Explore the mathematics of change and accumulation — limits, derivatives, integrals, and the powerful techniques that unify them.
Evaluate: \(\displaystyle\lim_{x \to 0} \frac{\sin 3x}{\tan 5x}\).
Step 1: Rewrite the expression by multiplying and dividing strategically: $$\frac{\sin 3x}{\tan 5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\tan 5x} \cdot \frac{3x}{5x} = \frac{\sin 3x}{3x} \cdot \frac{5x}{\tan 5x} \cdot \frac{3}{5}$$
Step 2: As \(x \to 0\), we use the standard limits: \(\lim_{u \to 0} \frac{\sin u}{u} = 1\) and \(\lim_{u \to 0} \frac{u}{\tan u} = 1\).
Step 3: Therefore: \(\frac{\sin 3x}{3x} \to 1\) and \(\frac{5x}{\tan 5x} \to 1\).
Step 4: The limit is \(1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}\).
Answer: \(\dfrac{3}{5}\)
Find \(\dfrac{dy}{dx}\) if \(y = \ln(\sin(e^{2x}))\).
Step 1: Outer layer: \(\frac{d}{dx}\ln(u) = \frac{1}{u} \cdot u'\), where \(u = \sin(e^{2x})\). So we get \(\frac{1}{\sin(e^{2x})} \cdot \frac{d}{dx}[\sin(e^{2x})]\).
Step 2: Middle layer: \(\frac{d}{dx}\sin(v) = \cos(v) \cdot v'\), where \(v = e^{2x}\). So \(\frac{d}{dx}[\sin(e^{2x})] = \cos(e^{2x}) \cdot \frac{d}{dx}[e^{2x}]\).
Step 3: Inner layer: \(\frac{d}{dx}[e^{2x}] = 2e^{2x}\).
Step 4: Combine all layers: $$\frac{dy}{dx} = \frac{1}{\sin(e^{2x})} \cdot \cos(e^{2x}) \cdot 2e^{2x} = \frac{2e^{2x}\cos(e^{2x})}{\sin(e^{2x})}$$
Answer: \(\dfrac{dy}{dx} = 2e^{2x}\cot(e^{2x})\)
Evaluate: \(\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\).
Step 1: Direct substitution: \(\frac{e^0 - 1 - 0}{0^2} = \frac{0}{0}\). This is indeterminate, so apply L'Hopital's Rule.
Step 2: First application: differentiate numerator and denominator: $$\lim_{x \to 0}\frac{e^x - 1}{2x} = \frac{0}{0}$$
Step 3: Still indeterminate. Apply L'Hopital's Rule again: $$\lim_{x \to 0}\frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}$$
Answer: \(\dfrac{1}{2}\)
Evaluate: \(\displaystyle\int x^2 e^x\,dx\).
Step 1: First integration by parts: let \(u = x^2\), \(dv = e^x\,dx\). Then \(du = 2x\,dx\), \(v = e^x\). $$\int x^2 e^x\,dx = x^2 e^x - \int 2x e^x\,dx$$
Step 2: Second integration by parts on \(\int 2x e^x\,dx\): let \(u = 2x\), \(dv = e^x\,dx\). Then \(du = 2\,dx\), \(v = e^x\). $$\int 2x e^x\,dx = 2x e^x - \int 2e^x\,dx = 2xe^x - 2e^x$$
Step 3: Combine: $$\int x^2 e^x\,dx = x^2 e^x - (2xe^x - 2e^x) + C = x^2 e^x - 2xe^x + 2e^x + C$$
Step 4: Factor: \(e^x(x^2 - 2x + 2) + C\).
Answer: \(e^x(x^2 - 2x + 2) + C\)
A spherical balloon is being inflated at a rate of \(100\;\text{cm}^3/\text{s}\). How fast is the radius increasing when the radius is 5 cm? (\(V = \frac{4}{3}\pi r^3\))
Step 1: Start with the volume formula: \(V = \frac{4}{3}\pi r^3\).
