Раздел III. Производная и дифференциал — § 15. Производные и дифференциалы высших порядков

Кудрявцев — задача 15.18

Л. Д. Кудрявцев, «Сборник задач по математическому анализу. Том 1». Условие, подробное решение и ответ.

Задача 15.18

Найти d3ydx3\dfrac{d^{3}y}{dx^{3}} для функций, заданных параметрически:

1) x=acos⁡t, y=asin⁡tx = a\cos t,\ y = a\sin t;
2) x=ach⁡t, y=ash⁡tx = a\ch t,\ y = a\sh t;
3) x=acos⁡3t, y=asin⁡3tx = a\cos^{3}t,\ y = a\sin^{3}t;
4) x=a(t−sin⁡t), y=a(1−cos⁡t)x = a(t-\sin t),\ y = a(1-\cos t);
5) x=e−tcos⁡t, y=e−tsin⁡tx = e^{-t}\cos t,\ y = e^{-t}\sin t;
6) x=cos⁡t−ln⁡ctg⁡t2, y=sin⁡tx = \cos t - \ln\ctg\dfrac{t}{2},\ y = \sin t.
Ответ:1) −3cos⁡ta2sin⁡5t;1)\ -\dfrac{3\cos t}{a^{2}\sin^{5}t};
2) 3ch⁡ta2sh⁡5t;2)\ \dfrac{3\ch t}{a^{2}\sh^{5}t};
3) cos⁡2t−4sin⁡2t9a2cos⁡7tsin⁡3t;3)\ \dfrac{\cos^{2}t - 4\sin^{2}t}{9a^{2}\cos^{7}t\sin^{3}t};
4) cos⁡t24a2sin⁡7t2;4)\ \dfrac{\cos\dfrac{t}{2}}{4a^{2}\sin^{7}\dfrac{t}{2}};
5) 4e2t(2sin⁡t−cos⁡t)(sin⁡t+cos⁡t)5;5)\ \dfrac{4e^{2t}(2\sin t - \cos t)}{(\sin t + \cos t)^{5}};
6) sin⁡t(1+3sin⁡2t)cos⁡7t6)\ \dfrac{\sin t(1 + 3\sin^{2}t)}{\cos^{7}t}

