s.

\(f(x) = \sqrt{x}\), \(f'(x) = \frac{1}{2\sqrt{x}}\). At \(a=4\): \(L(x) = 2 + \frac{1}{4}(x-4)\). \(L(4.1) = 2 + \frac{0.1}{4} = 2.025\).

Rewrite as \(\frac{x}{e^x}\) (\(\frac{\infty}{\infty}\) form). L'Hopital: \(\frac{1}{e^x} \to 0\) as \(x \to \infty\).

Since r = h/2: V = pi(h/2)^2*h/3 = pi*h^3/12. dV/dt = pi*h^2/4 * dh/dt. 3 = pi(16)/4 * dh/dt. dh/dt = 3/(4pi) m/min.

A positive second derivative means the first derivative is increasing, which geometrically means the graph is concave up (curves upward).

The particle changes direction when v(t) = 0 and velocity changes sign. t^2 - 4t + 3 = (t-1)(t-3) = 0 at t = 1, 3. Velocity changes sign at both.

After \(2h\): \(x = 160\), \(y = 120\), \(d = 200\). \(d^2 = x^2 + y^2\). \(2d\frac{dd}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}\). \(400\frac{dd}{dt} = 160(80) + 120(60) = 20000\). \(\frac{dd}{dt} = 100\) mph.

\(\frac{0}{0}\) form. First application: \(\frac{e^x - 1}{2x}\), still \(\frac{0}{0}\). Second application: \(\frac{e^x}{2} = \frac{1}{2}\) as \(x \to 0\).

V = pi*r^2*h = 1000, so h = 1000/(pi*r^2). S = 2pi*r^2 + 2pi*r*h = 2pi*r^2 + 2000/r. dS/dr = 4pi*r - 2000/r^2.

dV = 4pi*r^2 * dr = 4pi(100)(0.5) = 200pi cm^3. This is the differential approximation of the maximum error.

Related rates problems require differentiating an equation implicitly with respect to time so that rates of change like \(\frac{dr}{dt}\) appear, since the variables themselves are functions of time even though time is not written explicitly. Integrating would undo differentiation and does not connect rates of change, so 'Integration with respect to time' is wrong for this purpose. On the exam, always look for an equation linking quantities, then differentiate both sides with respect to \(t\) using the chain rule.

The differential \(dy = f'(x)\,dx\) uses the tangent line's slope to approximate the true change \(\Delta y\) in the function, which is accurate only for small \(dx\). It is not the exact change in \(y\) because the curve bends away from the tangent line, so 'The exact change in \(y\) for any \(dx\)' overstates its accuracy. Students should remember that differentials are linear approximations, most reliable near the point of tangency.

L'Hopital's rule applies directly to limits of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) because these are the two forms the rule's derivation covers. Forms like \(\infty - \infty\) must first be algebraically manipulated into a quotient before L'Hopital's rule can be used, so applying it 'without rewriting' is invalid. A key exam skill is recognizing when algebraic manipulation is needed before L'Hopital's rule becomes applicable.

Differentiating \(V = \frac{4}{3}\pi r^3\) with respect to \(t\) using the chain rule gives \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\), since the derivative of \(r^3\) with respect to \(r\) is \(3r^2\) and the \(\frac{1}{3}\) cancels with the \(3\). The choice \(\frac{4}{3}\pi r^3 \frac{dr}{dt}\) is wrong because it forgets to differentiate \(r^3\) at all and just multiplies the original volume formula by \(\frac{dr}{dt}\). Always differentiate the geometric formula fully before introducing the rate variables.

Because the tangent line matches the function's value and slope only at the point of tangency, the approximation error grows as you move away, so accuracy is best for points very close to that point. Estimating far from the point of tangency introduces larger error since the curve can bend significantly away from the line, making that choice incorrect. This principle underlies why linearization is only trusted for small perturbations \(\Delta x\).

The notation \(\frac{d\theta}{dt}\) specifically denotes how an angle \(\theta\) changes as time \(t\) progresses, which is standard in problems involving angles of elevation or rotating objects. It is not a distance rate, so 'The rate at which a distance changes over time' misidentifies the variable being differentiated. Recognizing variable names and their units is essential for correctly setting up related rates equations.

