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台灣大學 · 藥學系 · 轉學考考古題 · 民國113年（2024年）

113 年度 微積分(C)

台灣大學 · 藥學系 · 轉學考

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第 1 題15 分
For a point PP on a smooth plane curve CC, the osculating circle OO to CC at PP is defined to be the circle that satisfies two conditions: (1) the circle OO and the curve CC share the same tangent line at PP; (2) the rate of change of the slope of tangent of CC at PP equals that of OO at PP. Now consider the curve CC: y=1−x+tan⁡(x)y = 1 - x + \tan(x) for −π2<x<π2-\frac{\pi}{2} < x < \frac{\pi}{2}.
(a)5 分
Find dydx∣x=π4\frac{dy}{dx}|_{x=\frac{\pi}{4}} and d2ydx2∣x=π4\frac{d^2y}{dx^2}|_{x=\frac{\pi}{4}}.
(b)10 分
Find the center and the radius of the osculating circle OO to CC at the point whose xx-coordinate equals π4\frac{\pi}{4}.
第 2 題35 分
Consider f(x)=∫1x11+t3dtf(x) = \int_1^x \frac{1}{\sqrt{1+t^3}} dt for x≥1x \geq 1.
(a)5 分
Prove that f(x)<f(y)f(x) < f(y) whenever 1≤x<y1 \leq x < y.
(b)10 分
Prove that f(x)<23f(x) < \frac{2}{3} for all x≥1x \geq 1 and also show that f(4)>13f(4) > \frac{1}{3}.
(c)10 分
Let g(u)g(u), where 0<u<f(4)0 < u < f(4), be a function such that f(g(u))=uf(g(u)) = u. Find g′(u)g'(u) and g′′(u)g''(u) in terms of g(u)g(u).
(d)10 分
Let h(u)=eu−g(u)h(u) = e^u - g(u), where 0<u<f(4)0 < u < f(4). Prove that h(u)h(u) does not have a local minimum value.
第 3 題20 分
Let a∈Ra \in \mathbb{R} and f(x,y,z)f(x,y,z), g(x,y,z)g(x,y,z) be two smooth functions R3→R\mathbb{R}^3 \to \mathbb{R}. Consider the optimization problem: Maximize f(x,y,z)f(x,y,z) subject to g(x,y,z)=ag(x,y,z) = a. Suppose, for each a∈Ra \in \mathbb{R}, it is known that (1) the maximum value fmax(a)f_{max}(a) of f(x,y,z)f(x,y,z) is attained at r(a)=(x∗(a),y∗(a),z∗(a))r(a) = (x^*(a), y^*(a), z^*(a)), i.e. fmax(a)=f(r(a))f_{max}(a) = f(r(a)); (2) there exists λ(a)∈R\lambda(a) \in \mathbb{R} such that ∇f(r(a))=λ(a)⋅∇g(r(a))\nabla f(r(a)) = \lambda(a) \cdot \nabla g(r(a)). Answer the following questions.
(a)10 分
Prove that dfmaxda=λ(a)\frac{df_{max}}{da} = \lambda(a).
(b)10 分
It is known that a differentiable function f(x,y,z)f(x,y,z), when restricted to the surface z=x3+y3−3xy2+1z = x^3 + y^3 - 3xy^2 + 1, attains a global maximum value at (−1,1,4)(-1,1,4). Moreover, fz(−1,1,4)=20f_z(-1,1,4) = 20. Use linearization to estimate the change of the maximum value when f(x,y,z)f(x,y,z) is restricted to the surface z=x3+y3−3xy2+0.8z = x^3 + y^3 - 3xy^2 + 0.8 instead.
第 4 題30 分
Evaluate the following integrals.
(a)5 分
∫011x(1−x)dx\int_0^1 \frac{1}{\sqrt{x(1-x)}} dx.
(b)5 分
∫021x3⋅e−1/x2dx\int_0^2 \frac{1}{x^3} \cdot e^{-1/x^2} dx.
(c)10 分
∬Rx−x2−y2dA\iint_R \sqrt{x - x^2 - y^2} dA where RR is the region enclosed by x2+y2=2xx^2 + y^2 = 2x.
(d)10 分
∫01∫x1∫01−ysin⁡((z−1)4)dzdydx\int_0^1 \int_{\sqrt{x}}^1 \int_0^{1-y} \sin((z-1)^4) dz dy dx.
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