题目
Problem
The curve C C C
x = 2 3 t 3 2 − 2 t 1 2 , y = 2 t + 5 , 0 < t ≤ 3. x=\frac{2}{3}t^{\frac{3}{2}}-2t^{\frac{1}{2}},\qquad y=2t+5,\qquad 0<t\le3. x = 3 2 t 2 3 − 2 t 2 1 , y = 2 t + 5 , 0 < t ≤ 3. (a) Find the exact length of C C C
[5]
(b) Find the set of values of t t t d 2 y d x 2 > 0 \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}>0 d x 2 d 2 y > 0
[5]
题目中文翻译
曲线 C C C
x = 2 3 t 3 2 − 2 t 1 2 , y = 2 t + 5 , 0 < t ≤ 3 x=\frac{2}{3}t^{\frac{3}{2}}-2t^{\frac{1}{2}},\qquad y=2t+5,\qquad 0<t\le3 x = 3 2 t 2 3 − 2 t 2 1 , y = 2 t + 5 , 0 < t ≤ 3 (a) 求曲线 C C C
(5)
(b) 求满足 d 2 y d x 2 > 0 \dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}>0 d x 2 d 2 y > 0 t t t
(5)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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先求参数导数。把速度平方配成完全平方，再积分。虽然 t = 0 t=0 t = 0
答题过程
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Differentiating with respect to t t t x ′ = t − 1 / t x'=\sqrt t-1/\sqrt t x ′ = t − 1/ t y ′ = 2 y'=2 y ′ = 2
( x ′ ) 2 + ( y ′ ) 2 = ( t − 1 t ) 2 + 4 = t + 2 + 1 t = ( t + 1 t ) 2 . \begin{align*}
(x')^2+(y')^2=&\,\bigg(\sqrt t-\frac1{\sqrt t}\bigg)^2+4\\
=&\,t+2+\frac1t\\
=&\,\bigg(\sqrt t+\frac1{\sqrt t}\bigg)^2.
\end{align*} ( x ′ ) 2 + ( y ′ ) 2 = = = ( t − t 1 ) 2 + 4 t + 2 + t 1 ( t + t 1 ) 2 . For t > 0 t>0 t > 0 t + 1 / t > 0 \sqrt t+1/\sqrt t>0 t + 1/ t > 0 s = ∫ ( x ′ ) 2 + ( y ′ ) 2 d t s=\int\sqrt{(x')^2+(y')^2}\,\mathrm{d}t s = ∫ ( x ′ ) 2 + ( y ′ ) 2 d t
s = lim ε → 0 + ∫ ε 3 ( t + 1 t ) d t = [ 2 3 t 3 / 2 + 2 t 1 / 2 ] 0 3 = 4 3 . \begin{align*}
s=&\,\lim_{\varepsilon\to0^+}\int_\varepsilon^3\bigg(\sqrt t+\frac1{\sqrt t}\bigg)\,\mathrm{d}t\\
=&\,\bigg[\frac23t^{3/2}+2t^{1/2}\bigg]_0^3\\
=&\,\boxed{4\sqrt3}.
\end{align*} s = = = ε → 0 + lim ∫ ε 3 ( t + t 1 ) d t [ 3 2 t 3/2 + 2 t 1/2 ] 0 3 4 3 .
(b)
解法一
思路
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先算 d y d x \frac{\mathrm{d}y}{\mathrm{d}x} d x d y t t t x ′ x' x ′ t = 1 t=1 t = 1
答题过程
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For t ≠ 1 t\ne1 t = 1
d y d x = 2 t t − 1 , d d t ( 2 t t − 1 ) = ( t − 1 ) / t − 2 t ( t − 1 ) 2 = − t + 1 t ( t − 1 ) 2 . \begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}=&\,\frac{2\sqrt t}{t-1},\\
\frac{\mathrm{d}}{\mathrm{d}t}\bigg(\frac{2\sqrt t}{t-1}\bigg)
=&\,\frac{(t-1)/\sqrt t-2\sqrt t}{(t-1)^2}\\
=&\,-\frac{t+1}{\sqrt t(t-1)^2}.
\end{align*} d x d y = d t d ( t − 1 2 t ) = = t − 1 2 t , ( t − 1 ) 2 ( t − 1 ) / t − 2 t − t ( t − 1 ) 2 t + 1 . Since x ′ = ( t − 1 ) / t x'=(t-1)/\sqrt t x ′ = ( t − 1 ) / t
d 2 y d x 2 = − t + 1 ( t − 1 ) 3 . \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=-\frac{t+1}{(t-1)^3}. d x 2 d 2 y = − ( t − 1 ) 3 t + 1 . In the given domain 0 < t ≤ 3 0<t\le3 0 < t ≤ 3 − ( t + 1 ) -(t+1) − ( t + 1 ) t − 1 < 0 t-1<0 t − 1 < 0 0 < t < 1 \boxed{0<t<1} 0 < t < 1