L'Hopital's rule requires that \(f\) and \(g\) be differentiable in an interval around \(c\) (except possibly at \(c\)) and that the original limit produces an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). The requirement that 'the limit must exist and be finite before applying the rule' is backwards, since the rule is used precisely when the limit's value is unknown. Students must verify the indeterminate form first, otherwise applying L'Hopital's rule can produce an incorrect answer.

A negative second derivative means the slope \(f'(x)\) is decreasing, which geometrically corresponds to the curve bending downward, known as concave down. Concave up would require \(f''(x) > 0\), so that choice contradicts the given condition. This concavity test is often paired with linearization, since a concave down function lies below its tangent lines.

Since the limit is \(\frac{0}{0}\), L'Hopital's rule gives \(\lim_{x \to 0} \frac{\sin x}{1} = \sin(0) = 0\) by differentiating numerator and denominator separately. The answer is not \(1\), which is actually the value of \(\lim_{x\to0}\frac{\sin x}{x}\), a different but commonly confused limit. Careful differentiation of numerator and denominator independently, not as a quotient, is essential when applying L'Hopital's rule.

Linearization is defined as the tangent line to the curve at \(x = a\), which uses the point \((a, f(a))\) and slope \(f'(a)\) to model the function locally. A secant line, by contrast, connects two distinct points on the curve and generally has a different slope than the tangent, making that choice incorrect. Remembering that linearization is literally the tangent line helps connect the algebra of \(L(x) = f(a) + f'(a)(x-a)\) to its geometric meaning.

As the ladder slides, both the vertical height of the top and the horizontal distance of the base change over time, while the ladder's length stays fixed as it moves. The length of the ladder does not change, so treating it as the only time-dependent quantity is incorrect. Identifying which quantities are constant versus changing is the first key step in setting up any related rates equation.

Since the original limit is \(\frac{\infty}{\infty}\), L'Hopital's rule differentiates numerator and denominator separately, turning \(x\) into \(1\) and \(e^x\) into \(e^x\), giving \(\lim_{x\to\infty}\frac{1}{e^x}\). The option \(\frac{e^x}{1}\) incorrectly swaps which function was differentiated, which is a common algebra slip. Always differentiate the top and bottom independently rather than treating the fraction as a single quotient rule expression.

Because volume and radius are linked through the sphere volume formula, knowing \(\frac{dV}{dt}\) allows solving for the related rate \(\frac{dr}{dt}\) once a specific radius is known. The 'total volume at a specific time' is not itself a rate, so it does not match what related rates problems ask for. Related rates problems always pair one known rate with a request to find a connected unknown rate.

As \(x \to 0^+\), \(x \to 0\) while \(\ln x \to -\infty\), so the product has the indeterminate form \(0 \cdot \infty\), which must be rewritten as a quotient before applying L'Hopital's rule. It is not already \(\frac{0}{0}\), since that form applies only to quotients, not products, so that choice misclassifies the expression. Recognizing \(0\cdot\infty\) forms and converting them to \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) by rewriting as a fraction is a critical preliminary step.

The derivative \(\frac{ds}{dt}\) gives the instantaneous rate of change of position with respect to time, which by definition is velocity, measured in meters per second. Acceleration would instead require a second derivative, \(\frac{d^2s}{dt^2}\), so that choice describes a different quantity. This basic derivative-as-rate interpretation underlies every related rates problem in this unit.

Shadow-length problems rely on similar triangles formed by the streetlight, person, and shadow tip, giving a proportional equation between distances that can then be differentiated with respect to time, making this a classic related rates setup. It has nothing to do with estimating square roots via tangent lines, so the linearization choice is incorrect. Recognizing similar-triangle geometry is a key pattern-matching skill for related rates word problems.

Since \(f(x) = e^x\) has \(f(0) = 1\) and \(f'(0) = e^0 = 1\), the linearization is \(L(x) = 1 + 1\cdot(x - 0) = 1 + x\), matching the tangent line at \(x=0\). The choice \(x\) ignores the function's value at \(0\), which is \(1\), not \(0\), so it fails to match \(f(0)\). This standard linearization \(e^x \approx 1+x\) for small \(x\) is worth memorizing for quick estimates.

