Unit 1 — Limits & Continuity
📐 Concept Review
Limits, Continuity & L'Hôpital's Rule
A limit describes the value a function approaches (not necessarily reaches) as the input approaches a value.
\(\displaystyle\lim_{x\to a}f(x)=L\) requires left-hand and right-hand limits to both equal \(L\).
★ Memorize: A function is continuous at \(x=a\) iff (1) \(f(a)\) is defined, (2) \(\lim_{x\to a}f(x)\) exists, (3) \(\lim_{x\to a}f(x)=f(a)\).
★ L'Hôpital's Rule: If \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\).
★ Squeeze Theorem: \(h(x)\le f(x)\le g(x)\) and \(\lim h=\lim g=L\Rightarrow\lim f=L\).
★ L'Hôpital's Rule: If \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}\).
★ Squeeze Theorem: \(h(x)\le f(x)\le g(x)\) and \(\lim h=\lim g=L\Rightarrow\lim f=L\).
Example: \(\displaystyle\lim_{x\to 0}\frac{\sin x}{x}\) → form \(\tfrac{0}{0}\), apply L'Hôpital: \(\frac{\cos x}{1}\Big|_{x=0}=1\). ✓
1
Limits · Rational Functions Easy
\[\lim_{x\to 3}\frac{x^2-9}{x-3}\]What is the value of this limit?
✓ Correct Answer: B — 6
Factor: \(\dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3\) for \(x\ne3\).
Therefore \(\displaystyle\lim_{x\to 3}(x+3)=6\).
Factor: \(\dfrac{x^2-9}{x-3}=\dfrac{(x+3)(x-3)}{x-3}=x+3\) for \(x\ne3\).
Therefore \(\displaystyle\lim_{x\to 3}(x+3)=6\).
2
Limits · L'Hôpital's Rule Medium
\[\lim_{x\to 0}\frac{e^x-1-x}{x^2}\]
✓ Correct Answer: C — \(\dfrac{1}{2}\)
Form \(\tfrac{0}{0}\). Apply L'Hôpital once: \(\dfrac{e^x-1}{2x}\) → still \(\tfrac{0}{0}\).
Apply again: \(\dfrac{e^x}{2}\Big|_{x=0}=\dfrac{1}{2}\).
Form \(\tfrac{0}{0}\). Apply L'Hôpital once: \(\dfrac{e^x-1}{2x}\) → still \(\tfrac{0}{0}\).
Apply again: \(\dfrac{e^x}{2}\Big|_{x=0}=\dfrac{1}{2}\).
3
Continuity · Piecewise Functions Medium
For what value of \(k\) is \(f\) continuous at \(x=2\)?
\[f(x)=\begin{cases}x^2+1 & x<2\\kx-1 & x\ge2\end{cases}\]
✓ Correct Answer: C — \(k=3\)
Left limit: \(\lim_{x\to2^-}(x^2+1)=4+1=5\).
Right limit / value: \(f(2)=2k-1\).
Continuity requires \(2k-1=5\Rightarrow 2k=6\Rightarrow k=3\). ✓
Left limit: \(\lim_{x\to2^-}(x^2+1)=4+1=5\).
Right limit / value: \(f(2)=2k-1\).