Решение

x˙=−asin⁡t, y˙=acos⁡t\dot x = -a\sin t,\ \dot y = a\cos t.y′=y˙x˙=−ctg⁡t.y' = \dfrac{\dot y}{\dot x} = -\ctg t.Дифференцируем y′y' по tt и делим на x˙\dot x:ddt(y′)=1sin⁡2t,y′′=1/sin⁡2t−asin⁡t=−1asin⁡3t.\dfrac{d}{dt}(y') = \dfrac{1}{\sin^{2}t},\qquad
y'' = \dfrac{1/\sin^{2}t}{-a\sin t} = -\dfrac{1}{a\sin^{3}t}.Дифференцируем y′′y'' по tt:ddt ⁣(−1asin⁡3t)=−1a⋅(−3)sin⁡−4tcos⁡t=3cos⁡tasin⁡4t.\dfrac{d}{dt}\!\left(-\dfrac{1}{a\sin^{3}t}\right) = -\dfrac{1}{a}\cdot(-3)\sin^{-4}t\cos t = \dfrac{3\cos t}{a\sin^{4}t}.Делим на x˙=−asin⁡t\dot x = -a\sin t:y′′′=3cos⁡tasin⁡4t−asin⁡t=−3cos⁡ta2sin⁡5t.y''' = \dfrac{\dfrac{3\cos t}{a\sin^{4}t}}{-a\sin t} = -\dfrac{3\cos t}{a^{2}\sin^{5}t}.
x˙=ash⁡t, y˙=ach⁡t\dot x = a\sh t,\ \dot y = a\ch t.y′=y˙x˙=ach⁡tash⁡t=cth⁡t.y' = \dfrac{\dot y}{\dot x} = \dfrac{a\ch t}{a\sh t} = \cth t.ddt(cth⁡t)=−1sh⁡2t,y′′=−1/sh⁡2tash⁡t=−1ash⁡3t.\dfrac{d}{dt}(\cth t) = -\dfrac{1}{\sh^{2}t},\qquad
y'' = \dfrac{-1/\sh^{2}t}{a\sh t} = -\dfrac{1}{a\sh^{3}t}.Дифференцируем y′′y'':ddt ⁣(−1ash⁡3t)=−1a⋅(−3)sh⁡−4tch⁡t=3ch⁡tash⁡4t.\dfrac{d}{dt}\!\left(-\dfrac{1}{a\sh^{3}t}\right) = -\dfrac{1}{a}\cdot(-3)\sh^{-4}t\ch t = \dfrac{3\ch t}{a\sh^{4}t}.Делим на x˙=ash⁡t\dot x = a\sh t:y′′′=3ch⁡tash⁡4tash⁡t=3ch⁡ta2sh⁡5t.y''' = \dfrac{\dfrac{3\ch t}{a\sh^{4}t}}{a\sh t} = \dfrac{3\ch t}{a^{2}\sh^{5}t}.
x˙=−3acos⁡2tsin⁡t, y˙=3asin⁡2tcos⁡t\dot x = -3a\cos^{2}t\sin t,\ \dot y = 3a\sin^{2}t\cos t.y′=y˙x˙=3asin⁡2tcos⁡t−3acos⁡2tsin⁡t=−tg⁡t.y' = \dfrac{\dot y}{\dot x} = \dfrac{3a\sin^{2}t\cos t}{-3a\cos^{2}t\sin t} = -\tg t.y′′=ddt(−tg⁡t)/x˙=−sec⁡2t−3acos⁡2tsin⁡t=13acos⁡4tsin⁡t.y'' = \dfrac{d}{dt}(-\tg t) \Big/ \dot x = \dfrac{-\sec^{2}t}{-3a\cos^{2}t\sin t} = \dfrac{1}{3a\cos^{4}t\sin t}.Запишем y′′=13acos⁡−4tsin⁡−1ty'' = \dfrac{1}{3a}\cos^{-4}t\sin^{-1}t и продифференцируем:ddt(y′′)=13a[−4cos⁡−5t(−sin⁡t)sin⁡−1t+cos⁡−4t(−cos⁡t)sin⁡−2t]==13a[4sin⁡tcos⁡5tsin⁡t−cos⁡tcos⁡4tsin⁡2t]==13a[4cos⁡5t−1cos⁡3tsin⁡2t].\begin{aligned}
\dfrac{d}{dt}(y'') &= \dfrac{1}{3a}\bigl[ -4\cos^{-5}t(-\sin t)\sin^{-1}t + \cos^{-4}t(-\cos t)\sin^{-2}t \bigr]= \\
&= \dfrac{1}{3a}\bigl[ \dfrac{4\sin t}{\cos^{5}t\sin t} - \dfrac{\cos t}{\cos^{4}t\sin^{2}t} \bigr]= \\
&= \dfrac{1}{3a}\bigl[ \dfrac{4}{\cos^{5}t} - \dfrac{1}{\cos^{3}t\sin^{2}t} \bigr].
\end{aligned}Приводим к общему знаменателю cos⁡5tsin⁡2t\cos^{5}t\sin^{2}t:ddt(y′′)=13a⋅4sin⁡2t−cos⁡2tcos⁡5tsin⁡2t.\dfrac{d}{dt}(y'') = \dfrac{1}{3a}\cdot\dfrac{4\sin^{2}t - \cos^{2}t}{\cos^{5}t\sin^{2}t}.Делим на x˙=−3acos⁡2tsin⁡t\dot x = -3a\cos^{2}t\sin t:y′′′=13a⋅4sin⁡2t−cos⁡2tcos⁡5tsin⁡2t−3acos⁡2tsin⁡t=−4sin⁡2t−cos⁡2t9a2cos⁡7tsin⁡3t=cos⁡2t−4sin⁡2t9a2cos⁡7tsin⁡3t.y''' = \dfrac{\dfrac{1}{3a}\cdot\dfrac{4\sin^{2}t - \cos^{2}t}{\cos^{5}t\sin^{2}t}}{-3a\cos^{2}t\sin t}