Differentiating \(V = s^3\) with respect to time using the chain rule gives \(\frac{dV}{dt} = 3s^2\frac{ds}{dt}\), since the power rule brings down the exponent \(3\) and reduces the power by one. The option \(s^2\frac{ds}{dt}\) omits the necessary factor of \(3\) from differentiating \(s^3\), a common algebra error. This same power-rule-plus-chain-rule pattern applies to nearly every related rates problem involving a geometric power formula.

This formula is precisely the tangent line, or linear, approximation, which estimates function values near \(a\) using the point-slope form based on \(f(a)\) and \(f'(a)\). It is not the Mean Value Theorem, which instead guarantees the existence of a point where instantaneous rate equals average rate over an interval, so that choice describes a different theorem entirely. This approximation formula is the algebraic backbone of every linearization question on the AP exam.

Equations like \(x^2 + y^2 = c^2\) must be differentiated implicitly with respect to time, treating \(x\) and \(y\) as functions of \(t\), to correctly produce terms like \(2x\frac{dx}{dt}\). Substituting numerical values before differentiating is a common student mistake because it eliminates the variable needed to apply the chain rule correctly. Always differentiate symbolically first and substitute known numeric values only afterward.

Using \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\), substituting \(r = 4\) and \(\frac{dr}{dt} = 3\) gives \(4\pi (16)(3) = 192\pi\) cm\(^3\)/min. The option \(48\pi\) mistakenly uses \(r\) instead of \(r^2\) in the formula, undercounting the surface area factor. Always substitute known values only after differentiating the volume formula with respect to time.

Using similar triangles, \(\frac{10}{x} = \frac{2}{x - p}\) where \(x\) is the distance from the light to the shadow's tip and \(p\) is the person's distance, simplifying to \(x = 1.25p\), so differentiating gives \(\frac{dx}{dt} = 1.25 \frac{dp}{dt} = 1.25(1.5) = 1.875\) m/s. The choice \(1.5\) m/s is just the person's own walking speed and ignores the geometric scaling factor from similar triangles. This problem type highlights that the shadow's tip moves faster than the person due to the light's greater height.

Since \(f(0) = \sin(0) = 0\) and \(f'(0) = \cos(0) = 1\), the linearization is \(L(x) = 0 + 1\cdot x = x\), so \(L(0.1) = 0.1\), which closely approximates \(\sin(0.1) \approx 0.0998\). The choice \(0\) incorrectly ignores the slope term entirely and only uses the function's value at the base point. The small-angle approximation \(\sin x \approx x\) near \(x=0\) is a direct and useful application of linearization.

Applying L'Hopital's rule three times to the \(\frac{0}{0}\) form eventually yields \(\lim_{x\to0}\frac{2\sec^2x\tan x}{6} = \frac{0}{6}\) after simplification is checked more carefully with a Taylor series, giving the correct limit of \(\frac{1}{3}\). The answer \(0\) incorrectly stops the process too early, mistaking an intermediate indeterminate stage for the final answer. Repeated applications of L'Hopital's rule are sometimes necessary until the indeterminate form resolves into a determinate limit.

From \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\), solving for \(\frac{dr}{dt} = \frac{100}{4\pi(25)} = \frac{100}{100\pi} = \frac{1}{\pi}\) cm/s. The choice \(20\) cm/s comes from dividing \(100\) by \(5\) directly without accounting for the correct \(4\pi r^2\) factor from the volume derivative. Solving related rates for the unknown rate requires isolating it algebraically after differentiating, not dividing raw given values.

Since \(V = 8 \cdot 5 \cdot h = 40h\), differentiating gives \(\frac{dV}{dt} = 40\frac{dh}{dt}\), so \(\frac{dh}{dt} = \frac{2}{40} = 0.05\) m/min. The choice \(2\) m/min incorrectly equates the water level's rate with the volume rate, ignoring the fixed base area that scales the relationship. Constant cross-sectional area problems simplify related rates because the area factor stays fixed throughout the calculation.

Taking the natural log gives \(x\ln\left(1+\frac{3}{x}\right)\), a \(0\cdot\infty\) form rewritten as \(\frac{\ln(1+3/x)}{1/x}\), a \(\frac{0}{0}\) form to which L'Hopital's rule applies, yielding a limit of \(3\) for the log, so the original limit is \(e^3\). The choice \(e\) mistakenly treats the exponent's coefficient as \(1\) instead of correctly carrying through the factor of \(3\). This \(1^\infty\) pattern always requires logarithmic transformation before L'Hopital's rule can be applied directly.

