题目
Problem
The curve C C C
x = 8 ln ( tan 1 2 t ) − 3 cot t − 3 t , y = 6 ln ( tan 1 2 t ) + 4 cot t + 4 t , \begin{align*}
x =&\,8\ln\big(\tan\tfrac12t\big)-3\cot t-3t,\\
y =&\,6\ln\big(\tan\tfrac12t\big)+4\cot t+4t,
\end{align*} x = y = 8 ln ( tan 2 1 t ) − 3 cot t − 3 t , 6 ln ( tan 2 1 t ) + 4 cot t + 4 t , for 1 6 π ≤ t ≤ 1 3 π \frac{1}{6}\pi \le t \le \frac{1}{3}\pi 6 1 π ≤ t ≤ 3 1 π
(a) (i) Show that d x d t = 8 cosec t + 3 cot 2 t \frac{\mathrm{d}x}{\mathrm{d}t} = 8\cosec t + 3\cot^2 t d t d x = 8 cosec t + 3 cot 2 t
[1]
(ii) Find d y d t \frac{\mathrm{d}y}{\mathrm{d}t} d t d y
[1]
(iii) Find the exact value of the length of C C C
[6]
(b) Show that
d 2 y d x 2 = a cot t cosec t ( cosec 2 t + 1 ) ( b cosec t + c cot 2 t ) n , \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=
\frac{
a\cot t\cosec t(\cosec^2 t + 1)
}{
\big(b\cosec t + c\cot^2 t\big)^n
}, d x 2 d 2 y = ( b cosec t + c cot 2 t ) n a cot t cosec t ( cosec 2 t + 1 ) , where a a a b b b c c c n n n
[5]
题目中文翻译
曲线 C C C
x = 8 ln ( tan 1 2 t ) − 3 cot t − 3 t , y = 6 ln ( tan 1 2 t ) + 4 cot t + 4 t , \begin{align*}
x =&\,8\ln\big(\tan\tfrac12t\big)-3\cot t-3t,\\
y =&\,6\ln\big(\tan\tfrac12t\big)+4\cot t+4t,
\end{align*} x = y = 8 ln ( tan 2 1 t ) − 3 cot t − 3 t , 6 ln ( tan 2 1 t ) + 4 cot t + 4 t , 其中 1 6 π ≤ t ≤ 1 3 π \frac{1}{6}\pi \le t \le \frac{1}{3}\pi 6 1 π ≤ t ≤ 3 1 π
(a) (i) 证明 d x d t = 8 cosec t + 3 cot 2 t \frac{\mathrm{d}x}{\mathrm{d}t} = 8\cosec t + 3\cot^2 t d t d x = 8 cosec t + 3 cot 2 t
(1)
(ii) 求 d y d t \frac{\mathrm{d}y}{\mathrm{d}t} d t d y
(1)
(iii) 求 C C C
(6)
(b) 证明
d 2 y d x 2 = a cot t cosec t ( cosec 2 t + 1 ) ( b cosec t + c cot 2 t ) n , \frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=
\frac{
a\cot t\cosec t(\cosec^2 t + 1)
}{
\big(b\cosec t + c\cot^2 t\big)^n
}, d x 2 d 2 y = ( b cosec t + c cot 2 t ) n a cot t cosec t ( cosec 2 t + 1 ) , 其中 a a a b b b c c c n n n
(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)(i)
解法一
思路
展开
先把 ln ( tan ( t / 2 ) ) \ln(\tan(t/2)) ln ( tan ( t /2 )) csc t \csc t csc t csc 2 t − 1 = cot 2 t \csc^2t-1=\cot^2t csc 2 t − 1 = cot 2 t
答题过程
展开
Since
d d t ln ( tan t 2 ) = 1 2 sin ( t / 2 ) cos ( t / 2 ) = csc t , \frac{\mathrm{d}}{\mathrm{d}t}\ln\bigg(\tan\frac t2\bigg) =\frac{1}{2\sin(t/2)\cos(t/2)}=\csc t, d t d ln ( tan 2 t ) = 2 sin ( t /2 ) cos ( t /2 ) 1 = csc t , we have
d x d t = 8 csc t + 3 csc 2 t − 3 = 8 csc t + 3 cot 2 t . \frac{\mathrm{d}x}{\mathrm{d}t}=8\csc t+3\csc^2t-3=8\csc t+3\cot^2t. d t d x = 8 csc t + 3 csc 2 t − 3 = 8 csc t + 3 cot 2 t .
(a)(ii)
解法一
思路
展开
沿用对数项的导数，注意 cot t \cot t cot t
答题过程
展开
Differentiating gives
d y d t = 6 csc t − 4 csc 2 t + 4 = 6 csc t − 4 cot 2 t . \frac{\mathrm{d}y}{\mathrm{d}t} =6\csc t-4\csc^2t+4 =\boxed{6\csc t-4\cot^2t}. d t d y = 6 csc t − 4 csc 2 t + 4 = 6 csc t − 4 cot 2 t .