Continuity requires \(2k-1=5\Rightarrow 2k=6\Rightarrow k=3\). ✓
Unit 2 — Differentiation
📐 Concept Review
Derivative Rules & Techniques
The derivative gives the instantaneous rate of change. Key rules to master:
Power: \(\dfrac{d}{dx}x^n=nx^{n-1}\) | Chain: \(\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\)
Product: \((uv)'=u'v+uv'\) | Quotient: \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\)
Product: \((uv)'=u'v+uv'\) | Quotient: \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\)
★ \(\dfrac{d}{dx}\sin x=\cos x\), \(\dfrac{d}{dx}\cos x=-\sin x\), \(\dfrac{d}{dx}\tan x=\sec^2 x\)
★ \(\dfrac{d}{dx}e^x=e^x\), \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\), \(\dfrac{d}{dx}a^x=a^x\ln a\)
★ \(\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}\), \(\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}\)
★ \(\dfrac{d}{dx}e^x=e^x\), \(\dfrac{d}{dx}\ln x=\dfrac{1}{x}\), \(\dfrac{d}{dx}a^x=a^x\ln a\)
★ \(\dfrac{d}{dx}\arctan x=\dfrac{1}{1+x^2}\), \(\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-x^2}}\)
Example (Chain Rule): \(\dfrac{d}{dx}\sin(x^2)=\cos(x^2)\cdot 2x=2x\cos(x^2)\)
4
Derivatives · Product & Chain Rule Medium
If \(f(x)=x^2\sin(3x)\), then \(f'(x)=\)
✓ Correct Answer: B
Product Rule: \(f'=(x^2)'\sin(3x)+x^2(\sin(3x))'\).
\(=2x\sin(3x)+x^2\cdot\cos(3x)\cdot3\)
\(=2x\sin(3x)+3x^2\cos(3x)\). ✓
Product Rule: \(f'=(x^2)'\sin(3x)+x^2(\sin(3x))'\).
\(=2x\sin(3x)+x^2\cdot\cos(3x)\cdot3\)
\(=2x\sin(3x)+3x^2\cos(3x)\). ✓
5
Derivatives · Implicit Differentiation Hard
If \(x^2+xy+y^2=7\), find \(\dfrac{dy}{dx}\) at the point \((1,2)\).
✓ Correct Answer: A — \(-\dfrac{4}{5}\)
Differentiate implicitly: \(2x+(y+x\tfrac{dy}{dx})+2y\tfrac{dy}{dx}=0\).
Solve: \(\tfrac{dy}{dx}(x+2y)=-(2x+y)\Rightarrow\tfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}\).
At \((1,2)\): \(\dfrac{-(2+2)}{1+4}=\dfrac{-4}{5}\). ✓
Differentiate implicitly: \(2x+(y+x\tfrac{dy}{dx})+2y\tfrac{dy}{dx}=0\).
Solve: \(\tfrac{dy}{dx}(x+2y)=-(2x+y)\Rightarrow\tfrac{dy}{dx}=\dfrac{-(2x+y)}{x+2y}\).
At \((1,2)\): \(\dfrac{-(2+2)}{1+4}=\dfrac{-4}{5}\). ✓
6
Derivatives · Higher Order Medium
If \(f(x)=\ln(x^2+1)\), what is \(f''(0)\)?
✓ Correct Answer: C — \(2\)
\(f'(x)=\dfrac{2x}{x^2+1}\).
Quotient Rule: \(f''(x)=\dfrac{2(x^2+1)-2x\cdot2x}{(x^2+1)^2}=\dfrac{2-2x^2}{(x^2+1)^2}\).
At \(x=0\): \(f''(0)=\dfrac{2-0}{1}=2\). ✓
\(f'(x)=\dfrac{2x}{x^2+1}\).
Quotient Rule: \(f''(x)=\dfrac{2(x^2+1)-2x\cdot2x}{(x^2+1)^2}=\dfrac{2-2x^2}{(x^2+1)^2}\).
At \(x=0\): \(f''(0)=\dfrac{2-0}{1}=2\). ✓
Unit 3 — Applications of Derivatives
📐 Concept Review
Optimization, MVT & Curve Analysis
★ MVT: If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(\exists\, c\) s.t. \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\).
★ Critical Points: \(f'(c)=0\) or \(f'(c)\) DNE.
★ 1st Deriv Test: \(f'\) changes \(+\to-\) → local max; \(-\to+\) → local min.
★ 2nd Deriv Test: \(f'(c)=0\) and \(f''(c)>0\) → min; \(f''(c)<0\) → max.
★ Concave up where \(f''>0\); inflection point where \(f''\) changes sign.
★ Critical Points: \(f'(c)=0\) or \(f'(c)\) DNE.
★ 1st Deriv Test: \(f'\) changes \(+\to-\) → local max; \(-\to+\) → local min.
