Given a differential equation with an initial condition, students compute a second derivative at the initial point, write a second-degree Taylor polynomial, use the Lagrange error bound to bound the error of an approximation, and approximate a value using Euler's method.
Given a differential equation with an initial condition, students compute a second derivative at the initial point, write a second-degree Taylor polynomial, use the Lagrange error bound to bound the error of an approximation, and approximate a value using Euler's method.
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Original Tian2 solution. Each part identifies which rubric points are earned and why.
Given $\dfrac{dy}{dx}=(3-x)y^2$. Differentiate again with respect to $x$, treating $y$ as a function of $x$ (product rule on $(3-x)$ and $y^2$, chain rule on $y^2$): $$\frac{d^2y}{dx^2}=\frac{d}{dx}\big[(3-x)y^2\big]=-y^2+(3-x)\cdot 2y\frac{dy}{dx}.$$ First evaluate $\dfrac{dy}{dx}$ at $(1,-1)$: $$f'(1)=(3-1)(-1)^2=2.$$ Now substitute $x=1,\ y=-1,\ \dfrac{dy}{dx}=2$ into the second derivative: $$f''(1)=-(-1)^2+(3-1)(2)(-1)(2)=-1-8=-9.$$
3 points. P1: product rule, producing the -y^2 + (3-x)(...) structure. P2: chain rule on y^2, producing the 2y(dy/dx) factor. (An expression with the product rule but no chain rule, e.g. -y^2+(3-x)2y, earns P1 not P2; an expression -2y(dy/dx) with chain but no product rule earns P2 not P1.) P3: the correct value f''(1)=-9, eligible only after both P1 and P2; the unsimplified form -(-1)^2+(3-1)(2)(-1)(2) earns all three regardless of later arithmetic slips.
The second-degree Taylor polynomial about $x=1$ is $$P_2(x)=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2.$$ With $f(1)=-1$, $f'(1)=2$, $f''(1)=-9$: $$P_2(x)=-1+2(x-1)-\frac{9}{2}(x-1)^2.$$
2 points. P4: the constant and linear terms -1+2(x-1) (consistent with the f'(1) imported from part A). P5: the quadratic term -(9/2)(x-1)^2 (consistent with f''(1)). Including any degree-3+ term or '+ ...' forfeits P5. Presenting only an expanded form such as -(9/2)x^2+11x-15/2 earns P4 but not P5 (it is not in powers of (x-1)).
By the Lagrange (Taylor remainder) error bound, for the degree-2 polynomial centered at $x=1$, $$\big|f(1.1)-P_2(1.1)\big|\le \frac{\max_{1\le x\le 1.1}\big|f'''(x)\big|}{3!}\,\big|1.1-1\big|^3.$$ Using the given bound $|f'''(x)|\le 60$ on $[1,1.1]$ and $|1.1-1|=0.1$: $$\big|f(1.1)-P_2(1.1)\big|\le \frac{60}{6}(0.1)^3=10\cdot 0.001=0.01.$$ Therefore the approximation differs from $f(1.1)$ by at most $0.01$.
2 points. P6 (form of the bound): present either the symbolic Lagrange form max|f'''|/3! * |1.1-1|^3 or the numeric (60/6)(0.1)^3. P7 (analysis): having earned P6, explicitly connect the bound to 0.01 (e.g. 'Error <= 0.01' or 'error bound = 0.01'). Stating the error equals 0.01 (rather than is at most) does not earn P7.
Euler's method from $x=1$ to $x=1.4$ in two equal steps uses step size $h=0.2$, with $\dfrac{dy}{dx}=(3-x)y^2$. **Step 1** (from $(1,-1)$): $$f(1.2)\approx f(1)+h\cdot\frac{dy}{dx}\Big|_{(1,-1)}=-1+0.2\,(3-1)(-1)^2=-1+0.2(2)=-0.6.$$ **Step 2** (from $(1.2,-0.6)$): $$f(1.4)\approx f(1.2)+h\cdot\frac{dy}{dx}\Big|_{(1.2,-0.6)}=-0.6+0.2\,(3-1.2)(-0.6)^2=-0.6+0.2(1.8)(0.36)=-0.4704.$$ So $f(1.4)\approx -0.4704$ (about $-0.47$).
2 points. P8 (first step): correct initial condition, step size 0.2, and the (possibly imported) derivative expression, giving f(1.2)~-0.6. A correctly labeled table earns P8. P9 (answer with supporting work): the second step yielding f(1.4)~-0.4704 (~-0.47, or -294/625). The unsimplified form -0.6+0.2(3-1.2)(-0.6)^2 earns P9 regardless of later arithmetic; an answer imported consistently from an incorrect part-A f'(1) is also eligible.
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