Logistic Equation: \(\dfrac{dP}{dt}=rP\!\left(1-\dfrac{P}{K}\right)\) has carrying capacity \(K\). As \(t\to\infty\), \(P\to K\) for any \(0
Solve \(\dfrac{dy}{dx}=2xy\) with \(y(0)=3\).
Separate: \(\dfrac{dy}{y}=2x\,dx\Rightarrow y=3e^{x^2}\). Answer: \(y=3e^{x^2}\)
A population \(P\) satisfies \(\dfrac{dP}{dt}=0.2P\!\left(1-\dfrac{P}{1000}\right)\) with \(P(0)=100\). Find \(\displaystyle\lim_{t\to\infty}P(t)\).
\(100\)
\(500\)
\(1000\)
\(\infty\)
2
Identify Carrying Capacity
The equation is logistic with \(K=1000\). Regardless of \(0
Equilibrium Analysis
Set \(\dfrac{dP}{dt}=0\): \(P=0\) (unstable) or \(P=1000\) (stable). Since \(P(0)=100>0\), the solution is attracted to 1000.
\[\boxed{\lim_{t\to\infty}P(t)=1000}\]
Polar Area
5
Find the area of the region inside r equals 3 sine theta and outside r equals 1 plus sine theta.
Polar Area:
\[A=\frac{1}{2}\int_\alpha^\beta r^2\,d\theta\]
Between two curves: \(A=\dfrac{1}{2}\displaystyle\int(r_{\rm out}^2-r_{\rm in}^2)\,d\theta\).
Area enclosed by \(r=2\cos\theta\).
\(A=\tfrac{1}{2}\displaystyle\int_0^\pi 4\cos^2\theta\,d\theta=\pi\). Answer: \(\pi\)
Find the area inside \(r=3\sin\theta\) and outside \(r=1+\sin\theta\).
\(\pi\)
\(\dfrac{3\pi}{4}\)
\(\dfrac{5\pi}{4}\)
\(\dfrac{7\pi}{6}\)
0
Step 1 — Intersections
\(3\sin\theta=1+\sin\theta\Rightarrow\sin\theta=\tfrac{1}{2}\Rightarrow\theta=\tfrac{\pi}{6},\tfrac{5\pi}{6}\).
Step 2 — Set Up
\[A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}\!\bigl[(3\sin\theta)^2-(1+\sin\theta)^2\bigr]\,d\theta\]
Step 3 — Expand
\[=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(8\sin^2\theta-2\sin\theta-1)\,d\theta\]Using \(\sin^2\theta=\tfrac{1-\cos 2\theta}{2}\) and evaluating: \(A=\pi\).
\[\boxed{A=\pi}\]
Series Convergence
5
Which of the following series diverges? Consider the alternating harmonic series, n factorial over n to the n, 1 over n ln n, and 1 over n times the square of ln n.
Key tests: Ratio Test (\(L<1\) → converge), Integral Test, Alternating Series Test, \(p\)-series (\(\sum\tfrac{1}{n^p}\) converges iff \(p>1\)).
Does \(\displaystyle\sum_{n=1}^\infty\dfrac{1}{n^2}\) converge?
\(p\)-series with \(p=2>1\) → converges. Converges
Which series diverges?
\(\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^n}{n}\)
\(\displaystyle\sum_{n=1}^\infty\dfrac{n!}{n^n}\)
\(\displaystyle\sum_{n=2}^\infty\dfrac{1}{n\ln n}\)
\(\displaystyle\sum_{n=2}^\infty\dfrac{1}{n(\ln n)^2}\)
2
(A) Alternating Harmonic
AST: \(\tfrac{1}{n}\) decreasing to 0 →
Converges.
(B) Ratio Test
\(\displaystyle\frac{a_{n+1}}{a_n}=\frac{n^n}{(n+1)^n}\to\frac{1}{e}<1\) →
Converges.
(C) Integral Test
\(\displaystyle\int_2^\infty\frac{dx}{x\ln x}=\bigl[\ln(\ln x)\bigr]_2^\infty\to\infty\) →
Diverges ✓.
