函数导数gydF4y2Ba
返回gydF4y2Ba函数导数gydF4y2Ba的功能gydF4y2Ba
对于函数来说gydF4y2BaygydF4y2Ba=gydF4y2BaygydF4y2Ba(gydF4y2BaxgydF4y2Ba)gydF4y2Ba,在那里gydF4y2BaxgydF4y2Ba表示一个或多个自变量。如果gydF4y2BaDgydF4y2Ba
= functionalDerivative (gydF4y2BafgydF4y2Ba
,gydF4y2BaygydF4y2Ba
)gydF4y2BaygydF4y2Ba
是符号函数的向量,gydF4y2BafunctionalDerivativegydF4y2Ba
返回关于函数的函数导数的向量gydF4y2BaygydF4y2Ba
,其中所有功能都在gydF4y2BaygydF4y2Ba
必须依赖于相同的自变量。gydF4y2Ba
求给出的函数的函数导数gydF4y2Ba
对于函数来说gydF4y2BaygydF4y2Ba
。gydF4y2Ba
y(x) f = y* siny;D = functionalDerivative (f, y)gydF4y2Ba
D(x) = siny (x) + cos(y(x))*y(x)gydF4y2Ba
求给出的函数的函数导数gydF4y2Ba
关于函数gydF4y2BaugydF4y2Ba
和gydF4y2BavgydF4y2Ba
。gydF4y2Ba
syms u(x) v(x) H = u²*diff(v,x)+v*diff(u,x,x);D =函数导数(H,[u v])gydF4y2Ba
D (x) = 2 * u (x) * diff (v (x) x) + diff (v (x), x, x) diff (u (x), x, x) - 2 * u (x) * diff (u (x), x)gydF4y2Ba
functionalDerivativegydF4y2Ba
返回符号函数的向量,其中包含函数的导数gydF4y2BaHgydF4y2Ba
关于gydF4y2BaugydF4y2Ba
和gydF4y2BavgydF4y2Ba
,分别。gydF4y2Ba
首先找出有质量弹簧的拉格朗日量gydF4y2Ba米gydF4y2Ba
和弹簧常数gydF4y2BakgydF4y2Ba
,然后推导出欧拉-拉格朗日方程。拉格朗日是动能之差gydF4y2BaTgydF4y2Ba
和潜在的能量gydF4y2BaVgydF4y2Ba
哪些是位移的函数gydF4y2Bax (t)gydF4y2Ba
。gydF4y2Ba
syms m k x(t) t = sym(1)/2*m*diff(x,t)^2;V =符号(1)/ 2 * k * x ^ 2;L = T - VgydF4y2Ba
L(t) = (m*diff(x(t), t)²)/2 - (k*x(t)²)/2gydF4y2Ba
通过求函数导数求欧拉-拉格朗日方程gydF4y2BalgydF4y2Ba
关于gydF4y2BaxgydF4y2Ba
,并将其等同于gydF4y2Ba0gydF4y2Ba
。gydF4y2Ba
函数导数(L,x) == 0gydF4y2Ba
eqn(t) = - m*diff(x(t), t, t) - k*x(t) == 0gydF4y2Ba
diff (x (t), t, t)gydF4y2Ba
是加速度。这个方程gydF4y2BaeqngydF4y2Ba
表示描述弹簧运动的期望微分方程。gydF4y2Ba
解决gydF4y2BaeqngydF4y2Ba
使用gydF4y2BadsolvegydF4y2Ba
。通过假设质量得到解的期望形式gydF4y2Ba米gydF4y2Ba
和弹簧常数gydF4y2BakgydF4y2Ba
是积极的。gydF4y2Ba
假设(k,'正')xSol = dsolve(eqn,x(0) == 0gydF4y2Ba
xSol = c3 * sin ((k ^ (1/2) * t) / m ^ (1/2))gydF4y2Ba
为进一步计算提供清晰的假设。gydF4y2Ba
假设([k m],“明确的”)gydF4y2Ba
一个问题是在重力作用下找到最快的下降路径。物体沿曲线运动的时间gydF4y2Bay (x)gydF4y2Ba
在重力下为gydF4y2Ba
在哪里gydF4y2BaggydF4y2Ba是重力加速度。gydF4y2Ba
通过最小化找到最快的路径gydF4y2BafgydF4y2Ba
相对于路径而言gydF4y2BaygydF4y2Ba
。求最小值的条件是gydF4y2Ba
计算此条件,得到描述臂弯问题的微分方程。使用gydF4y2Ba简化gydF4y2Ba
将解简化为期望的形式。gydF4y2Ba
信谊g y (x)假设g, "正面" f =√((1 + diff (y) ^ 2) / (2 * g * y));函数导数(f,y) == 0;eqn =简化(eqn)gydF4y2Ba
eqn (x) = diff (y (x), x) ^ 2 + 2 * y (x) * diff (y (x) x, x) = = 1gydF4y2Ba
这个方程是臂弯问题的标准微分方程。gydF4y2Ba
如果函数gydF4y2BaugydF4y2Ba(gydF4y2BaxgydF4y2Ba,gydF4y2BaygydF4y2Ba)gydF4y2Ba在三维空间中描述曲面,然后用函数法求出曲面的表面积gydF4y2Ba
在哪里gydF4y2BaugydF4y2BaxgydF4y2Ba和gydF4y2BaugydF4y2BaygydF4y2Ba的偏导数是gydF4y2BaugydF4y2Ba关于gydF4y2BaxgydF4y2Ba和gydF4y2BaygydF4y2Ba。gydF4y2Ba
找出描述由函数描述的三维曲面的最小曲面的方程gydF4y2Bau (x, y)gydF4y2Ba
通过求函数的导数gydF4y2BafgydF4y2Ba
关于gydF4y2BaugydF4y2Ba
。gydF4y2Ba
信谊u (x, y) f =√6 (1 + diff (u, x) ^ 2 + diff (u, y) ^ 2);D = functionalDerivative (f, u)gydF4y2Ba
D (x, y) = (diff (u (x, y), y) ^ 2 * diff (u (x, y), x, x)……+ diff(u(x, y), x)^2*diff(u(x, y), y, y)…- 2 * diff (u (x, y), x) * diff (u (x, y), y) * diff (u (x, y), x, y)……+ diff(u(x, y), x, x)…+ diff(u(x, y), y, y))/(diff(u(x, y), x)^2…+ diff(u(x, y), y)²+ 1)²(3/2)gydF4y2Ba
这个方程万博 尤文图斯的解gydF4y2BaDgydF4y2Ba
描述三维空间中的极小表面,如肥皂泡。gydF4y2Ba