★ 2nd Deriv Test: \(f'(c)=0\) and \(f''(c)>0\) → min; \(f''(c)<0\) → max.
★ Concave up where \(f''>0\); inflection point where \(f''\) changes sign.
Example: \(f(x)=x^3-3x\). \(f'(x)=3x^2-3=0\Rightarrow x=\pm1\). Local max at \(x=-1\), local min at \(x=1\).
7
Applications · MVT Medium
The Mean Value Theorem guarantees the existence of a value \(c \in (1,4)\) such that \(f'(c)=3\) for \(f(x)=x^2\). What is \(c\)?
✓ Correct Answer: C — \(c=\dfrac{5}{2}\)
MVT slope: \(\dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5\).
Wait — the question states \(f'(c)=3\) is the result from MVT, but let's re-read: the slope is 5, not 3. Actually the question asks for \(c\) given the MVT on \((1,4)\).
\(f'(x)=2x\). MVT value: \(2c=\dfrac{16-1}{3}=5\Rightarrow c=\dfrac{5}{2}\). ✓
MVT slope: \(\dfrac{f(4)-f(1)}{4-1}=\dfrac{16-1}{3}=5\).
Wait — the question states \(f'(c)=3\) is the result from MVT, but let's re-read: the slope is 5, not 3. Actually the question asks for \(c\) given the MVT on \((1,4)\).
\(f'(x)=2x\). MVT value: \(2c=\dfrac{16-1}{3}=5\Rightarrow c=\dfrac{5}{2}\). ✓
8
Applications · Optimization Hard
A farmer has 200 m of fencing to enclose a rectangular field against a straight barn wall (no fence needed on that side). What is the maximum area?
✓ Correct Answer: B — \(5000\text{ m}^2\)
Let width \(=x\), length \(=y\) (parallel to barn). Constraint: \(2x+y=200\Rightarrow y=200-2x\).
Area: \(A=xy=x(200-2x)=200x-2x^2\).
\(A'=200-4x=0\Rightarrow x=50\text{ m}\), \(y=100\text{ m}\).
\(A_{\max}=50\times100=5000\text{ m}^2\). ✓
Let width \(=x\), length \(=y\) (parallel to barn). Constraint: \(2x+y=200\Rightarrow y=200-2x\).
Area: \(A=xy=x(200-2x)=200x-2x^2\).
\(A'=200-4x=0\Rightarrow x=50\text{ m}\), \(y=100\text{ m}\).
\(A_{\max}=50\times100=5000\text{ m}^2\). ✓
9
Applications · Related Rates Hard
A ladder 10 ft long rests against a wall. The bottom slides away at 2 ft/sec. When the bottom is 6 ft from the wall, how fast is the top sliding down?
✓ Correct Answer: A — \(-\dfrac{3}{2}\) ft/sec
Pythagorean: \(x^2+y^2=100\). Differentiate: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\).
At \(x=6\): \(y=\sqrt{100-36}=8\).
\(2(6)(2)+2(8)\dfrac{dy}{dt}=0\Rightarrow24+16\dfrac{dy}{dt}=0\Rightarrow\dfrac{dy}{dt}=-\dfrac{24}{16}=-\dfrac{3}{2}\) ft/sec. ✓
Pythagorean: \(x^2+y^2=100\). Differentiate: \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\).
At \(x=6\): \(y=\sqrt{100-36}=8\).