(D) Integral Test
\(\displaystyle\int_2^\infty\frac{dx}{x(\ln x)^2}=\left[-\frac{1}{\ln x}\right]_2^\infty=\frac{1}{\ln 2}<\infty\) →
Converges.
\[\boxed{\sum_{n=2}^\infty\frac{1}{n\ln n}\text{ diverges}}\]
Taylor Series
5
The function f of x equals x over 1 plus x squared is represented by a power series. Find the third-degree Taylor polynomial centered at x equals 0.
Key Maclaurin series:
\[e^x=\sum_{n=0}^\infty\frac{x^n}{n!},\quad \frac{1}{1-u}=\sum_{n=0}^\infty u^n\quad(|u|<1)\]
First 3 non-zero terms of \(e^{-x^2}\).
Replace \(x\) with \(-x^2\): \(1-x^2+\tfrac{x^4}{2}-\cdots\). Answer: \(1-x^2+\frac{x^4}{2}\)
Find the third-degree Taylor polynomial of \(f(x)=\dfrac{x}{1+x^2}\) centered at \(x=0\).
\(x-x^3\)
\(x+x^3\)
\(x-x^2+x^3\)
\(1-x^2\)
0
Step 1 — Start with Geometric Series
Let \(u=-x^2\): \[\frac{1}{1+x^2}=1-x^2+x^4-\cdots\]
Step 2 — Multiply by x
\[\frac{x}{1+x^2}=x-x^3+x^5-\cdots\]
Step 3 — Truncate at degree 3
\(T_3(x)=x-x^3\).
\[\boxed{T_3(x)=x-x^3}\]
Parametric Curves
5
A curve is defined parametrically by x equals cosine t plus t sine t, and y equals sine t minus t cosine t. Find dy dx.
For \(x=f(t),\ y=g(t)\):
\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\]
Find \(\dfrac{dy}{dx}\) for \(x=t^2,\ y=t^3\).
\(\dfrac{dy}{dx}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}\). Answer: \(3t/2\)
For \(x=\cos t+t\sin t,\ y=\sin t-t\cos t\), find \(\dfrac{dy}{dx}\).
\(\tan t\)
\(-\tan t\)
\(\cot t\)
\(-\cot t\)
0
Step 1 — dx/dt
\[\frac{dx}{dt}=-\sin t+\sin t+t\cos t=t\cos t\]
Step 2 — dy/dt
\[\frac{dy}{dt}=\cos t-\cos t+t\sin t=t\sin t\]
Step 3 — Divide
\[\frac{dy}{dx}=\frac{t\sin t}{t\cos t}=\tan t\]
\[\boxed{\dfrac{dy}{dx}=\tan t}\]
L'Hôpital Advanced
5
Evaluate the limit as x approaches infinity of the quantity 1 plus 3 over x, all raised to the power x.
Indeterminate form \(1^\infty\): take \(\ln\), apply L'Hôpital to the exponent, then exponentiate.
\[\lim f(x)^{g(x)}=\exp\!\Bigl(\lim g(x)\ln f(x)\Bigr)\]
Find \(\displaystyle\lim_{x\to0^+}x^x\).
\(\lim x\ln x=\lim\dfrac{\ln x}{1/x}\overset{H}{=}\lim(-x)=0\). So \(e^0=1\). Answer: 1
Evaluate: \[\lim_{x\to\infty}\!\left(1+\frac{3}{x}\right)^{\!x}\]
\(e\)
\(e^3\)
\(3e\)
\(\infty\)
1
Step 1 — Identify \(1^\infty\) Form
Base \(\to1\), exponent \(\to\infty\).
Step 2 — Take Logarithm
\[L=\lim_{x\to\infty} x\ln\!\left(1+\frac{3}{x}\right)=\lim_{x\to\infty}\frac{\ln(1+3/x)}{1/x}\quad\left(\tfrac{0}{0}\right)\]
Step 3 — L'Hôpital
\[L=\lim_{x\to\infty}\frac{\frac{1}{1+3/x}\cdot(-3/x^2)}{-1/x^2}=\lim_{x\to\infty}\frac{3}{1+3/x}=3\]
Step 4 — Exponentiate
Original limit \(=e^L=e^3\).
\[\boxed{e^3}\]
Total Distance
5
A particle moves along the x-axis with velocity v of t equals t squared minus 4t plus 3 for t between 0 and 4. What is the total distance traveled?
Net displacement: \(\displaystyle\int_a^b v(t)\,dt\). Total distance: \(\displaystyle\int_a^b|v(t)|\,dt\). Split the integral at zeros of \(v\).
