Given tabular values of f', students work with a function defined by an arc-length integral (finding its derivative at a point and interpreting what the integral represents about the graph of f), approximate a value using Euler's method, and evaluate an integral using integration by parts.
Given tabular values of f', students work with a function defined by an arc-length integral (finding its derivative at a point and interpreting what the integral represents about the graph of f), approximate a value using Euler's method, and evaluate an integral using integration by parts.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
$h$ is defined by an arc-length-type integral $h(x)=\int_0^x \sqrt{1+(f'(t))^2}\,dt$. By the Fundamental Theorem of Calculus, the derivative is the integrand evaluated at the upper limit: $$h'(x)=\sqrt{1+(f'(x))^2}.$$ At $x=\pi$, the table gives $f'(\pi)=6$, so $$h'(\pi)=\sqrt{1+(f'(\pi))^2}=\sqrt{1+6^2}=\sqrt{37}.$$ Thus $h'(\pi)=\mathbf{\sqrt{37}}$ (approximately $6.083$).
2 points. 1 point for applying the Fundamental Theorem of Calculus to get h'(x)=sqrt(1+(f'(x))^2); 1 point for the answer sqrt(37) (using f'(pi)=6). The answer need not be simplified to a decimal; sqrt(1+36) is acceptable. Differentiating incorrectly (e.g., forgetting the FTC and trying to evaluate the integral) forfeits both points.
The integrand $\sqrt{1+(f'(x))^2}$ is exactly the arc-length density for the graph of a function. Therefore $$\int_0^\pi \sqrt{1+(f'(x))^2}\,dx$$ represents the **arc length** (the length) of the graph of $f$ over the interval $\mathbf{0\le x\le \pi}$ (from $x=0$ to $x=\pi$).
2 points. 1 point for identifying the integral as the arc length (length) of the graph of f; 1 point for specifying the interval [0, pi] (from x=0 to x=pi). Saying 'distance traveled' or 'area' instead of arc length of the curve does not earn the first point; omitting the interval loses the second.
Euler's method approximates $f$ using the tangent-line update $f_{\text{new}}=f_{\text{old}}+f'\cdot\Delta x$. Going from $x=0$ to $x=2\pi$ in two equal steps gives step size $\Delta x=\pi$. **Step 1** (from $x=0$ to $x=\pi$), using $f(0)=0$ and $f'(0)=5$: $$f(\pi)\approx f(0)+f'(0)\,\Delta x=0+5\pi=5\pi.$$ **Step 2** (from $x=\pi$ to $x=2\pi$), using $f'(\pi)=6$: $$f(2\pi)\approx f(\pi)+f'(\pi)\,\Delta x=5\pi+6\pi=11\pi.$$ Therefore $f(2\pi)\approx \mathbf{11\pi}$.
2 points. 1 point for demonstrating two correct Euler steps with the right slopes (f'(0)=5 then f'(pi)=6) and step size pi, with at most one error; 1 point for the answer 11pi. If there is one slope/arithmetic error in the first step, the wrong f(pi) must be carried into the second step to keep the method point, but the answer point is then lost. The final answer may be reported as the point (2pi, 11pi).
Evaluate $\displaystyle\int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt$ by integration by parts. Let $$u=t+5,\quad dv=\cos\!\left(\tfrac{t}{4}\right)dt\ \Rightarrow\ du=dt,\quad v=4\sin\!\left(\tfrac{t}{4}\right).$$ Then $$\int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt = 4(t+5)\sin\!\left(\tfrac{t}{4}\right)-\int 4\sin\!\left(\tfrac{t}{4}\right)dt.$$ Since $\int 4\sin\!\left(\tfrac{t}{4}\right)dt = -16\cos\!\left(\tfrac{t}{4}\right)$, $$\int (t+5)\cos\!\left(\tfrac{t}{4}\right)dt = 4(t+5)\sin\!\left(\tfrac{t}{4}\right)+16\cos\!\left(\tfrac{t}{4}\right)+C.$$ So the antiderivative is $\boxed{4(t+5)\sin\!\left(\tfrac{t}{4}\right)+16\cos\!\left(\tfrac{t}{4}\right)+C}$ (equivalently $4t\sin(t/4)+20\sin(t/4)+16\cos(t/4)+C$). Differentiating confirms the result returns $(t+5)\cos(t/4)$.
3 points. 1 point for a correct choice of u and dv (u=t+5, dv=cos(t/4)dt, giving v=4sin(t/4)); 1 point for correctly applying the parts formula uv - integral v du, i.e. 4(t+5)sin(t/4) - integral 4 sin(t/4) dt; 1 point for the correct final antiderivative 4(t+5)sin(t/4)+16cos(t/4)+C. The tabular method is acceptable for the first two points. Omitting +C is not penalized here, but a sign error in v or in the second integral loses the corresponding point.
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