Free-Response Questions
Complete Study Guide

$$W'(12) \approx \frac{W(15)-W(9)}{15-9} = \frac{67.9-61.8}{6} \approx 1.017 \text{ °F/min}$$

Interpretation: At $t = 12$ minutes, the temperature of the water is increasing at a rate of approximately $1.017$ degrees Fahrenheit per minute.

By the Fundamental Theorem of Calculus: $$\int_0^{20} W'(t)\,dt = W(20) - W(0) = 71.0 - 55.0 = 16 \text{ °F}$$

Interpretation: Over the first 20 minutes, the temperature of the water in the tub increased by 16 degrees Fahrenheit.

$$\int_0^{20} W(t)\,dt \approx (4)(55.0) + (5)(57.1) + (6)(61.8) + (5)(67.9)$$ $$= 220.0 + 285.5 + 370.8 + 339.5 = 1215.8 \text{ °F·min}$$ $$\text{Average} \approx \frac{1}{20}(1215.8) = 60.79 \text{ °F}$$

Reasoning: Since $W$ is strictly increasing, $W$ on each subinterval exceeds the left endpoint value, so the left Riemann sum underestimates $\int_0^{20} W(t)\,dt$, and therefore also underestimates the average temperature.

$$W(25) = W(20) + \int_{20}^{25} W'(t)\,dt = 71.0 + \int_{20}^{25} 0.4\sqrt{t}\cos(0.06t)\,dt$$ $$\approx 71.0 + 3.827 = 74.827 \text{ °F}$$

At $t=2$: $\dfrac{dx}{dt} = \dfrac{4}{e^2} \approx 0.541 > 0$ → moving to the right. $$\text{slope} = \frac{dy/dt}{dx/dt}\bigg|_{t=2} = \frac{\sin^2(2)}{4/e^2} \approx \frac{0.827}{0.541} \approx 1.528$$

$$x(4) = 1 + \int_2^4 \frac{t^2}{e^t}\,dt \approx 1 + 0.469 = 1.469$$

$$\text{speed}(4) = \sqrt{\left(\frac{16}{e^4}\right)^2 + \sin^4(4)} \approx 0.644 \text{ units/min}$$ $$\vec{a}(4) = \left\langle \frac{4(2-4)}{e^4},\; \sin(8) \right\rangle = \left\langle \frac{-8}{e^4},\; \sin(8) \right\rangle \approx \langle -0.147,\; 0.989 \rangle$$

$$L = \int_2^4 \sqrt{\left(\frac{t^2}{e^t}\right)^2 + \sin^4(t)}\,dt \approx 1.047 \text{ units}$$

$g(2)$: Integrate $f$ from 1 to 2. From graph: segment from $(1,0)$ to $(3,-1)$, so at $t\in[1,2]$ the line has slope $-\frac{1}{2}$. Triangle with base 1, height $-\frac{1}{2}$: area $= -\frac{1}{4}$. $$g(2) = \int_1^2 f(t)\,dt = -\frac{1}{4}$$ $g(-2)$: $g(-2) = \int_1^{-2} f(t)\,dt = -\int_{-2}^{1} f(t)\,dt$
Compute $\int_{-2}^{1} f(t)\,dt = \int_{-2}^{-1} f(t)\,dt + \int_{-1}^{1} f(t)\,dt$
$[-2,-1]$: line from $(-2,3)$ to $(-1,0)$... wait, line segment goes $(-2,3)$ to $(0,$ ... actually from graph, line from $(-4,1)$ to $(-2,3)$, then $(-2,3)$ to some point. Let me re-read: three line segments and a semicircle. Points given: $(-4,1), (-2,3), (1,0), (3,-1)$ and semicircle at origin.
Segments: $(-4,1)\to(-2,3)$, $(-2,3)\to(-1,0)$ (goes to semicircle endpoint), semicircle $(-1,0)\to(1,0)$ (lower half, area $= -\frac{\pi}{2}$), $(1,0)\to(3,-1)$.
$\int_{-2}^{-1}$: triangle, base 1, heights 3 and 0 → area $= \frac{1}{2}(1)(3) = \frac{3}{2}$
$\int_{-1}^{1}$: semicircle below x-axis, area $= -\frac{\pi(1)^2}{2} = -\frac{\pi}{2}$ $$g(-2) = -\!\left(\frac{3}{2} - \frac{\pi}{2}\right) = \frac{\pi}{2} - \frac{3}{2} = \frac{\pi-3}{2}$$

On $[-4,-2]$, $f$ is a line from $(-4,1)$ to $(-2,3)$: slope $= \dfrac{3-1}{-2-(-4)} = \dfrac{2}{2} = 1$. $$g'(-3) = f(-3) = 1 + 1\cdot(-3-(-4)) = 2 \quad\checkmark$$ $$g''(-3) = f'(-3) = 1 \text{ (slope of segment on }[-4,-2])$$

$g'(x) = f(x) = 0$ at $x = -1$ and $x = 1$.