= -\dfrac{4\sin^{2}t - \cos^{2}t}{9a^{2}\cos^{7}t\sin^{3}t}
= \dfrac{\cos^{2}t - 4\sin^{2}t}{9a^{2}\cos^{7}t\sin^{3}t}.
x˙=a(1−cos⁡t)=2asin⁡2t2, y˙=asin⁡t=2asin⁡t2cos⁡t2\dot x = a(1-\cos t) = 2a\sin^{2}\dfrac{t}{2},\ \dot y = a\sin t = 2a\sin\dfrac{t}{2}\cos\dfrac{t}{2}.y′=y˙x˙=sin⁡t1−cos⁡t=ctg⁡t2.y' = \dfrac{\dot y}{\dot x} = \dfrac{\sin t}{1-\cos t} = \ctg\dfrac{t}{2}.ddt ⁣(ctg⁡t2)=−12csc⁡2t2,y′′=−12csc⁡2t22asin⁡2t2=−14asin⁡4t2.\dfrac{d}{dt}\!\left(\ctg\dfrac{t}{2}\right) = -\dfrac{1}{2}\csc^{2}\dfrac{t}{2},\qquad
y'' = \dfrac{-\dfrac{1}{2}\csc^{2}\dfrac{t}{2}}{2a\sin^{2}\dfrac{t}{2}} = -\dfrac{1}{4a\sin^{4}\dfrac{t}{2}}.Дифференцируем y′′y'':ddt ⁣(−14asin⁡4t2)=−14a⋅(−4)sin⁡−5t2⋅cos⁡t2⋅12=cos⁡t22asin⁡5t2.\dfrac{d}{dt}\!\left(-\dfrac{1}{4a\sin^{4}\dfrac{t}{2}}\right) = -\dfrac{1}{4a}\cdot(-4)\sin^{-5}\dfrac{t}{2}\cdot\cos\dfrac{t}{2}\cdot\dfrac{1}{2}
= \dfrac{\cos\dfrac{t}{2}}{2a\sin^{5}\dfrac{t}{2}}.Делим на x˙=2asin⁡2t2\dot x = 2a\sin^{2}\dfrac{t}{2}:y′′′=cos⁡t22asin⁡5t22asin⁡2t2=cos⁡t24a2sin⁡7t2.y''' = \dfrac{\dfrac{\cos\dfrac{t}{2}}{2a\sin^{5}\dfrac{t}{2}}}{2a\sin^{2}\dfrac{t}{2}} = \dfrac{\cos\dfrac{t}{2}}{4a^{2}\sin^{7}\dfrac{t}{2}}.
x˙=−e−t(cos⁡t+sin⁡t), y˙=e−t(cos⁡t−sin⁡t)\dot x = -e^{-t}(\cos t + \sin t),\ \dot y = e^{-t}(\cos t - \sin t).y′=y˙x˙=e−t(cos⁡t−sin⁡t)−e−t(cos⁡t+sin⁡t)=−cos⁡t−sin⁡tcos⁡t+sin⁡t=sin⁡t−cos⁡tcos⁡t+sin⁡t.y' = \dfrac{\dot y}{\dot x} = \dfrac{e^{-t}(\cos t - \sin t)}{-e^{-t}(\cos t + \sin t)} = -\dfrac{\cos t - \sin t}{\cos t + \sin t} = \dfrac{\sin t - \cos t}{\cos t + \sin t}.Дифференцируем y′y'. Пусть u=sin⁡t−cos⁡t, v=cos⁡t+sin⁡tu = \sin t - \cos t,\ v = \cos t + \sin t.u′=cos⁡t+sin⁡t=v,v′=−sin⁡t+cos⁡t=−u.u' = \cos t + \sin t = v,\qquad v' = -\sin t + \cos t = -u.ddt ⁣(uv)=u′v−uv′v2=v⋅v−u⋅(−u)v2=v2+u2v2=(cos⁡t+sin⁡t)2+(sin⁡t−cos⁡t)2(cos⁡t+sin⁡t)2.\dfrac{d}{dt}\!\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^{2}} = \dfrac{v\cdot v - u\cdot(-u)}{v^{2}} = \dfrac{v^{2} + u^{2}}{v^{2}} = \dfrac{(\cos t + \sin t)^{2} + (\sin t - \cos t)^{2}}{(\cos t + \sin t)^{2}}.(cos⁡t+sin⁡t)2+(sin⁡t−cos⁡t)2=cos⁡2t+sin⁡2t+2sin⁡tcos⁡t+sin⁡2t+cos⁡2t−2sin⁡tcos⁡t=2.(\cos t + \sin t)^{2} + (\sin t - \cos t)^{2} = \cos^{2}t + \sin^{2}t + 2\sin t\cos t + \sin^{2}t + \cos^{2}t - 2\sin t\cos t = 2.Следовательно, ddt(y′)=2(cos⁡t+sin⁡t)2\dfrac{d}{dt}(y') = \dfrac{2}{(\cos t + \sin t)^{2}}.