Step 2: Differentiate both sides with respect to time \(t\): $$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$$
Step 3: Substitute known values: \(\frac{dV}{dt} = 100\) cm\(^3\)/s and \(r = 5\) cm: $$100 = 4\pi (5)^2 \cdot \frac{dr}{dt} = 100\pi \cdot \frac{dr}{dt}$$
Step 4: Solve: \(\frac{dr}{dt} = \frac{100}{100\pi} = \frac{1}{\pi}\).
Answer: The radius is increasing at \(\dfrac{1}{\pi} \approx 0.318\) cm/s
Find the area enclosed between \(y = x^2\) and \(y = 2x + 3\).
Step 1: Find intersections: \(x^2 = 2x + 3 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0\). So \(x = -1\) and \(x = 3\).
Step 2: Between \(x = -1\) and \(x = 3\), the line \(y = 2x+3\) is above the parabola \(y = x^2\) (check: at \(x = 0\), line gives 3, parabola gives 0).
Step 3: Area \(= \int_{-1}^{3} [(2x+3) - x^2]\,dx = \int_{-1}^{3} (2x + 3 - x^2)\,dx\).
Step 4: Integrate: \(\left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3}\).
Step 5: At \(x = 3\): \(9 + 9 - 9 = 9\). At \(x = -1\): \(1 - 3 + \frac{1}{3} = -\frac{5}{3}\).
Step 6: Area \(= 9 - (-\frac{5}{3}) = 9 + \frac{5}{3} = \frac{32}{3}\).
Answer: \(\dfrac{32}{3}\) square units
If \(x^3 + y^3 = 6xy\), find \(\dfrac{dy}{dx}\) and the equation of the tangent line at \((3, 3)\).
Step 1: Differentiate implicitly: \(3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}\).
Step 2: Collect \(\frac{dy}{dx}\) terms: \(3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2\).
Step 3: Factor: \(\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2\), so: $$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$$
Step 4: At \((3, 3)\): \(\frac{dy}{dx} = \frac{2(3) - 9}{9 - 6} = \frac{-3}{3} = -1\).
Step 5: Tangent line: \(y - 3 = -1(x - 3)\), i.e., \(y = -x + 6\).
Answer: \(\dfrac{dy}{dx} = \dfrac{2y - x^2}{y^2 - 2x}\); tangent line at \((3,3)\) is \(y = -x + 6\)
Determine whether \(\displaystyle\int_1^{\infty} \frac{1}{x^2}\,dx\) converges. If so, find its value.
Step 1: Replace the improper bound with a limit: $$\int_1^{\infty} \frac{1}{x^2}\,dx = \lim_{b \to \infty} \int_1^b x^{-2}\,dx$$
Step 2: Integrate: \(\int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac{1}{x}\).
Step 3: Evaluate: \(\left[-\frac{1}{x}\right]_1^b = -\frac{1}{b} - \left(-\frac{1}{1}\right) = -\frac{1}{b} + 1\).
Step 4: Take the limit: \(\lim_{b \to \infty}\left(-\frac{1}{b} + 1\right) = 0 + 1 = 1\).
Answer: The integral converges, and its value is \(1\)
Find the first four nonzero terms of the Maclaurin series for \(f(x) = e^x \sin x\).