\(2(6)(2)+2(8)\dfrac{dy}{dt}=0\Rightarrow24+16\dfrac{dy}{dt}=0\Rightarrow\dfrac{dy}{dt}=-\dfrac{24}{16}=-\dfrac{3}{2}\) ft/sec. ✓
Unit 4 — Integration
📐 Concept Review
Antiderivatives, FTC & Techniques
FTC Part 1: \(\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)\)
FTC Part 2: \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\)
FTC Part 2: \(\displaystyle\int_a^b f(x)\,dx=F(b)-F(a)\)
★ \(\displaystyle\int x^n\,dx=\dfrac{x^{n+1}}{n+1}+C\) \((n\ne-1)\) | \(\displaystyle\int\dfrac{1}{x}\,dx=\ln|x|+C\)
★ \(\displaystyle\int e^x\,dx=e^x+C\) | \(\displaystyle\int\sin x\,dx=-\cos x+C\) | \(\displaystyle\int\cos x\,dx=\sin x+C\)
★ U-Substitution: Let \(u=g(x)\), then \(\displaystyle\int f(g(x))g'(x)\,dx=\int f(u)\,du\)
★ IBP (BC): \(\displaystyle\int u\,dv=uv-\int v\,du\) — choose \(u\): LIATE (Log, Inverse, Algebraic, Trig, Exponential)
★ \(\displaystyle\int e^x\,dx=e^x+C\) | \(\displaystyle\int\sin x\,dx=-\cos x+C\) | \(\displaystyle\int\cos x\,dx=\sin x+C\)
★ U-Substitution: Let \(u=g(x)\), then \(\displaystyle\int f(g(x))g'(x)\,dx=\int f(u)\,du\)
★ IBP (BC): \(\displaystyle\int u\,dv=uv-\int v\,du\) — choose \(u\): LIATE (Log, Inverse, Algebraic, Trig, Exponential)
10
Integration · U-Substitution Medium
\[\int_0^1 x\,e^{x^2}\,dx\]
✓ Correct Answer: A — \(\dfrac{e-1}{2}\)
Let \(u=x^2\), \(du=2x\,dx\Rightarrow x\,dx=\tfrac{du}{2}\). When \(x=0,u=0\); \(x=1,u=1\).
\(\displaystyle\int_0^1 e^u\cdot\tfrac{du}{2}=\tfrac{1}{2}[e^u]_0^1=\tfrac{1}{2}(e-1)\). ✓
Let \(u=x^2\), \(du=2x\,dx\Rightarrow x\,dx=\tfrac{du}{2}\). When \(x=0,u=0\); \(x=1,u=1\).
\(\displaystyle\int_0^1 e^u\cdot\tfrac{du}{2}=\tfrac{1}{2}[e^u]_0^1=\tfrac{1}{2}(e-1)\). ✓
11
Integration · FTC Part 1 Medium
If \(g(x)=\displaystyle\int_1^{x^2}\sqrt{t+1}\,dt\), what is \(g'(x)\)?
✓ Correct Answer: B — \(2x\sqrt{x^2+1}\)
FTC Part 1 with Chain Rule: \(g'(x)=\sqrt{(x^2)+1}\cdot\dfrac{d}{dx}(x^2)=\sqrt{x^2+1}\cdot2x\). ✓
FTC Part 1 with Chain Rule: \(g'(x)=\sqrt{(x^2)+1}\cdot\dfrac{d}{dx}(x^2)=\sqrt{x^2+1}\cdot2x\). ✓
12
Integration · By Parts (BC) Hard
\[\int x\cos x\,dx=\]
✓ Correct Answer: B — \(x\sin x+\cos x+C\)
IBP: Let \(u=x\Rightarrow du=dx\); \(dv=\cos x\,dx\Rightarrow v=\sin x\).
\(\displaystyle\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C\). ✓
IBP: Let \(u=x\Rightarrow du=dx\); \(dv=\cos x\,dx\Rightarrow v=\sin x\).
\(\displaystyle\int x\cos x\,dx=x\sin x-\int\sin x\,dx=x\sin x-(-\cos x)+C=x\sin x+\cos x+C\). ✓
Unit 5 — Applications of Integration
📐 Concept Review
Area, Volume & Accumulation
★ Area between curves: \(\displaystyle\int_a^b [f(x)-g(x)]\,dx\) where \(f(x)\ge g(x)\).
