题目
Problem
The curve C C C
x = e t − 1 3 t 3 , y = 4 e t / 2 ( t − 2 ) , for 0 ≤ t ≤ 2. \begin{align*}
x=&\,\mathrm{e}^t-\frac13t^3,\\
y=&\,4\mathrm{e}^{t/2}(t-2),\qquad\text{for }0\le t\le2.
\end{align*} x = y = e t − 3 1 t 3 , 4 e t /2 ( t − 2 ) , for 0 ≤ t ≤ 2. Find, in terms of e \mathrm{e} e C C C
(6)
题目中文翻译
曲线 C C C
x = e t − 1 3 t 3 , y = 4 e t / 2 ( t − 2 ) , 其中 0 ≤ t ≤ 2. \begin{align*}
x=&\,\mathrm{e}^t-\frac13t^3,\\
y=&\,4\mathrm{e}^{t/2}(t-2),\qquad\text{其中 }0\le t\le2.
\end{align*} x = y = e t − 3 1 t 3 , 4 e t /2 ( t − 2 ) , 其中 0 ≤ t ≤ 2. 求曲线 C C C e \mathrm{e} e
(6)
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
解法一
思路
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对参数方程求导。把两个导数的平方相加后，可配成 ( e t + t 2 ) 2 (e^t+t^2)^2 ( e t + t 2 ) 2
答题过程
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Differentiating gives x ′ = e t − t 2 x'=e^t-t^2 x ′ = e t − t 2
y ′ = 4 e t / 2 ( 1 2 ( t − 2 ) + 1 ) = 2 t e t / 2 . y'=4e^{t/2}\bigg(\frac12(t-2)+1\bigg)=2te^{t/2}. y ′ = 4 e t /2 ( 2 1 ( t − 2 ) + 1 ) = 2 t e t /2 . The squared speed is
( x ′ ) 2 + ( y ′ ) 2 = ( e t − t 2 ) 2 + 4 t 2 e t = e 2 t + 2 t 2 e t + t 4 = ( e t + t 2 ) 2 . \begin{align*}
(x')^2+(y')^2=&\,(e^t-t^2)^2+4t^2e^t\\
=&\,e^{2t}+2t^2e^t+t^4\\
=&\,(e^t+t^2)^2.
\end{align*} ( x ′ ) 2 + ( y ′ ) 2 = = = ( e t − t 2 ) 2 + 4 t 2 e t e 2 t + 2 t 2 e t + t 4 ( e t + t 2 ) 2 . Using s = ∫ 0 2 ( x ′ ) 2 + ( y ′ ) 2 d t s=\int_0^2\sqrt{(x')^2+(y')^2}\,\mathrm{d}t s = ∫ 0 2 ( x ′ ) 2 + ( y ′ ) 2 d t e t + t 2 > 0 e^t+t^2>0 e t + t 2 > 0
∫ 0 2 ( e t + t 2 ) d t = [ e t + t 3 3 ] 0 2 = e 2 + 5 3 . \int_0^2(e^t+t^2)\,\mathrm{d}t
=\bigg[e^t+\frac{t^3}{3}\bigg]_0^2=\boxed{e^2+\frac53}. ∫ 0 2 ( e t + t 2 ) d t = [ e t + 3 t 3 ] 0 2 = e 2 + 3 5 . 