(a)(iii)
解法一
思路
展开
将两个导数分别平方相加，交叉项恰好抵消。再用 cot 2 t = csc 2 t − 1 \cot^2t=\csc^2t-1 cot 2 t = csc 2 t − 1
答题过程
展开
The square of the speed is
( d x d t ) 2 + ( d y d t ) 2 = 100 csc 2 t + 25 cot 4 t = 25 [ 4 csc 2 t + ( csc 2 t − 1 ) 2 ] = 25 ( csc 2 t + 1 ) 2 . \begin{align*}
\bigg(\frac{\mathrm{d}x}{\mathrm{d}t}\bigg)^2+
\bigg(\frac{\mathrm{d}y}{\mathrm{d}t}\bigg)^2
=&\,100\csc^2t+25\cot^4t\\
=&\,25\big[4\csc^2t+(\csc^2t-1)^2\big]\\
=&\,25(\csc^2t+1)^2.
\end{align*} ( d t d x ) 2 + ( d t d y ) 2 = = = 100 csc 2 t + 25 cot 4 t 25 [ 4 csc 2 t + ( csc 2 t − 1 ) 2 ] 25 ( csc 2 t + 1 ) 2 . The speed is therefore 5 ( csc 2 t + 1 ) 5(\csc^2t+1) 5 ( csc 2 t + 1 )
L = 5 ∫ π / 6 π / 3 ( csc 2 t + 1 ) d t = 5 [ t − cot t ] π / 6 π / 3 = 5 [ π 3 − 1 3 − π 6 + 3 ] = 5 π 6 + 10 3 3 . \begin{align*}
L=&\,5\int_{\pi/6}^{\pi/3}(\csc^2t+1)\,\mathrm{d}t\\
=&\,5[t-\cot t]_{\pi/6}^{\pi/3}\\
=&\,5\bigg[\frac\pi3-\frac1{\sqrt3}-\frac\pi6+\sqrt3\bigg]\\
=&\,\boxed{\frac{5\pi}{6}+\frac{10\sqrt3}{3}}.
\end{align*} L = = = = 5 ∫ π /6 π /3 ( csc 2 t + 1 ) d t 5 [ t − cot t ] π /6 π /3 5 [ 3 π − 3 1 − 6 π + 3 ] 6 5 π + 3 10 3 .
(b)
解法一
思路
展开
先求 y ′ / x ′ y'/x' y ′ / x ′ t t t x ′ x' x ′
答题过程
展开
Let u = csc t u=\csc t u = csc t v = cot 2 t v=\cot^2t v = cot 2 t x ′ = 8 u + 3 v x'=8u+3v x ′ = 8 u + 3 v y ′ = 6 u − 4 v y'=6u-4v y ′ = 6 u − 4 v u ′ = − u cot t u'=-u\cot t u ′ = − u cot t v ′ = − 2 u 2 cot t v'=-2u^2\cot t v ′ = − 2 u 2 cot t d y d x = ( 6 u − 4 v ) / ( 8 u + 3 v ) \frac{\mathrm{d}y}{\mathrm{d}x}=(6u-4v)/(8u+3v) d x d y = ( 6 u − 4 v ) / ( 8 u + 3 v ) t t t
H = ( 6 u ′ − 4 v ′ ) ( 8 u + 3 v ) − ( 6 u − 4 v ) ( 8 u ′ + 3 v ′ ) = 18 u ′ v − 32 u v ′ − 18 u v ′ + 32 u ′ v = 50 ( u ′ v − u v ′ ) = 50 u cot t ( 2 u 2 − v ) . \begin{align*}
H=&\,(6u'-4v')(8u+3v)\\
&\,\hspace{2pt}-(6u-4v)(8u'+3v')\\
=&\,18u'v-32uv'-18uv'+32u'v\\
=&\,50(u'v-uv')\\
=&\,50u\cot t\,(2u^2-v).
\end{align*} H = = = = ( 6 u ′ − 4 v ′ ) ( 8 u + 3 v ) − ( 6 u − 4 v ) ( 8 u ′ + 3 v ′ ) 18 u ′ v − 32 u v ′ − 18 u v ′ + 32 u ′ v 50 ( u ′ v − u v ′ ) 50 u cot t ( 2 u 2 − v ) . Dividing by x ′ = 8 u + 3 v x'=8u+3v x ′ = 8 u + 3 v
d 2 y d x 2 = H ( 8 u + 3 v ) 3 = 50 u cot t ( 2 u 2 − v ) ( 8 u + 3 v ) 3 = 50 cot t csc t ( csc 2 t + 1 ) ( 8 csc t + 3 cot 2 t ) 3 . \begin{align*}
\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
=&\,\frac{H}{(8u+3v)^3}\\
=&\,\frac{50u\cot t\,(2u^2-v)}{(8u+3v)^3}\\
=&\,\frac{50\cot t\csc t\,(\csc^2t+1)}{(8\csc t+3\cot^2t)^3}.
\end{align*} d x 2 d 2 y = = = ( 8 u + 3 v ) 3 H ( 8 u + 3 v ) 3 50 u cot t ( 2 u 2 − v ) ( 8 csc t + 3 cot 2 t ) 3 50 cot t csc t ( csc 2 t + 1 ) . The last step uses v = u 2 − 1 v=u^2-1 v = u 2 − 1 ( a , b , c , n ) = ( 50 , 8 , 3 , 3 ) (a,b,c,n)=(50,8,3,3) ( a , b , c , n ) = ( 50 , 8 , 3 , 3 )