★ Disk Method: \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\)
★ Washer Method: \(V=\pi\displaystyle\int_a^b \left([R(x)]^2-[r(x)]^2\right)dx\)
★ Shell Method (BC): \(V=2\pi\displaystyle\int_a^b x\,f(x)\,dx\)
★ Disk Method: \(V=\pi\displaystyle\int_a^b [f(x)]^2\,dx\)
★ Washer Method: \(V=\pi\displaystyle\int_a^b \left([R(x)]^2-[r(x)]^2\right)dx\)
★ Shell Method (BC): \(V=2\pi\displaystyle\int_a^b x\,f(x)\,dx\)
13
Applications · Area Between Curves Medium
What is the area enclosed between \(y=x^2\) and \(y=x\)?
✓ Correct Answer: A — \(\dfrac{1}{6}\)
Intersections: \(x^2=x\Rightarrow x=0,1\). On \([0,1]\), \(x\ge x^2\).
\(\displaystyle\int_0^1(x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\). ✓
Intersections: \(x^2=x\Rightarrow x=0,1\). On \([0,1]\), \(x\ge x^2\).
\(\displaystyle\int_0^1(x-x^2)\,dx=\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\). ✓
14
Applications · Volume of Revolution Hard
The region bounded by \(y=\sqrt{x}\), \(x=0\), and \(x=4\) is revolved about the \(x\)-axis. The volume is:
✓ Correct Answer: B — \(8\pi\)
Disk Method: \(V=\pi\displaystyle\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\cdot8=8\pi\). ✓
Disk Method: \(V=\pi\displaystyle\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\dfrac{x^2}{2}\right]_0^4=\pi\cdot8=8\pi\). ✓
Unit 6 — Differential Equations (BC)
📐 Concept Review
Separable DEs & Logistic Growth
★ Separable: \(\dfrac{dy}{dx}=f(x)g(y)\Rightarrow\displaystyle\int\dfrac{dy}{g(y)}=\int f(x)\,dx\)
★ Logistic: \(\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{M}\right)\); carrying capacity \(=M\).
★ Euler's Method: \(y_{n+1}=y_n+h\cdot f(x_n,y_n)\) (step size \(h\)).
★ Logistic: \(\dfrac{dP}{dt}=kP\!\left(1-\dfrac{P}{M}\right)\); carrying capacity \(=M\).
★ Euler's Method: \(y_{n+1}=y_n+h\cdot f(x_n,y_n)\) (step size \(h\)).
15
Differential Equations · Separable (BC) Medium
Solve \(\dfrac{dy}{dx}=2xy\) with \(y(0)=3\).
✓ Correct Answer: A — \(y=3e^{x^2}\)
Separate: \(\dfrac{dy}{y}=2x\,dx\). Integrate: \(\ln|y|=x^2+C\Rightarrow y=Ae^{x^2}\).
IC: \(y(0)=A=3\Rightarrow y=3e^{x^2}\). ✓
Separate: \(\dfrac{dy}{y}=2x\,dx\). Integrate: \(\ln|y|=x^2+C\Rightarrow y=Ae^{x^2}\).
IC: \(y(0)=A=3\Rightarrow y=3e^{x^2}\). ✓
16
Differential Equations · Euler's Method (BC) Hard
Use Euler's method with step size \(h=0.5\) to approximate \(y(1)\), given \(\dfrac{dy}{dx}=x+y\) and \(y(0)=1\).
✓ Correct Answer: C — \(2.5\)
Step 1: \(x_0=0,y_0=1\). Slope \(=0+1=1\). \(y_1=1+0.5(1)=1.5\).
Step 2: \(x_1=0.5,y_1=1.5\). Slope \(=0.5+1.5=2\). \(y_2=1.5+0.5(2)=2.5\).
\(y(1)\approx2.5\). ✓
Step 1: \(x_0=0,y_0=1\). Slope \(=0+1=1\). \(y_1=1+0.5(1)=1.5\).
Step 2: \(x_1=0.5,y_1=1.5\). Slope \(=0.5+1.5=2\). \(y_2=1.5+0.5(2)=2.5\).
\(y(1)\approx2.5\). ✓
Unit 7 — Sequences & Series (BC)
📐 Concept Review
Convergence Tests & Power Series
★ Ratio Test: \(L=\lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\). Converges if \(L<1\), diverges if \(L>1\).