Делим на x˙=−e−t(cos⁡t+sin⁡t)\dot x = -e^{-t}(\cos t + \sin t):y′′=2(cos⁡t+sin⁡t)2−e−t(cos⁡t+sin⁡t)=−2et(cos⁡t+sin⁡t)3.y'' = \dfrac{\dfrac{2}{(\cos t + \sin t)^{2}}}{-e^{-t}(\cos t + \sin t)} = -\dfrac{2e^{t}}{(\cos t + \sin t)^{3}}.Теперь дифференцируем y′′y''. Запишем y′′=−2et(cos⁡t+sin⁡t)−3y'' = -2e^{t}(\cos t + \sin t)^{-3}.ddt(y′′)=−2et(cos⁡t+sin⁡t)−3−2et⋅(−3)(cos⁡t+sin⁡t)−4⋅(−sin⁡t+cos⁡t)==2et(cos⁡t+sin⁡t)−4[−(cos⁡t+sin⁡t)+3(cos⁡t−sin⁡t)]==2et(cos⁡t+sin⁡t)−4(−cos⁡t−sin⁡t+3cos⁡t−3sin⁡t)==2et(cos⁡t+sin⁡t)−4(2cos⁡t−4sin⁡t)==4et(cos⁡t−2sin⁡t)(cos⁡t+sin⁡t)4.\begin{aligned}
\dfrac{d}{dt}(y'') &= -2e^{t}(\cos t + \sin t)^{-3} - 2e^{t}\cdot(-3)(\cos t + \sin t)^{-4}\cdot(-\sin t + \cos t)= \\
&= 2e^{t}(\cos t + \sin t)^{-4}\bigl[ -(\cos t + \sin t) + 3(\cos t - \sin t) \bigr] =\\
&= 2e^{t}(\cos t + \sin t)^{-4}( -\cos t - \sin t + 3\cos t - 3\sin t) =\\
&= 2e^{t}(\cos t + \sin t)^{-4}(2\cos t - 4\sin t) =\\
&= \dfrac{4e^{t}(\cos t - 2\sin t)}{(\cos t + \sin t)^{4}}.
\end{aligned}Делим на x˙=−e−t(cos⁡t+sin⁡t)\dot x = -e^{-t}(\cos t + \sin t):y′′′=4et(cos⁡t−2sin⁡t)(cos⁡t+sin⁡t)4−e−t(cos⁡t+sin⁡t)=−4e2t(cos⁡t−2sin⁡t)(cos⁡t+sin⁡t)5=4e2t(2sin⁡t−cos⁡t)(sin⁡t+cos⁡t)5.y''' = \dfrac{\dfrac{4e^{t}(\cos t - 2\sin t)}{(\cos t + \sin t)^{4}}}{-e^{-t}(\cos t + \sin t)}
= -\dfrac{4e^{2t}(\cos t - 2\sin t)}{(\cos t + \sin t)^{5}}
= \dfrac{4e^{2t}(2\sin t - \cos t)}{(\sin t + \cos t)^{5}}.
x˙=cos⁡2tsin⁡t, y˙=cos⁡t\dot x = \dfrac{\cos^{2}t}{\sin t},\ \dot y = \cos t.y′=cos⁡tcos⁡2t/sin⁡t=tg⁡t,y′′=sec⁡2tcos⁡2t/sin⁡t=sin⁡tcos⁡4t.y' = \dfrac{\cos t}{\cos^{2}t/\sin t} = \tg t,\qquad
y'' = \dfrac{\sec^{2}t}{\cos^{2}t/\sin t} = \dfrac{\sin t}{\cos^{4}t}.Дифференцируем y′′y'':ddt ⁣(sin⁡tcos⁡4t)=cos⁡t⋅cos⁡4t−sin⁡t⋅4cos⁡3t(−sin⁡t)cos⁡8t==cos⁡5t+4sin⁡2tcos⁡3tcos⁡8t=cos⁡2t+4sin⁡2tcos⁡5t.\begin{aligned}
\dfrac{d}{dt}\!\left(\dfrac{\sin t}{\cos^{4}t}\right)
&= \dfrac{\cos t\cdot\cos^{4}t - \sin t\cdot4\cos^{3}t(-\sin t)}{\cos^{8}t} =\\
&= \dfrac{\cos^{5}t + 4\sin^{2}t\cos^{3}t}{\cos^{8}t}
= \dfrac{\cos^{2}t + 4\sin^{2}t}{\cos^{5}t}.
\end{aligned}Делим на x˙=cos⁡2tsin⁡t\dot x = \dfrac{\cos^{2}t}{\sin t}:y′′′=cos⁡2t+4sin⁡2tcos⁡5tcos⁡2tsin⁡t=cos⁡2t+4sin⁡2tcos⁡5t⋅sin⁡tcos⁡2t=sin⁡t(cos⁡2t+4sin⁡2t)cos⁡7t.y''' = \dfrac{\dfrac{\cos^{2}t + 4\sin^{2}t}{\cos^{5}t}}{\dfrac{\cos^{2}t}{\sin t}}
= \dfrac{\cos^{2}t + 4\sin^{2}t}{\cos^{5}t}\cdot\dfrac{\sin t}{\cos^{2}t}
= \dfrac{\sin t(\cos^{2}t + 4\sin^{2}t)}{\cos^{7}t}.Учитывая cos⁡2t+4sin⁡2t=1+3sin⁡2t\cos^{2}t + 4\sin^{2}t = 1 + 3\sin^{2}t, окончательноy′′′=sin⁡t(1+3sin⁡2t)cos⁡7t.y''' = \dfrac{\sin t(1 + 3\sin^{2}t)}{\cos^{7}t}.
Кудрявцев № 15.18 — решение и ответ | Высшая математика — Профиматика