With \(L^2 = 60^2 + x^2\), at \(x=80\), \(L = 100\), and differentiating gives \(2L\frac{dL}{dt} = 2x\frac{dx}{dt}\), so \(\frac{dL}{dt} = \frac{80(5)}{100} = 4\) ft/s. The choice \(5\) ft/s wrongly assumes the string length changes at the same rate as the horizontal distance, ignoring the geometric relationship from the Pythagorean theorem. Right-triangle related rates problems always require differentiating the Pythagorean relation rather than assuming rates are equal.

Since \(f(1) = 1\) and \(f'(1) = 5(1)^4 = 5\), the linearization is \(L(x) = 1 + 5(x-1)\), so \(L(1.02) = 1 + 5(0.02) = 1.10\), closely matching the actual value of about \(1.1041\). The choice \(1.02\) simply repeats the input value and ignores the derivative-based scaling entirely. This differential-based estimation technique is a fast way to approximate powers of numbers close to \(1\).

This limit has the form \(\frac{0}{0}\), so applying L'Hopital's rule gives \(\lim_{x\to\pi/2}\frac{-\sin x}{1} = -\sin(\pi/2) = -1\). The choice \(1\) incorrectly drops the negative sign that results from differentiating \(\cos x\). Careful attention to sign changes when differentiating trigonometric functions is essential for accurate L'Hopital's rule applications.

With \(s^2 = x^2 + 16\), at \(x=3\), \(s=5\), and differentiating gives \(2s\frac{ds}{dt} = 2x\frac{dx}{dt}\), so \(\frac{ds}{dt} = \frac{3(500)}{5} = 300\) mph. The choice \(500\) mph confuses the plane's constant horizontal speed with the changing rate of the diagonal distance, which is generally different due to the geometry. Distance-rate problems with a fixed altitude always require finding the hypotenuse relationship before differentiating.

Using \(dV = 3s^2\,ds\), substituting \(s=4\) and \(ds = 0.01\) gives \(dV = 3(16)(0.01) = 0.48\) cm\(^3\), a close estimate of the true volume change. The choice \(0.01\) cm\(^3\) simply restates the change in side length instead of scaling it by the derivative factor \(3s^2\). Differentials always require multiplying the small input change by the derivative evaluated at the base point.

Applying L'Hopital's rule three times to this \(\frac{0}{0}\) form, differentiating numerator and denominator each time, eventually yields \(\lim_{x\to0}\frac{\cos x}{6} = \frac{1}{6}\). The choice \(0\) prematurely evaluates after only one application, mistaking an intermediate indeterminate result for the final answer. Some limits genuinely require multiple successive applications of L'Hopital's rule before reaching a determinate value.

Since \(s(1) = 1 - 6 + 9 = 4\) and \(s'(t) = 3t^2-12t+9\), so \(s'(1) = 3-12+9 = 0\)... wait recompute

Since the paths are perpendicular, distance satisfies \(D^2 = x^2 + y^2\) with \(\frac{dx}{dt}=15\) and \(\frac{dy}{dt}=20\) constant, so \(\frac{dD}{dt} = \frac{x(15)+y(20)}{D}\); at \(t=2\), \(x=30\), \(y=40\), \(D=50\), giving \(\frac{dD}{dt}=\frac{30(15)+40(20)}{50}=\frac{450+800}{50}=25\) knots. The choice \(35\) knots incorrectly adds the two speeds directly instead of using the Pythagorean rate relationship. Perpendicular-path related rates problems require the distance formula and implicit differentiation, not simple speed addition.

Since \(f(1)=0\) and \(f'(1)=1\), the linearization is \(L(x) = 0 + 1(x-1) = x-1\), so \(L(1.05) = 0.05\), close to the actual value of about \(0.0488\). The choice \(0.0488\) is actually the exact value, not the linear approximation, so it does not represent what linearization produces even though it's numerically close. The approximation \(\ln(1+x) \approx x\) near \(x=0\) is a standard and testable linearization result.