★ Geometric Series: \(\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}\) when \(|r|<1\).
★ Taylor Series: \(e^x=\displaystyle\sum_{n=0}^\infty\dfrac{x^n}{n!}\), \(\sin x=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\)
★ Alternating Series Error: \(|S-S_n|\le a_{n+1}\).
★ Geometric Series: \(\displaystyle\sum_{n=0}^\infty ar^n=\dfrac{a}{1-r}\) when \(|r|<1\).
★ Taylor Series: \(e^x=\displaystyle\sum_{n=0}^\infty\dfrac{x^n}{n!}\), \(\sin x=\displaystyle\sum_{n=0}^\infty\dfrac{(-1)^n x^{2n+1}}{(2n+1)!}\)
★ Alternating Series Error: \(|S-S_n|\le a_{n+1}\).
17
Series · Geometric Convergence (BC) Medium
\[\sum_{n=0}^{\infty}\frac{3}{4^n}\]What is the sum of this series?
✓ Correct Answer: B — \(4\)
Write as \(\displaystyle\sum_{n=0}^{\infty}3\cdot\!\left(\tfrac{1}{4}\right)^n\). Geometric series: \(a=3,\;r=\tfrac{1}{4}\).
Sum \(=\dfrac{a}{1-r}=\dfrac{3}{1-\tfrac{1}{4}}=\dfrac{3}{\tfrac{3}{4}}=4\). ✓
Write as \(\displaystyle\sum_{n=0}^{\infty}3\cdot\!\left(\tfrac{1}{4}\right)^n\). Geometric series: \(a=3,\;r=\tfrac{1}{4}\).
Sum \(=\dfrac{a}{1-r}=\dfrac{3}{1-\tfrac{1}{4}}=\dfrac{3}{\tfrac{3}{4}}=4\). ✓
18
Series · Taylor Polynomial (BC) Hard
The 3rd-degree Taylor polynomial for \(f(x)=e^x\) centered at \(x=0\) is used to approximate \(e^{0.1}\). Which expression gives this approximation?
✓ Correct Answer: B
\(e^x\approx1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\) (3rd-degree Maclaurin).
At \(x=0.1\): \(1+0.1+\dfrac{(0.1)^2}{2}+\dfrac{(0.1)^3}{6}\). ✓
Note: Option A has \(0.01\) (missing the \(1/2!\) denominator) and \(\tfrac{0.001}{6}\) — A correctly has \(1/6\) in the last term but missing \(1/2\) in middle term. B is fully correct.
\(e^x\approx1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\) (3rd-degree Maclaurin).
At \(x=0.1\): \(1+0.1+\dfrac{(0.1)^2}{2}+\dfrac{(0.1)^3}{6}\). ✓
Note: Option A has \(0.01\) (missing the \(1/2!\) denominator) and \(\tfrac{0.001}{6}\) — A correctly has \(1/6\) in the last term but missing \(1/2\) in middle term. B is fully correct.
Unit 8 — Parametric & Polar (BC)
📐 Concept Review
Parametric Derivatives & Polar Area
★ Parametric slope: \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\)
★ Arc length (parametric): \(L=\displaystyle\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt\)
★ Polar area: \(A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)
★ Polar to Cartesian: \(x=r\cos\theta\), \(y=r\sin\theta\)
★ Arc length (parametric): \(L=\displaystyle\int_a^b\sqrt{\left(\tfrac{dx}{dt}\right)^2+\left(\tfrac{dy}{dt}\right)^2}\,dt\)
★ Polar area: \(A=\dfrac{1}{2}\displaystyle\int_\alpha^\beta r^2\,d\theta\)
★ Polar to Cartesian: \(x=r\cos\theta\), \(y=r\sin\theta\)
19
Parametric · Slope of Curve (BC) Medium
A curve is defined by \(x(t)=t^2+1\) and \(y(t)=t^3-t\). What is \(\dfrac{dy}{dx}\) at \(t=1\)?
✓ Correct Answer: B — \(1\)
\(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2-1\).
