Entry rate: $r(t) = 44\!\left(\dfrac{t}{100}\right)^3\!\left(1-\dfrac{t}{300}\right)^7$ for $0\le t\le 300$, else $0$. Exit rate: constant $0.7$ people/sec. Initial: $L(0)=20$.
People in line = initial + (entered) − (exited). Always set up as $L(t)=L(0)+\int_0^t [r(s)-0.7]\,ds$.
This asks only for people who enter — just integrate $r(t)$, not the net change.
- Home → New Document → Calculator
- Press menu → 4: Calculus → 3: Integral
- Enter: ∫(44*(t/100)^3*(1-t/300)^7, t, 0, 300)
- Press enter → Read result
- Alternatively: type directly
∫(44*(t/100)^3*(1-t/300)^7,t,0,300)
The question asks how many enter, not the net change. Only integrate $r(t)$.
Apply the Net Change Theorem over the entire interval:
- ∫(44*(t/100)^3*(1-t/300)^7 - 0.7, t, 0, 300)
- Add 20 to result: ans + 20
For $t>300$, $r(t)=0$, so people only leave at rate $0.7$. Line size decreases linearly:
Minimize $L(t)$. Since $L'(t) = r(t) - 0.7$, find where $L'(t)$ changes sign from $-$ to $+$ (i.e., $r(t)=0.7$).
Also check endpoints $t=0$ and $t=300$.
- Menu → 3: Algebra → 1: Solve
- Enter: solve(44*(t/100)^3*(1-t/300)^7=0.7, t)|0≤t≤300
- Find the critical value in $(0,300)$
- Then compute $L(t^*) = 20 + \int_0^{t^*}[r(t)-0.7]\,dt$
- State that $L'(t)=r(t)-0.7$
- Identify the critical point $t^*$ where $r(t^*)=0.7$
- Show $L'$ changes from negative to positive at $t^*$ → minimum
- Compare with $L(0)=20$ and $L(300)$
Since $r(t)<0.7$ for $0
$$L(33.051) = 20 + \int_0^{33.051}[r(t)-0.7]\,dt \approx \boxed{\textbf{≈ 6.873} \approx \textbf{7 people}}$$
$p(h)=0.2h^2e^{-0.0025h^2}$ for $0\le h\le30$; cross-sectional area = 3 m²
Differentiate $p(h)=0.2h^2e^{-0.0025h^2}$:
- Define: p(h):=0.2*h^2*e^(-0.0025*h^2)
- Derivative at 25: d/dh(p(h))|h=25 or menu → Calculus → Derivative
Units of $p'$ are (millions of cells/m³) per meter = millions of cells/m⁴. Forgetting units costs a point.
Cells in thin slice at depth $h$: density × volume = $p(h)\cdot 3\,dh$. Integrate:
- 3*∫(0.2*h^2*e^(-0.0025*h^2), h, 0, 30)
For $h\ge30$: $0\le f(h)\le u(h)$ and $\int_{30}^\infty u(h)\,dh=105$. So:
Why $\le 2000$ million:
Since $f(h)\le u(h)$ for all $h\ge30$: $$3\int_{30}^K f(h)\,dh \le 3\int_{30}^\infty u(h)\,dh = 3\times105 = 315$$ Total $\le 3\int_0^{30}p(h)\,dh + 315 \approx 485 + 315 = \mathbf{800} \le 2000$. ✓
- Make sure calculator is in Radian mode: menu → 5: Settings → Angle: Radian
- ∫(√((662*sin(5*t))^2+(880*cos(6*t))^2), t, 0, 1)
Always verify your TI-Nspire is in Radian mode before computing trig integrals on the AP exam.
$g=f'$. Piecewise linear for $-5\le x<3$; $g(x)=2(x-4)^2$ for $3\le x\le6$. $f(1)=3$.
- $f$ increasing ↔ $g>0$
- $f$ decreasing ↔ $g<0$
- $f$ concave up ↔ $g'>0$ ↔ $g$ increasing
- $f$ concave down ↔ $g'<0$ ↔ $g$ decreasing
- Inflection point of $f$ ↔ $g$ changes sign from + to − or − to +
Read $\int_{-5}^1 g(x)\,dx$ as the signed area under the piecewise-linear graph of $g$.
From the graph, count triangles and rectangles on $[-5,1]$:
*Read exact areas from graph: triangle from $x=-5$ to $x=0$ has base 5, height −3: area $= -7.5$. Triangle from $x=0$ to $x=1$ approximate from graph: area $≈+1$.*
$f(-5)=f(1)-\int_{-5}^1 g$, NOT $f(-5)=f(1)+\int_{-5}^1 g$. Use FTC: $f(1)-f(-5)=\int_{-5}^1 g$.
From graph: $\int_1^3 g\,dx = $ area of rectangle(s) under piecewise linear $g$.
For $3\le x\le6$: $g(x)=2(x-4)^2$, compute analytically:
$$\int_1^6 g(x)\,dx = \underbrace{\int_1^3 g\,dx}_{\text{from graph}} + 6$$ Final answer depends on exact graph reading. With $\int_1^3 g\,dx = 3$: $$= 3+6 = \boxed{\textbf{9}}$$
- $f$ increasing: $g(x)>0$
- $f$ concave up: $g'(x)>0$ (i.e., $g$ is increasing)
From graph: $g>0$ on approximately $(0,3)\cup(3,6)$ ... check where $g$ is also increasing.
$\boxed{\textbf{(4,6)}}$ — on this interval $g(x)=2(x-4)^2>0$ and $g'(x)=4(x-4)>0$ for $x>4$.