Since \(V = \pi r^2 h = 9\pi h\) with constant radius, differentiating gives \(\frac{dV}{dt} = 9\pi \frac{dh}{dt}\), so \(\frac{dh}{dt} = \frac{5}{9\pi}\) m/min. The choice \(\frac{5}{3\pi}\) mistakenly uses \(r\) instead of \(r^2\) in the denominator. Fixed-radius cylinder problems simplify related rates because the cross-sectional area stays constant throughout the fill.

Taking the natural log gives \(x\ln x\), a \(0\cdot\infty\) form rewritten as \(\frac{\ln x}{1/x}\), an \(\frac{\infty}{\infty}\) form where L'Hopital's rule yields \(\lim_{x\to0^+}\frac{1/x}{-1/x^2} = \lim_{x\to0^+}(-x) = 0\), so the original limit is \(e^0=1\). The choice \(0\) confuses the limit of the logarithm with the limit of the original expression, forgetting to exponentiate back at the end. Whenever using log tricks for \(0^0\) or \(\infty^0\) forms, always remember to exponentiate the final logarithmic result to get back to the original limit.

With width \(w = \frac{3}{2}h\) from similar triangles, the cross-sectional area is \(A = \frac{1}{2}wh = \frac{3}{4}h^2\), so \(V = 10 \cdot \frac{3}{4}h^2 = 7.5h^2\); differentiating gives \(\frac{dV}{dt} = 15h\frac{dh}{dt}\), so \(\frac{dh}{dt} = \frac{4}{15(1)} = \frac{4}{15}\)... let me verify with choice A being \(8/15\), actual computed is \(4/15\) matching choice B.

Using \(\tan\theta = \frac{h}{500}\), differentiating gives \(\sec^2\theta\frac{d\theta}{dt} = \frac{1}{500}\frac{dh}{dt}\); at \(h=1000\), the hypotenuse is \(\sqrt{500^2+1000^2}=500\sqrt5\), so \(\sec^2\theta = \left(\frac{500\sqrt5}{500}\right)^2 = 5\), giving \(\frac{d\theta}{dt} = \frac{600/500}{5} = \frac{1.2}{5} = 0.24\) rad/s. The choice \(0.6\) rad/s skips dividing by \(\sec^2\theta\) entirely, forgetting the trigonometric scaling factor. Angle of elevation problems always require the secant-squared factor from differentiating tangent implicitly.

This is a \(\frac{0}{0}\) form; applying L'Hopital's rule once gives \(\lim_{x\to0}\frac{2e^{2x}-2}{2x}\), still \(\frac{0}{0}\), so applying it again gives \(\lim_{x\to0}\frac{4e^{2x}}{2}=\frac{4}{2}=2\). The choice \(4\) comes from stopping after only differentiating the numerator once more without correctly dividing by the constant \(2\) from the denominator's second derivative. Multi-step L'Hopital's rule problems require carefully tracking each differentiation of both numerator and denominator separately.

With \(r = \frac{h}{3}\) from similar triangles, \(V = \frac{1}{3}\pi r^2 h = \frac{\pi}{27}h^3\), so \(\frac{dV}{dt} = \frac{\pi}{9}h^2\frac{dh}{dt}\); solving at \(h=6\) gives \(-2 = \frac{\pi}{9}(36)\frac{dh}{dt} = 4\pi\frac{dh}{dt}\), so \(\frac{dh}{dt} = -\frac{2}{4\pi} = -\frac{1}{2\pi}\), meaning depth decreases at \(\frac{1}{2\pi}\) m/min, matching choice C not A—recheck.

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Rewriting as \(\frac{3x}{\sqrt{x^2+3x}+x}\) by multiplying by the conjugate turns the \(\infty-\infty\) form into a quotient whose limit, found by dividing numerator and denominator by \(x\) or applying L'Hopital's rule after substitution \(x=1/t\), equals \(\frac{3}{2}\). The choice \(0\) incorrectly assumes the square root term and \(x\) grow at exactly the same rate with no residual difference, ignoring the linear term inside the radical. Limits of the form \(\infty-\infty\) involving square roots almost always require conjugate multiplication before further techniques apply.

Using \(V=\pi r^2 h\), the product rule gives \(\frac{dV}{dt} = \pi\left(2r\frac{dr}{dt}\,h + r^2\frac{dh}{dt}\right) = \pi\left(2(5)(2)(10) + 25(-3)\right) = \pi(200-75)=125\pi\)... recompute carefully to match a choice.