Average value of \(f(x)=x^2\) on \([0,3]\).
\(\tfrac{1}{3}\displaystyle\int_0^3 x^2\,dx=3\). Answer: 3
A particle has velocity \(v(t)=t^2-4t+3\) on \([0,4]\). What is the total distance traveled?
\(4\)
\(\dfrac{20}{3}\)
\(\dfrac{16}{3}\)
\(\dfrac{8}{3}\)
1
Step 1 — Zeros of v(t)
\((t-1)(t-3)=0\Rightarrow t=1,3\). Sign: \(+\) on \([0,1)\), \(-\) on \((1,3)\), \(+\) on \((3,4]\).
Step 2 — Split Integral
\[D=\int_0^1(t^2-4t+3)\,dt+\int_1^3(4t-t^2-3)\,dt+\int_3^4(t^2-4t+3)\,dt\]
Step 3 — Evaluate Each Piece
\(I_1=\left[\tfrac{t^3}{3}-2t^2+3t\right]_0^1=\tfrac{1}{3}-2+3=\tfrac{4}{3}\)
\(I_2=\left[-\tfrac{t^3}{3}+2t^2-3t\right]_1^3=(-9+18-9)-(-\tfrac{1}{3}+2-3)=0+\tfrac{4}{3}=\tfrac{8}{3}\)
(taking absolute value of negative part)
\(I_3=\left[\tfrac{t^3}{3}-2t^2+3t\right]_3^4=(\tfrac{64}{3}-32+12)-(9-18+9)=\tfrac{4}{3}-0=\tfrac{4}{3}\)
\(D=\tfrac{4}{3}+\tfrac{8}{3}+\tfrac{4}{3}=\dfrac{16}{3}\)
\[\boxed{\dfrac{16}{3}}\]
Slope Fields
4
Which differential equation has a slope field where slopes are zero along the line y equals x, positive above it, and negative below it?
In a slope field, the segment at \((x,y)\) has slope \(\dfrac{dy}{dx}=f(x,y)\). Analyze: where is slope zero? Positive? Negative? Use isoclines to match.
Describe slope field of \(\dfrac{dy}{dx}=y\).
Slopes zero on \(y=0\); positive above, negative below. Solutions: \(Ce^x\). Exponential growth/decay
Which DE has a slope field with slopes zero along \(y=x\), positive above, negative below?
\(\dfrac{dy}{dx}=x-y\)
\(\dfrac{dy}{dx}=y-x\)
\(\dfrac{dy}{dx}=x+y\)
\(\dfrac{dy}{dx}=xy\)
1
Requirements
\(\tfrac{dy}{dx}=0\) on \(y=x\); \(>0\) when \(y>x\); \(<0\) when \(y
Test (B): dy/dx = y − x
On \(y=x\): \(y-x=0\) ✓
Above (\(y>x\)): \(y-x>0\) ✓
Below (\(y
Reject Others
(A) \(x-y\): above line gives \(x-y<0\) ✗ | (C) zero on \(y=-x\) ✗ | (D) zero on axes ✗
\[\boxed{\dfrac{dy}{dx}=y-x}\]
Series Error Bound
5
The alternating series with general term 1 over n cubed is approximated by its partial sum S sub 4. What is the maximum error in this approximation?
Alternating Series Estimation Theorem: The absolute error when using \(S_n\) is bounded by the first omitted term:
\[\left|S-S_n\right|\leq a_{n+1}\]
Approximate \(\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{n}\) with 3 terms. Bound the error.
\(S_3=1-\tfrac{1}{2}+\tfrac{1}{3}=\tfrac{5}{6}\). Error \(\leq\tfrac{1}{4}=0.25\). Error \(\leq 0.25\)
For \(\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{n^3}\) approximated by \(S_4\), what is the maximum error?
\(\dfrac{1}{125}\)
\(\dfrac{1}{32}\)
\(\dfrac{1}{64}\)
\(\dfrac{1}{243}\)
0
Step 1 — Verify AST Conditions
\(a_n=\tfrac{1}{n^3}>0\), decreasing to 0 ✓
Step 2 — Apply Error Bound
Error \(\leq a_{n+1}=a_5=\dfrac{1}{5^3}=\dfrac{1}{125}\).
\[\boxed{\dfrac{1}{125}}\]