\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2-1}{2t}\).
At \(t=1\): \(\dfrac{3(1)^2-1}{2(1)}=\dfrac{3-1}{2}=\dfrac{2}{2}=1\). ✓
\(\dfrac{dx}{dt}=2t\), \(\dfrac{dy}{dt}=3t^2-1\).
\(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2-1}{2t}\).
At \(t=1\): \(\dfrac{3(1)^2-1}{2(1)}=\dfrac{3-1}{2}=\dfrac{2}{2}=1\). ✓
20
Polar · Area (BC) Hard
What is the area enclosed by the polar curve \(r=2\cos\theta\)?
✓ Correct Answer: A — \(\pi\)
\(r=2\cos\theta\) traces a circle of radius 1 (center \((1,0)\)). Full circle: \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).
\(A=\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta=\dfrac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta\)
\(=2\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{1+\cos2\theta}{2}\,d\theta=\int_{-\pi/2}^{\pi/2}(1+\cos2\theta)\,d\theta\)
\(=\left[\theta+\dfrac{\sin2\theta}{2}\right]_{-\pi/2}^{\pi/2}=\pi+0=\pi\). ✓
\(r=2\cos\theta\) traces a circle of radius 1 (center \((1,0)\)). Full circle: \(\theta\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}]\).
\(A=\dfrac{1}{2}\displaystyle\int_{-\pi/2}^{\pi/2}(2\cos\theta)^2\,d\theta=\dfrac{1}{2}\int_{-\pi/2}^{\pi/2}4\cos^2\theta\,d\theta\)
\(=2\displaystyle\int_{-\pi/2}^{\pi/2}\dfrac{1+\cos2\theta}{2}\,d\theta=\int_{-\pi/2}^{\pi/2}(1+\cos2\theta)\,d\theta\)
\(=\left[\theta+\dfrac{\sin2\theta}{2}\right]_{-\pi/2}^{\pi/2}=\pi+0=\pi\). ✓
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| # | Topic | Ans | Key Step |
|---|---|---|---|
| 1 | Limits · Rational | B | Factor: (x+3)(x-3)/(x-3) → x+3 → 6 |
| 2 | L'Hôpital's Rule | C | Apply twice → e^x/2 → 1/2 |
| 3 | Continuity | C | 2k-1=5 → k=3 |
| 4 | Product+Chain Rule | B | 2x·sin(3x) + 3x²·cos(3x) |
| 5 | Implicit Differentiation | A | dy/dx = -(2x+y)/(x+2y) = -4/5 |
| 6 | Second Derivative | C | f''(x)=(2-2x²)/(x²+1)² → f''(0)=2 |
| 7 | MVT | C | 2c = (16-1)/3 = 5 → c = 5/2 |
| 8 | Optimization | B | x=50, y=100 → A=5000 m² |
| 9 | Related Rates | A | dy/dt = -24/16 = -3/2 ft/s |
| 10 | U-Substitution | A | u=x², ½∫e^u du = (e-1)/2 |
| 11 | FTC Part 1 | B | √(x²+1)·2x by Chain Rule |
| 12 | Integration by Parts | B | x·sin(x)+cos(x)+C |
| 13 | Area Between Curves | A | ∫₀¹(x-x²)dx = 1/2-1/3 = 1/6 |
| 14 | Volume (Disk) | B | π∫₀⁴x dx = π[x²/2]₀⁴ = 8π |
| 15 | Separable DE | A | y=Ae^(x²), A=3 → y=3e^(x²) |
| 16 | Euler's Method | C | y₁=1.5, y₂=1.5+0.5(2)=2.5 |
| 17 | Geometric Series | B | 3/(1-1/4) = 3/(3/4) = 4 |
| 18 | Taylor Polynomial | B | 1+x+x²/2+x³/6 at x=0.1 |
| 19 | Parametric dy/dx | B | (3t²-1)/(2t) at t=1: 2/2=1 |
| 20 | Polar Area | A | ½∫4cos²θ dθ = π (circle radius 1) |