Reason: $f'(x)=g(x)>0$ (increasing) and $f''(x)=g'(x)>0$ (concave up) on $(4,6)$.
Inflection point of $f$ occurs where $f''=g'$ changes sign, i.e., where $g$ changes from increasing to decreasing (or vice versa), OR where $g$ itself changes sign (if $f''=g'$ passes through zero by a sign change).
More precisely: inflection of $f$ where $g'=f''$ changes sign.
An inflection point of $f$ requires $g'$ (i.e., $f''$) to change sign, not just equal zero. A local max/min of $g$ is a change in concavity of $f$.
$x = \mathbf{-3}$: $g$ changes from decreasing to increasing → $g'$ changes $-$ to $+$ → inflection of $f$.
$x = \mathbf{1}$: $g$ changes from increasing to decreasing → inflection of $f$.
$x = \mathbf{4}$: $g(x)=2(x-4)^2$ has minimum at $x=4$ → $g'$ changes $-$ to $+$ → inflection of $f$.
Reason: At each $x$, $f''=g'$ changes sign, so $f$ changes concavity.
| $t$ (years) | 2 | 3 | 5 | 7 | 10 |
|---|---|---|---|---|---|
| $H(t)$ (meters) | 1.5 | 2 | 6 | 11 | 15 |
Use the values on either side of $t=6$: nearest data points are $t=5$ and $t=7$.
Always state three things: (1) $H$ is differentiable (given), (2) compute the average rate of change, (3) cite MVT.
The average rate of change is: $$\frac{H(10)-H(2)}{10-2} = \frac{15-1.5}{8} = \frac{13.5}{8} = 1.6875$$ Hmm — that's not 2. Instead, check the interval $[3,7]$: $$\frac{H(7)-H(3)}{7-3} = \frac{11-2}{4} = \frac{9}{4} = 2.25$$ And $[2,7]$: $\frac{11-1.5}{5}=1.9$. And $[3,10]$: $\frac{15-2}{7}\approx1.86$...
Better: check $[5,7]$: $\frac{11-6}{2}=2.5>2$ and $[2,10]$: $1.6875<2$.
By IVT applied to $H'$ (since $H$ is twice-differentiable, $H'$ is continuous): $H'$ takes the value 2.5 somewhere on $(5,7)$ and must also be less than 2 somewhere.
Standard Answer: Since $H$ is differentiable on $(2,10)$ and continuous on $[2,10]$, by the Mean Value Theorem: $$\frac{H(10)-H(2)}{10-2}=\frac{13.5}{8}\ne 2$$ Check subintervals until we find one where average rate = 2, OR argue via IVT on $H'$: $H'(c)=2.5$ on $(5,7)$ and some interval where $H'<2$, so by IVT $H'=2$ somewhere.
Average height $= \dfrac{1}{10-2}\int_2^{10}H(t)\,dt \approx \dfrac{1}{8}\cdot T$ where $T$ is the trapezoidal sum.
Subinterval widths are 1, 2, 2, 3 — not equal! Each trapezoid has its own width. Never assume equal spacing from a table.
Differentiate $G$: use Quotient Rule.
When $G(x)=50$: $\dfrac{100x}{1+x}=50 \Rightarrow 100x=50+50x \Rightarrow 50x=50 \Rightarrow x=1$.
You need the value of $x$ (diameter) when $H=50$, not just plug in $H=50$ directly. Solve $G(x)=50$ to get $x=1$ first.
Curves intersect at $\theta=\pi/3$ and $\theta=5\pi/3$. The shaded region $R$ is on the left side where $r=4$ is outer and $r=3+2\cos\theta$ is inner.
The region is where $r=4$ is outer. Bounds must match where $r=4$ is actually larger — check with the graph. Use $[\pi/3,\, 5\pi/3]$ (going around the left side).
At $\theta=\pi/2$: $r=3+2\cos(\pi/2)=3+0=3$, $r'=-2\sin(\pi/2)=-2$.
Distance from origin = $r$. Given $\dfrac{dr}{dt}=3$. Since $r=3+2\cos\theta$:
At $\theta=\pi/3$: $\sin(\pi/3)=\sqrt{3}/2$.
Always state units of $d\theta/dt$ = radians per second on the AP exam.
Ratio test: converges when $|x|/3 < 1$, i.e., $|x|<3$, i.e., $-3
At $x=3$: converges (alternating harmonic series, AST).
At $x=-3$: diverges (harmonic series).
$$\text{Interval of convergence: } \boxed{-3 < x \le 3}$$
For an alternating series satisfying AST conditions, the error $\le |a_{n+1}|$ (the first omitted term).
$P_4(x)$ includes terms up to $x^4$. The error is bounded by the first omitted term (degree 5 term) evaluated at $x=2$.
$P_4(x)$ includes terms up through $x^4$. First omitted term at $n=4$ (which gives $x^5$):
- Terms alternate in sign ✓
- Terms decrease in absolute value for $x=2$ ✓ (since $|x|=2<3$)
- Terms → 0 ✓
$P_4$ contains terms of degree 2, 3, 4 (no degree 0 or 1 terms!). The first omitted term is the degree-5 term. General term index: check whether your general term is indexed by $n$ starting at 1 (giving $x^{n+1}$), so $n=4$ gives $x^5$.
- menu → 4 → 3 Definite integral
- menu → 4 → 1 Derivative at a point
- menu → 3 → 1 Solve equation
- menu → 5 → 1 → Angle: Radian Set radian mode
- ctrl + enter Approximate (decimal) result
- ctrl + T Toggle CAS/approximate