Step 1: Write the series: \(e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots\) and \(\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\)
Step 2: Multiply and collect by powers of \(x\):
\(x^1\) term: \(1 \cdot x = x\)
\(x^2\) term: \(x \cdot x = x^2\)
\(x^3\) term: \(\frac{x^2}{2} \cdot x + 1 \cdot \left(-\frac{x^3}{6}\right) = \frac{x^3}{2} - \frac{x^3}{6} = \frac{x^3}{3}\)
\(x^4\) term: \(\frac{x^3}{6} \cdot x + x \cdot \left(-\frac{x^3}{6}\right) = \frac{x^4}{6} - \frac{x^4}{6} = 0\)
\(x^5\) term: \(\frac{x^4}{24} \cdot x + \frac{x^2}{2}\left(-\frac{x^3}{6}\right) + 1 \cdot \frac{x^5}{120} = \frac{x^5}{24} - \frac{x^5}{12} + \frac{x^5}{120} = -\frac{x^5}{30}\)
Answer: \(e^x\sin x = x + x^2 + \dfrac{x^3}{3} - \dfrac{x^5}{30} + \cdots\)
A farmer wants to fence a rectangular area of \(1800\;\text{m}^2\) and divide it into three equal pens with fences parallel to one side. What dimensions minimize the total fencing?
Step 1: Let \(x\) = width, \(y\) = length. Area constraint: \(xy = 1800\). Total fencing: \(F = 2y + 4x\) (two lengths + four widths for the three pens).
Step 2: Substitute \(y = \frac{1800}{x}\): $$F(x) = 2 \cdot \frac{1800}{x} + 4x = \frac{3600}{x} + 4x$$
Step 3: Differentiate and set to zero: \(F'(x) = -\frac{3600}{x^2} + 4 = 0\).
Step 4: Solve: \(\frac{3600}{x^2} = 4 \Rightarrow x^2 = 900 \Rightarrow x = 30\) m.
Step 5: Then \(y = \frac{1800}{30} = 60\) m. Verify minimum: \(F''(x) = \frac{7200}{x^3} > 0\), confirming a minimum.
Step 6: Total fencing: \(F = 2(60) + 4(30) = 120 + 120 = 240\) m.
Answer: Width = 30 m, Length = 60 m (minimum fencing = 240 m)
Evaluate: \(\displaystyle\int \frac{dx}{x^2\sqrt{x^2 + 9}}\).
Step 1: Substitute \(x = 3\tan\theta\): \(dx = 3\sec^2\theta\,d\theta\), \(x^2 = 9\tan^2\theta\), \(\sqrt{x^2+9} = 3\sec\theta\).
Step 2: The integral becomes: $$\int\frac{3\sec^2\theta\,d\theta}{9\tan^2\theta \cdot 3\sec\theta} = \int\frac{\sec\theta}{9\tan^2\theta}\,d\theta = \frac{1}{9}\int\frac{\cos\theta}{\sin^2\theta}\,d\theta$$
Step 3: Let \(u = \sin\theta\), \(du = \cos\theta\,d\theta\): $$\frac{1}{9}\int\frac{du}{u^2} = \frac{1}{9}\left(-\frac{1}{u}\right) = -\frac{1}{9\sin\theta}$$
Step 4: Convert back: from \(x = 3\tan\theta\), we get \(\sin\theta = \frac{x}{\sqrt{x^2+9}}\).
Step 5: So \(-\frac{1}{9\sin\theta} = -\frac{\sqrt{x^2+9}}{9x}\).
Answer: \(-\dfrac{\sqrt{x^2+9}}{9x} + C\)
Solve the differential equation \(\dfrac{dy}{dx} + 2y = e^{-x}\) with initial condition \(y(0) = 1\).
Step 1: The ODE is \(y' + 2y = e^{-x}\). The integrating factor is \(\mu = e^{\int 2\,dx} = e^{2x}\).
Step 2: Multiply both sides by \(e^{2x}\): $$e^{2x}y' + 2e^{2x}y = e^{2x} \cdot e^{-x} = e^{x}$$
Step 3: The left side is \(\frac{d}{dx}[e^{2x}y]\). So: $$\frac{d}{dx}[e^{2x}y] = e^x$$
Step 4: Integrate both sides: \(e^{2x}y = e^x + C\).
Step 5: Solve for \(y\): \(y = e^{-x} + Ce^{-2x}\).
Step 6: Apply \(y(0) = 1\): \(1 = 1 + C \Rightarrow C = 0\).
Answer: \(y = e^{-x}\)