THOMAS’ CALCULUS Exercises

最新推荐文章于 2025-07-06 05:00:00 发布

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本文探讨了两种洗衣机工作原理的数学证明过程，并展示了它们与微分方程的关系。通过积分和导数运算，作者揭示了两个方法的等价性，最后给出了两种方法的通用形式和具体解例。

6.2 (45) 在这里插入图片描述

Proof:
$W(t)′=π((f−1(f(t)))2−a2)f′(t)=π(t2−a2)f′(t)W'_{(t)}=\pi( (f^{-1}(f(t)))^2 - a^2)f'(t)=\pi(t^2-a^2)f'(t)$

$S(t)=2πf(t)∫atxdx−2π∫atxf(x)dx=πf(t)[x2]at−2π∫atxf(x)dx=πf(t)t2−πf(t)a2−2π∫atxf(x)dxS_{(t)}=2\pi f(t)\int^t_a{x}{\rm d}x -2\pi\int^t_axf(x){\rm d}x = \pi f(t)[x^2]^t_a - 2\pi\int^t_axf(x){\rm d}x=\pi f(t)t^2-\pi f(t)a^2- 2\pi\int^t_axf(x){\rm d}x$
==>
$S(t)′=πf′(t)t2+2πtf(t)−πa2f′(t)−2πtf(t)=π(t2−a2)f′(t)S'_{(t)}=\pi f'(t)t^2+2\pi tf(t) - \pi a^2f'(t)-2\pi tf(t)=\pi(t^2-a^2)f'(t)$
==>
$W'_{(t)}=S'_{(t)}$ .

$W (a) = S (a) = 0$

Therefore, $W (t) = S (t)$ for all t [a, b].

I found another washer method which can be proof been equivalent to the above two methods.

$W(t)=∫f(a)f(t)π(x2−a2)dyW(t)=\int^{f(t)}_{f(a)}\pi (x^2-a^2){\rm d}y$

$y = f (x)$ => $d y = f^{'} (x) d x$
=>
$W(t)==∫atπ(x2−a2)f′(x)dxW(t)==\int^t_a\pi (x^2-a^2)f'(x){\rm d}x$
=>
$W(t)′=π(t2−a2)f′(t)W'_{(t)}=\pi(t^2-a^2)f'(t)$

7.2 - 32
在这里插入图片描述
Solution:

$dydt=r−ky\frac{dy}{dt}=r-ky$ => $1r−kydy=dt\frac{1}{r-ky}dy=dt$ => $∫1r−kydy=∫dt\int{\frac{1}{r-ky}}{\rm d}y=\int{\rm d}t$ => $1kln(r−ky)=t+C\frac{1}{k}ln(r-ky)=t+C$ => $y=1k(r−ce−kt)y=\frac{1}{k}(r-ce^{-kt})$

Alternatively:

Let $z = r - k y$ . Then $dzdt=−kdydt=−k(r−ky)=−kz\frac{dz}{dt}=-k\frac{dy}{dt}=-k(r-ky)=-kz$ . The equation $dzdt=−kz\frac{dz}{dt}=-kz$ has solution $z=ce^{-kt}$ , so $r-ky=ce^{-kt}$ and $y=1k(r−ce−kt)y=\frac{1}{k}(r-ce^{-kt})$