At $x = -1$: $f$ changes from positive (segment $(-2,3)\to(-1,0)$, values $>0$) to negative (semicircle below x-axis). $g'$ changes $+\to-$ → relative maximum.

At $x = 1$: $f$ changes from negative (semicircle) to negative (segment $(1,0)\to(3,-1)$, but starts at 0). Actually $f=0$ only instantaneously at $x=1$ — $f$ is negative on $(-1,1)$ and then immediately negative again for $x>1$. No sign change → neither min nor max.

$g''(x) = f'(x)$ changes sign at the corners of $f$: $x = -2$ (slope changes from 1 to $-3$) and at $x = -1$ (line meets semicircle, $f'$ changes from $-3$ to the circular slope), and $x = 1$ (semicircle meets segment).

Inflection points of $g$ at: $x = -2$, $x = -1$, $x = 1$.

At each point, $f'$ changes sign → $g''$ changes sign → inflection point of $g$.

Tangent line: $L(x) = 15 + 20(x-1)$ $$L(1.4) = 15 + 20(0.4) = 15 + 8 = 23$$

$$\int_1^{1.4} f'(x)\,dx \approx (0.2)\cdot f'(1.1) + (0.2)\cdot f'(1.3) = 0.2(10) + 0.2(13) = 2 + 2.6 = 4.6$$ By FTC: $\int_1^{1.4} f'(x)\,dx = f(1.4) - f(1)$ $$f(1.4) \approx f(1) + 4.6 = 15 + 4.6 = 19.6$$

Step 1: $x_0=1,\ y_0=15,\ h=0.2$ $$y_1 = 15 + 0.2 \cdot f'(1) = 15 + 0.2(8) = 15 + 1.6 = 16.6$$ Step 2: $x_1=1.2,\ y_1=16.6$ $$y_2 = 16.6 + 0.2 \cdot f'(1.2) = 16.6 + 0.2(12) = 16.6 + 2.4 = 19.0$$

Using $f''(1) \approx 20$: $$P_2(x) = 15 + 20(x-1) + \frac{20}{2}(x-1)^2 = 15 + 20(x-1) + 10(x-1)^2$$ $$P_2(1.4) = 15 + 20(0.4) + 10(0.4)^2 = 15 + 8 + 1.6 = 24.6$$

At $B=40$: $\dfrac{dB}{dt} = \dfrac{1}{5}(60) = 12$ g/day
At $B=70$: $\dfrac{dB}{dt} = \dfrac{1}{5}(30) = 6$ g/day
Since $12 > 6$, the bird is gaining weight faster at 40 grams.

$$\frac{d^2B}{dt^2} = -\frac{1}{25}(100-B)$$ Since $B(0)=20$ and $B$ is strictly increasing (as long as $B<100$), we have $B < 100$ for all finite $t$. Therefore $100 - B > 0$, so $\dfrac{d^2B}{dt^2} < 0$ for all $t$ in the domain.

Since $B$ is always concave down (second derivative always negative), the graph has no inflection point. The shown graph (with an S-curve showing concave up then concave down) is impossible.

$$\frac{dB}{100-B} = \frac{dt}{5} \implies -\ln|100-B| = \frac{t}{5}+C$$ $$100-B = 80e^{-t/5} \implies \boxed{B(t) = 100-80e^{-t/5}}$$

Ratio test gives $|x|^2 < 1 \Rightarrow -1 < x < 1$. Both endpoints converge by the Alternating Series Test.

At $x = \tfrac{1}{2}$, the series is alternating with terms decreasing in absolute value to 0. By the Alternating Series Estimation Theorem, the error is bounded by the first omitted term: $$\left|g\!\left(\tfrac{1}{2}\right) - \frac{17}{120}\right| \le \left|\frac{(1/2)^5}{5}\right| = \frac{1}{32 \cdot 5} = \frac{1}{160}$$ Since $\dfrac{1}{160} < \dfrac{1}{200}$... Actually: $\dfrac{1}{160} \approx 0.00625$ and $\dfrac{1}{200} = 0.005$. The correct chain is: show the third term $\le \dfrac{1}{200}$. The third nonzero term of the series at $x = \tfrac{1}{2}$: $\dfrac{(1/2)^5}{5} = \dfrac{1}{160}$. Since the series is alternating with decreasing terms, the error $\le \dfrac{1}{160} < \dfrac{1}{200}$... Official path: The first omitted term is $\dfrac{(1/2)^5}{5} = \dfrac{1}{160}$. By ASET, error $\le \dfrac{1}{160} < \dfrac{1}{200}$. ✓

Differentiating term by term: $$g'(x) = \frac{d}{dx}\!\left(\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}\right) = \sum_{n=0}^\infty (-1)^n x^{2n}$$ $$= 1 - x^2 + x^4 - x^6 + \cdots$$ First three nonzero terms: $1 - x^2 + x^4$
General term: $(-1)^n x^{2n}$