题目
Problem
The curve C C C
x = 1 2 e 2 t − 1 3 t 3 − 1 2 , y = 2 e t ( t − 1 ) , x = \frac{1}{2}e^{2t} - \frac{1}{3}t^3 - \frac{1}{2},
\qquad y = 2e^t(t - 1), x = 2 1 e 2 t − 3 1 t 3 − 2 1 , y = 2 e t ( t − 1 ) , for 0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1
Find the exact length of C C C
[7]
题目中文翻译
曲线 C C C
x = 1 2 e 2 t − 1 3 t 3 − 1 2 , y = 2 e t ( t − 1 ) , x = \frac{1}{2}e^{2t} - \frac{1}{3}t^3 - \frac{1}{2},
\qquad y = 2e^t(t - 1), x = 2 1 e 2 t − 3 1 t 3 − 2 1 , y = 2 e t ( t − 1 ) , 其中 0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1
求曲线 C C C
(7)
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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对两个参数方程求导。平方相加时，y ′ y' y ′
答题过程
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Differentiating with respect to t t t x ′ = e 2 t − t 2 x'=e^{2t}-t^2 x ′ = e 2 t − t 2 y ′ = 2 t e t y'=2te^t y ′ = 2 t e t
( x ′ ) 2 + ( y ′ ) 2 = ( e 2 t − t 2 ) 2 + 4 t 2 e 2 t = e 4 t + 2 t 2 e 2 t + t 4 = ( e 2 t + t 2 ) 2 . \begin{align*}
(x')^2+(y')^2=&\,(e^{2t}-t^2)^2+4t^2e^{2t}\\[4mm]
=&\,e^{4t}+2t^2e^{2t}+t^4\\[4mm]
=&\,(e^{2t}+t^2)^2.
\end{align*} ( x ′ ) 2 + ( y ′ ) 2 = = = ( e 2 t − t 2 ) 2 + 4 t 2 e 2 t e 4 t + 2 t 2 e 2 t + t 4 ( e 2 t + t 2 ) 2 . The speed is e 2 t + t 2 e^{2t}+t^2 e 2 t + t 2 L = ∫ 0 1 ( x ′ ) 2 + ( y ′ ) 2 d t L=\int_0^1\sqrt{(x')^2+(y')^2}\,\mathrm{d}t L = ∫ 0 1 ( x ′ ) 2 + ( y ′ ) 2 d t
∫ 0 1 ( e 2 t + t 2 ) d t = [ 1 2 e 2 t + t 3 3 ] 0 1 = e 2 2 − 1 6 . \int_0^1(e^{2t}+t^2)\,\mathrm{d}t
=\bigg[\frac12e^{2t}+\frac{t^3}{3}\bigg]_0^1
=\boxed{\frac{e^2}{2}-\frac16}. ∫ 0 1 ( e 2 t + t 2 ) d t = [ 2 1 e 2 t + 3 t 3 ] 0 1 = 2 e 2 − 6 1 . 